RBSE Solutions for Class 11 Maths Chapter 12 Conic Section

RBSE Solutions for Class 11 Maths Chapter 12 Conic Section

Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.1

Question 1.
Find the equation of the circle whose:
(i) Centre (- 2, 3) and radius is 4
(ii) Centre (a, b) and radius is a – b
Solution:
(i) According to question,
Centre of circle (h, k) = (- 2, 3)
and Radius of circle r = 4 unit
Then, equation of circle
From formula (x – h)² + (y- k)² = r²
[x – (- 2)]² + (y – 3)² = 42
⇒ (x + 2)² + (y – 3)² = 16
⇒ x² + 4x + 4 + y² – 6y + 9 = 16
⇒ x² + y² + 4x – 6y = 16 – 9 – 4
⇒ x² + y² + 4x – 6y = 3
⇒ x² + y² + 4x – 6y – 3 = 0
Thus, required equation of circle
x² + y² + 4x – 6y – 3 = 0.

(ii) According to question,
Centre of circle (h, k) = (a, b)
and Radius of circle r = (a – b) unit
Then, equation of circle
From formula (x – h)² + (y – k)² = r²
⇒ (x – a)² + (y – b)² = (a – b)²
⇒ x² + a² – 2ax + y² + b² – 2 by = a² + b² – 2ab
⇒ x² + y² – 2ax – 2by + a² + b² – a² – b² + 2ab = 0
⇒ x² + y² – lax – 2by + 2ab = 0.

Question 2.
Find the coordinates of centre and radius of the following circles :
(i) x(x + y – 6) = y (x – y + 8)
(ii) √1+k2(x² + y²) = 2ax + 2aky
(iii) 4(x² + y²) = 1
Solution :
(i) x(x + y – 6) = y(x – y + 8)
⇒ x² + xy – 6x = xy – y² + 8y
⇒ x² + y² + xy – 6x – xy – 8y
⇒ x² + y² – 6x – 8y = 0 …(i)
General equation of circle,
ax² + by² + 2gx + 2fy + c = 0 …(ii)
Comparing equation (i) and (ii),
2g = -6 ⇒ g = -3
2f = -8 ⇒ f = – 4
c = 0
Thus centre of circle

Standard equation of circle,
(x – h)² + (y – k)² = a² …(ii)
Comparing equation (i) and (ii),
h = 0, k = 0, a = 1/2
Thus centre of circle is (0, 0) and radius is 1/2.

Question 3.
Find the equation of a circle which touches y-axis and cuts an intercept of length 2/on x-axis.
Solution:
In the standard of equation of circle put

Question 4.
Find the equation of the circle which cuts x- axis at a distance +3 from origin and cuts on intercept at y-axis of length 6 units.
Solution:
Let circle touches x-axis at point E and cuts AB intercepts at y-axis.
According to questions,

Let C is centre of the circle. Draw CD ⊥AB. Then CD = OE = 3 and
AD = AB/2 = 6/2 = 3
In right angled triangle ACD,
CA² = AD² + CD²
= 3² + 3² = 9 + 9 = 18
= 2 × 9 = 3√2
∴ Radius of circle a = CA = 3√2
whereas CE = CA = 3√2
Thus, centre of circle will be (3, 3√2)
Equation of circle,
(x – 3)² – (y – 3√2)² = (3√2)²
⇒ x² + 9 – 6x + y² + 18 – 6√2 y = 18
⇒ x² + y² – 6x – 6√2 y + 9 = 0
Similarly circle will be in IInd, IIIrd and IVth quadrant whose centres will be
(- 3, 3√2), (- 3, – 3√2), (3, – 3√2)
x² + y² + 6x – 6√2y + 9 = 0
x² + y² + 6x + 6√2y + 9 = 0
and x² + y² – 6x + 6√2 + 9 = 0
Thus, total four circles are possible whose equations is
x² + y² ± 6x ± 6√2y + 9 = 0

Question 5.
Find the centre and radius of circle
x² + y² – 8x + 10y – 12 = 0.
Solution:
According to question, equation of circle
x² + y² – 8x + 10y – 12 = 0
⇒ x² – 8x + y² + 10y – 12 = 0
⇒ x² – 8x + 16 – 16 + y²
+ 10y + 25 – 25 – 12 = 0
(On completing square)
⇒ (x – 4)² + (y + 5)² – 53 = 0
⇒ (x – 4)² + [y – (- 5)]² = 53

Question 6.
Find the centre and radius of the circle
2x² + 2y² – x = 0.
Solution:
According to question, equation of circle

Question 7.
Find the equation of the circle passing through the points (2, 3) and (- 1, 1) and whose centre lie on line x – 3y – 11 = 0.
Solution:
The equation of circle
(x – h)² + (y – k)² = r² …(i)
According to question, circle passes through the point (2, 3) and (-1,1)
Thus (2 – h)² + (3 – k)² = r²
⇒ 4 + h² – 4h + 9 + k² – 6k = r²
and (- 1 -h)² + (1 – k)² = r²
⇒ 1 + h² + 2h + 1 + k² – 2k = r²
Again h² + k² – 4h – 6k + 13 = r² …(ii)
and h² + k² + 2h – 2k + 2 – r² …(iii)
Since centre of circle (h, k) lie on line
x – 3y – 11 = 0
Thus h – 3k – 11 – 0
⇒ h – 3k – 11 …(iv)
Subtracting eq. (iii) from (ii),

Question 8.
Find the equation of circle of radius 5, whose centre lies on x-axis and which passes through point (2, 3).
Solution:
According to question, centre of circle lies on x-axis. Let centre of circle (h, 0) and radius = 5 unit.
Then equation of circle
(x – h)² + (y – 0)² = 52
⇒ (x – h)² + y² = 25
But circle passes through point (2, 3).
Then (2 – h)² + 3² = 25
⇒ 4 + h² – 4h + 9 = 25
⇒ h² -4h + 13 – 25 = 0
⇒ h² – 4h – 12 = 0
⇒ h² – (6 – 2)h – 12 = 0
⇒ h² – 6h + 2h – 12 = 0
⇒ h(h – 6) + 2(h – 6) = 0
⇒ (h – 6) (h + 2) = 0
⇒ h = 6 or h = -2
Then centre of each (6, 0) or (- 2, 0)
Put h = 6 in equation (1), the equation of circle
(x – 6)² + y² = 25
or x² + 36 – 12x + y² = 25
or x² + y² – 12x + 36 – 25 = 0
or x² + y² – 12x + 11 = 0
Again, putting h = – 2 in eqn. (i), equation of circle
[x – (- 2)²]² + y² = 25
⇒ (x + 2)² + y² = 25
⇒ x² + 4x + 4 + y² = 25
⇒ x² + y² + 4x + 4 – 25 = 0
⇒ x² + y² + 4x – 21 = 0
Thus, required equation of circle
X² + y² – 12x + 11 = 0
⇒ x² + y² + 4x – 21 = 0

Question 9.
Find the equation of circle that passes through point (0, 0) and cuts intercepts a and b at axis.
Solution:
According to question, circle passes through (0, 0) and cut intercepts a and b at x and y-axis.
Intersection points of circle with x-axis will be (a, 0) and with y-axis will be (0, b).

Thus, circle will passed through three points (0, 0), (a, 0) and (0, b).
Let equation of circle,
(x – h)² + (y – k)² = r² …(1)
Circle passes through (0, 0), (a, 0) and (0, b).
Thus, (0 – h)² + (0 – k)² = r² …(i)
(a – h)² + (0 – k)² = r² …(ii)
(0 – h)² + (b – k)² = r² …(iii)
h² + k² = r² …(2)
From eqⁿ. (i),
From eqⁿ. (ii),
a² + h² – 2 ah + k² = r²
⇒ h² + k² – 2 ah + a² = r² …(3)
From eqⁿ. (iii)
h² + b² + k² – 2bk = r²
⇒ h² + k² – 2bk + b² = r² …(4)
From eq11. (2) and (3)
r² – 2ah + a² = r²
⇒ -2 ah + a² – r² – r²
⇒ a² – 2h = 0
⇒ a(a -2h) = 0
Then a ≠ 0 and a – 2h = 0
⇒ h = a/2
Similarly, from eqⁿ. (2) and (4)
r² – 2bk + b² = r²
⇒ – 2bk + b² = r² – r²
⇒ b² – 2bk = 0
⇒ b(b – 2k) = 0

Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.2

Question 1.
Find the point of intersection of circle x² + y² = 25 and line 4x + 3y = 12 and also find length of intersecting chord.
Solution:
Equation of circle,
x² + y² = 25 …(i)
Equation of line,
4x + 3y = 12 …(ii)
⇒ 3y = 12 – 4x
⇒ y = 4 – 4/3 x …(iii)
From equation (i) and (iii),

Question 2.
If circle x² +y² = a² cuts an intercept of length 2l at straight line mx + c, then prove that
c² = (1 + m²)(a² – l²).
Solution:
Equation of circle,
x² + y². = a²
and equation of line,
y = mx + c
Length of intercept = 2l
Formula for length of intercept,

Question 3.
Find the length of intercept cut by circle x² + y² = c² at line x/a + y/b = 1.
Solution:
Equation of circle,
x² + y² = c²

Question 4.
For what value of A, line 3x + 4y = k touches the circle x² + y² = 10x
Solution:
Equation of line,
3x + 4y = k
⇒ 4y = – 3x + k

Question 5.
Find the condition, when
(i) Line y = mx + c, touches the circle
(x – a)² + (y – b)² = r²
(ii) Line lx + my + n = 0 touches the circle
x² + y² = a²
Solution :
(i) Equation of line
y = mx + c …(i)
Equation of circle (x – a)² + (y – b)² = r²
Centre of circle = (a, b) and Radius = r
Line will touch (i) and (ii), if
Radius of circle = length of ± drawn from centre to line

⇒ (1 + m²)r² = m²a² + b² + c² – 2mab – 2bc + 2mac
⇒ r² + m²r² = m²a² + b² + c² – 2mab – 2bc + 2mac
⇒ m² – a² – m²r² + 2mac – 2mab + b² + c² – 2bc = r²
⇒ (a² – r²)m² + (2ac – 2ab)m + b² + c² – 2bc = r²
⇒ (a² – r²)m² + 2a(c – b)m + (b – c)² = r²
⇒ m²(a² – r²) + 2ma(c – b) + (c – b)² = r²
(ii) Equation of given line
lx + my + n = 0
⇒ my = – lx – n

Question 6.
(i) Find the equation of tangent of the circle x² + y² = 64 which passes through point (4, 7).
(ii) Find the equation of tangent of the circle x² + y² – 4 which makes an angle of 60° with x-axis.
Solution:
(i) Equation of circle
x² – y² = 64
⇒ x² + y² = (8)²
Equation of tangent passing through (4, 7)
(y – y₁) = m(x – x₁)
⇒ (y – 7) = m(x – 4)
⇒ y – 7 = mx – 4m
⇒ mx – y – 4m + 7 = 0 …(i)
∵ line of Eq. (i) touches the circle
∴ Perpendicular drawn from centre (0, 0) of circle to the tangent will be equal to radius of circle,

Question 7.
Find the value of c, where line y = c, touches the circle x² + y² – 2x + 2y – 2 = 0 at point (1, 1).
Solution:
Equation of circle,
x² + y² – 2x + 2y – 2 = 0 …(i)
General equation of circle,
x² + y² + 2gx + 2fy + c = 0 …(ii)
Comparing equation (i) and (ii),
2g = -2 ⇒ g = -1
2f = 2 ⇒ f = 1.
c = -2

Question 8.
Find the equation of tangent at point (5, 12) and (12, – 5) at circle x² + y⁷ = 169. Prove that they will be perpendicular to each other. Also find the coordinates of intersection point.
Solution:
Equation of circle
x² + y² = 169 …(i)
At point (x₁.y₁) equation of tangent line passing through the circle x² + y² = a²
xx₁ + yy₁ = a²
At point (5, 12)
x × 5 + y × 12 = 169
⇒ 5x + 12y – 169 = 0 …(ii)
⇒ Gradient of this line,

Thus, line (ii) and (iii) are ⊥^r to each other.
For intersection point, solving equation (ii) and (iii).
Eqⁿ. (ii) × 5 and eqⁿ. (iii) × 12
(5x + 12y = 169) × 5
(12x -Sy= 169) × 12
⇒ 25x + 60y = 845
144x – 60y = 2028
169x = 2873
x = 2873/169 = 17
Put value of x in eqⁿ. (ii),
5 × 17 + 12y – 169 = 0
85 – 169 + 12y = 0
-84 + 12y = 0
⇒ y = 84/12 = 7
Thus, intersection point is (17, 7).

Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.3

Question 1.
Find the equation of parabola whose
(i) Focus (2, 3) and directrix x – 4y + 3 = 0
(ii) Focus (- 3, 0) and directri x + 5 = 0
Solution:
(i) Let P(h, k) is any point on parabola, then According to parabola,
SP = PM or SP² = PM²

= 1/17 (h² + 16k² + 9 – 8hk – 24k + 6h)
⇒ 17h² – 68h + 68 + 17k² – 102k + 153 – h² – 16k²
-9 + 8hk + 24k – 6h = 0
⇒ 16h² + 8hk + k² – 74h – 78k + 212 = 0
Thus locus of point P(h. k)
16x² + 8xy + y² – 74x – 78y + 212 = 0, which is required equation of parabola.
(ii) Let P(h, ft) be any variable at Parabola.
According to definition of parabola,
SP = PM or SP² = PM²

Thus, locus of point P(h, k) is y² = 4x + 16, which is required.

Question 2.
Find the vertex, axis, focus and latus rectum for following parabola:
(i) y² = 8x + 8y
(ii) x² + 2y = 8x – 7
Solution:
(i) y² = 8x + 8y
⇒ y² – 8y = 8x
⇒ y² – 2 × 4 × y + 4² = 8x + 4²
⇒ (y – 4)² = 8x + 16
⇒ (y – 4)² = 8(x + 2) …(i)
For replacing origin at point (4, – 2), put x + 2 and y – 4 = Y
Y² = 8W
⇒ Y² = 4.2.X …(i)
Which is of the form of parbola y² = 4ax where a = 2 and axis X = 0
Coordinates of vertex = (0, 0), coordinates of focus = (0,-1)
Length of Latus Rectum = 4 × 2 = 8
for given parabola (i), put value of X and Y in results.
X = 0 ⇒ x + 2 – 0 ⇒ x = -2
Y = 0 ⇒ y – 4 = 0 ⇒ y = 4
Thus coordinates of vertex = (- 2, 4)
Coordinates of focus
X = 0 ⇒ x + 2 = 2, x = 0
Y = 0 ⇒ y – 4 = 0, y = 4
Thus, coordinates of focus = (0, 4)
Axis Y = 0 ⇒ y – 4 = 0 ⇒ y = 4
Latus rectum = 4a = 4 × 2 = 8
(ii) x² + 2y = 8x – 7
x² – 8x = – 2y – 7
⇒ x² – 2 × 4 × x + 42 = -2y – 7 + 4²
⇒ (x – 4)² = – 2y – 7 + 16
⇒ (x – 4)² = – 2y + 9

In parabola X² = 4ay
Axis X = 0, then x – 4 = 0 ⇒ x = 4
Vertex (0, 0), then
x – 4 = 0 ⇒ x = 4

Question 3.
Length of double ordinate of parabola y² = 4ax is 8a. Prove that lines joining the origin to double ordinate will be perpendicular to each other.
Solution:
Equation of parabola,
y² = 4ax
Length of double ordinate = 8a
Let double ordinate is PP’
Given PP’ = 8a
Let OP = x₁ and OP’ = x₁

Then coordinates of P and P’ will be (x₁, 4a) and (x₁, – 4a)
Since point P lies on parabola, y² = 4ax
∴ (4a)² = 4ax₁ ⇒ x₁ = 4a
So, coordinates of P and Q = (4a, 4a) and (4a, – 4a)
∴ m₁ = Slope of line OP

Question 4.
If vertex and focus of parabola are at a distance of a and a’ from origin to x-axis, then prove that equation of parabola will be y²= 4(a’ – a)(x – a).
Solution:
According to question, coordinates of x-axis of vertex of parbola is at a distance a.

Thus vertex of parabola will be (a, 0) and focus (a’, 0). Vertex of parabola (a, 0) and length of latus rectum
= 4(Distance between focus and vertex)
= 4(a’ – a)
Since parabola is symmetric to x – axis.
Then from equation of parabola Y² = 4AX
y = 4 × (a’ – a) × (x – a)
⇒ y = 4(a’ – a) (x – a)
This is required equation.

Question 5.
PQ is double ordinate of a parabola. Find the locus of its trisection point.
Solution:
Given, equation of parabola
y² = 4ax
Let R, S trisects double ordinate PQ.
Let double ordinate PQ meets A-axis at point A.

Let coordinates of point R are (h, k)
then OA – h, AR = k
∴ RS = RA + AS = k + k – 2k
⇒ PR = RS = SQ = 2k
⇒ PA =RA+PR = k + 2k = 3k
∴ Coordinates of point P are (h, 3k),
∵ Point P lies on parabola
∴ (3k)² = 4ah ⇒ 9k² = 4ah
Thus, required locus is 9y² = 4ax.

Question 6.
Prove that locus of mid points of all the chords passing through the vertex of parabola y² = 4ax is a parabola y² = 2ax.
Solution:
Given, equation of parabola
y² = 4ax
Let (h, k) be coordinates of mid point P of chord OA passing through vertex O(0, 0) of parabola.
Let (at², 2at) be coordinates of point A.

Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.4

Question 1.
Find the coordinates of the points of intersection where straight line 4y + 3x + 6 = 0 cuts the parabola 2y2 = 9x.
Solution:
Equation of straight line,
4y + 3x + 6 = 0
⇒ 3x = -4y – 6 …(i)
Equation of parabola,
2y² = 9x …(ii)
From eqⁿ. (i) and (ii),
2y² = 3(-4y – 6)
⇒ 2y² = – 12y – 18
⇒ 2y² + 12y + 18 = 0
⇒ y² ± 6y + 9 = 0
⇒ (y + 3)² = 0 ⇒ y = -3
Putting y = – 3 in eqⁿ. (ii),
9x = 2 × (-3)² = 2 × 9
⇒ x = 2
Thus coordinates of intersection point of (2, – 3).

Question 2.
Find the length of chord cut by line 4y – 8 at parabola y² = 8x.
Solution:
Equation of parabola,
y² = 8x …(i)
Equation of line,
4y – 3x = 8
⇒ 4y = 3x + 8

Thus, coordinates of ends of chord are (8,8) and (8/9,8/3) Thus, length of chord

Thus, length of chord is 80/9 units.

Question 3.
Prove that straight line x + y = 1 touches parabola y = x – x².
Solution:
Equation of parabola,
y = x – x² …(i)
Equation of straight line,
x + y = 1 …(ii)
On solving eq. (i) and (ii),
⇒ x + x – x² = 1
⇒ x² – 2x + 1 = 0
This is a quadratic equation, its roots will be equal and coincide
⇒ B² – 4AC = 0
⇒ (- 2)² – 4(1) (1) = 0
⇒ 4 – 4 = 0
⇒ 0 = 0
Hence, straight line x + y = 1 will touch the parabola
y = x – x².

Question 4.
Find the condition that line lx + my + n = 0 touches the parabola y² = 4ax.
Solution:
Equation of parabola,
y² = 4 ax
Equation of straight line,
lx + my + n = 0
⇒ my = -lx – n

Question 5.
Prove that the length of focal chord of parabola y² = 4ax making an angle a with x-axis will be 4a cosec² a.
Solution:

coordinates of ends P and Q of chord PQ, making an angle α with x-axis and passing through focus S (a, 0) of parabola y² = 4 ax.
From point Q, draw a line parallel to x – axis and draw PR ⊥ from point P to that line.


PQ – 2a . 2 cosec α . cosec α
= 4a cosec² α
Thus length of chord PQ = 4a cosec² α.

Question 6.
Find the condition that line x cos α + y sin α – P touches the parabola y² = 4ax
Solution:
Equation of straight line,
x cos α + y sin α = p
⇒ y sin α = – x cos α + p

By the condition of tangency of parabola y² = 4ax to straight line y = mx + c

Question 7.
Find the equation of tangents at the following parabolas:
(i) y² = dx, which is parallel to line 2x – 3y = 4

(ii) y² = 8x, which is perpendicular to line 2x – y + 1 = 0
Solution:
(i) Equation of parabola,
y² = 6x
Equation of straight line,
2x – 3y = 4
⇒ 2x – 3y – 4 = 0
Equation of line parallel to this line,
2x – 3y + k = 0
Line (i) will touch parabola y² = 6x if equation

has equal roots, for which condition,
B² = 4AC
⇒ (4k – 54)² = 4 × 4 × k²
⇒ (4k – 54)² – (4k)² = 0
⇒ (4k – 54 – 4k) (4k – 54 + 4k) = 0
⇒ (-54) (8k – 54) = 0

This is required equation.
(ii) Equation of parabola
y² = 8x
Equation of line,
2x – y + 1 = 0
Equation of line perpendicular to this line,
x + 2y – 2k = 0 …(i)
Line (i) will touch the parabola y² = 8x, if

⇒ 4k² + x² – 4kx – 32x = 0
⇒ x² +(-4k – 32)x + 4k² = 0
Its roots will be same, if
B² = 4AC
⇒ (-4k – 32)² = 4 × 1 × 4k²
⇒ (-4k – 32)² – (4k)² = 0
⇒ (- 4k – 32 – 4k)(-4k – 32 + 4k) = 0
⇒ 8k = -32
⇒ k = -4
Putting value of k in equation (i)
x + 2y – 4 = 0
This is required equation.

Question 8.
For which value of k line 2x – 3y-k will touch parabola y² = 6x ?
Solution:
Equation of line,
2x – 3y = k
⇒ 3y = 2x – k

Question 9.
Find the equation of tangents which are drawn from point (4, 10) to parabola y² = 8x.
Solution:
From point (x₁, y₁) two tangents can be drawn at parabola whose combined equation can be find by equation SS’ = T²
Given Point – (4, 0)
Equation of parabola,
⇒ y² = 8x where a = 2
then S = y² – 4ax₁
⇒ S = y² – 8x …(i)
S’ = y1² – 4ax²
⇒ S’ = (10)² – 4 × 2 × 4
⇒ S’ = 100 – 32 = 68 …(ii)
T = yy1 – 2a(x + x₁)
⇒ T = y × 10 – 2 × 2 × (x + 4)
= T = 10y – 4(x + 4)
T² = {10y – 4(x + 4)}²
⇒ T² = 100 y² + 16(x + 4)² – 80(x + 4)y
⇒ T² = 100y² + 16x² + 256 + 128x – 80xy- 320y …(iii)
Put the values from eqn. (i), (ii), (iii) in SS’ – T²

(y² – 8x) (68) = 16x² + 100y² – 80xy + 128x – 320y + 256
⇒ 68y² – 744x – 16x² – 100y²
+ 80xy – 128x+ 320y – 256 = 0
⇒ – 16x² – 32y² + 80xy – 672x + 320y – 256 = 0
⇒ x² + 2y² – 5xy + 42x – 20y + 16 = 0
This is required equation.

Question 10.
Find the equation of normal at the following parabolas:
(i) At point (2, 4) on y² = 8x
(ii) Upper side of latus rectum of
y² + 12x = 0
Solution:
(i) Equation of parabola,
y² = 8x
Point (2, 4), equation of normal at point (x₁, y₁) on parabola y² = 4 ax

This is required equation.
(ii) y² + 12x = 0
⇒ y² = -12x
⇒ y² = 4(-3)x
Comparing it by y² = 4ax
a = – 3
Equation of latus rectum x = a
⇒ x = – 3
Putting x = – 3 in the equation of parabola

Question 11.
Find the equation of normal at following parabolas :
(i) y² = 4x which is parallel to line y – 2x + 5 = 0
(ii) y² = 4x which is perpendicular to line
x + 3y – 1 = 0
Solution:
(i) Equation of parabola,
y² = 4x
Comparing with y² = 4ax
a = 1
Equation of line y – 2x + 5 = 0
⇒ y = 2x – 5
Comparing with y = mx + c
m = 2, c = – 5
At parabola y² = 4ax parallel to line y = mx + c, eqⁿ. of normal from y = mx – 2am – am³
y = 2x – 2 × 1 × 2 – 1 × (2)³
⇒ y = 2x – 4 – 8
⇒ 7 – 2x + 12 = 0
⇒ 2x – y – 12 =0
This is required equation.
(ii) Equation of parabola,
y² = 4x
Comparing with y² = 4ax.
a = 1

Equation of normal of parabola y² = 4ax, perpendicular to line 7 = mx + c

⇒ y = 3x – 2 × 1 × 3 – 1 × (3)³
⇒ y = 3x – 6 – 27
⇒ y = 3x – 33
⇒ 3x – y – 33 = 0
This is required equation.

Question 12.
Prove that line 2x + y – 12a = 0 is normal chord at parabola y² = 4ax and its length is 55a−−√
Solution:
Equation of line,
2x + y – 12a = 0
⇒ y = – 2x + 12a …(i)
Equation of parabola,
y² = 4ax …(ii)
Put value of y from (i) in (ii),
(-2x + 12a)² = 4ax
⇒ 4x² + 144a² + 2(-2x) (12a) – 4ax = 0
⇒ 4x² + 144a² – 48ax – 4ax = 0
⇒ 4x² – 52ax + 144a² = 0
⇒ x² – 13ax + 36a² = 0
⇒ x² – 9ax – 4ax + 36a² = 0
⇒ x(x – 9a) – 4a(x – 9a) = 0
⇒ (x – 9a) (x – 4a) = 0
x = 4a, 9a
Put these value of x in equation (i),
y² = 4a × 4a ⇒ y = 4a
y² = 4a × 9a ⇒ y = 6a
Thus coordinates of ends of latus rectum are (4a, 4a) and (9a, – 6a)
∵ Coordinates will lie in 1st and 4th quadrant will be (+, +) and (+, -).
Thus length of latus rectum

Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.5

Question 1.
Find the equation of ellipse whose :
(i) Focus (- 1, 1), Directrix x – y + 4 = 0 and eccentricity is e – 1 / 5–√
(ii) Focus (- 2,3), Directrix 3x + 4y = 1 and eccentricity is e = 1/3
Solution:
(i) Let (h, k) be any point on ellipse, then according to definition,
Distance of P from focus = e(Distance of P from Directrix)
⇒ PS = e(PM)
⇒ (PS)² = e²(PM)²

Thus locus of point P(h, k), 9x² + 9y² + xy – 16x – 16y + 4 = 0 which is required equation of ellipse.
(ii) Let P(h, k) be any point on ellipse, then according to definition
Distance of P from focus = e(Distance of P from directrix)
⇒ PS = e(PM)
⇒ (PS)² = e²(PM)²

⇒ 225h² + 225k² + 900h – 1350k + 2925
– 9h² – 16k² – 1 – 24hk + 81 + 6h
⇒ 216h² + 209k² – 24hk + 906h – 1342k + 2924 = 0
At point (x, y).
216x² + 209y² – 24xy + 906x – 1342y + 2924 = 0
This is required equation.

Question 2.
Find the eccentricity, latus rectum and focus of the following ellipse :
(i) 4x² + 9y² = 1,
(ii) 25x² + 4y² = 100,
(iii) 3x² + 4y² – 12x – 8y + 4 = 0
Solution:
(i) Equation of ellipse,
4x² + 9y² = 1




Coordinate of focus coordinates of focus of ellipse will be (± ae, 0).

Question 3.
Find the equation of ellipse whose axis are coordinate axis and passes through points (6, 2) and (4, 3).
Solution:
Standard equation of ellipse

It passes through point (6, 2).

It also passes through point (4, 3).

Multiply eqⁿ. (i) by 9 and eqⁿ. (ii) by 4 then subtracting

Put the value of a² in equation (i),

Question 4.
Find the eccentricity of ellipse whose latus rectum is half of its minor axis.
Solution:
Let equation of ellipse
x2/a2 + y2/b2 = 1

Question 5.
Find the locus of a point which moves such that sum of its distances from point (1, 0) and (- 1, 0) remains 3. Which curve is this locus ?
Solution:
Let P(h, k) is any point such that sum of whose distance from A(1, 0) and B(- 1, 0) remains 3.
According to questions,

Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.6

Question 1.
Prove that liney = x+ √(5/6)touches ellipse 2x² + 3y² = 1. Also find the coordinates of tangent point.
Solution:

Question 2.
Show that line x – 3y – 4 = 0 touches ellipse 3x² + 4y² = 20.
Solution:
Equation of line, x – 3y – 4 =0
⇒ x = 3y + 4
Equation of ellipse,
3x² + 4y² = 20
Putting value of x from (i) in eqn. (ii),
3(3y + 4)² + 4y² = 20
⇒ 3(9y² + 24y + 16) + 4y² – 20 = 0
⇒ 27y² + 72y + 48 + 4y² – 20 = 0
⇒ 31y² + 72y + 28 = 0
Line will touch eqⁿ. (ii) if
(72)² – 4(31 )(28) = 0
∴ (72)² – 4 × 31 × 28 = 5184 – 3472
= 1712 ≠ 0
Thus, given line will not be touch given ellipse.
∴ Given equation is wrong.

Question 3.
For which value of k, the line 3x – 4y = k touches the ellipse 5x² + 4y² = 20.
Solution:
Equation of line,
3x – 4y = k
⇒ 3x – k = 4y
⇒ 4y = 3x – k

Question 4.
Prove that line

touches ellipse

Also find the coordinates of tangent point.
Solution:

Question 5.
Find the condition that line lx + my = n touches the ellipse

Solution:

Question 6.
Find the equations of tangent for ellipse 4x² + 3K² = 5 which makes equation angle of 60° with x-axis. Also find the coordinates of tangent point.
Solution:
Equation of ellipse
4x² + 3y² = 5

Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Ex 12.7

Question 1.
Find the length of axis, focus eccentricity latus rectum and equation of directrix of the hyperbola
9x² – 16y² = 144.
Solution:
Given equation,
9x² – 16y² = 144.

Question 2.
Find the equations of hyperbola whose :
(i) Focus is (2, 1), directrix x + 2y – 1 = 0 and eccentricity is 2
(ii) Focus is (1,2), directrix 2x + y = 1 and eccentricity is √3.
Solution:
(i) Let (x, y) be any point at hyperbola then by definition, distance of point (x, y) from focus
= e × Distance of directrix from (x, y)

This is required equation
(ii) Let (x, y) be any point at hyperbola then by definition, distance of point (x, y) from focus
= e × Distance of point from (x, y) from focus

Question 3.
Find the vertex, focus, latus rectum and eccentricity of hyperbola
x² – 6x – 4y² – 16y – 11 = 0.
Solution:
Equation of hyperbola,
x² – 6x – 4y² – 16y – 11 = 0
⇒ (x² – 2 × 3 × x + 3² – 3²) – 4(y² + 4y + 2² – 2²) – 11 = 0
⇒ (x – 3)² – 3² – 4 (y + 2)² + 16 – 11 = 0
⇒ (x – 3)² – 4 (y + 2)² + 16 – 11 – 9 = 0

Question 4.
Find the equation of hyperbola whose :
(i) Length of latus rectum is 8 and conjugate axis = 1/2 (Distance between focus)
(ii) Distance bewteen focus is 16 and conjugate axis is √2
(iii) Length of conjugate axis is 7 and passes through point (3, – 2).
Solution:
(i) Let equation of hyperbola,


(ii) Distance bewteen focus, 2ae = 16 conjugate axis 2b = √2


(iii) Conjugate axis 2b = 7

Question 5.
Prove that intersecting point of straight lines
x/a – y/b = m and x/a + y/b = 1/m is hyperbola.
a b a b m
Solution:
Given straight lines,
x/a – y/b = m ….(i)
and x/a + y/b = 1/m ….(ii)
Multiplying equation (i) and (ii)
(x/a – y/b) (x/a + y/b) = m × 1/m

This is equation of hyperbola.
Thus, locus of intersection point of two lines is hyperbola

Question 6.
Find the common point of hyperbola
5x² – 9y² = 45 and line y = x + 2.
Solution:
Equation of hyperbola,
5x² – 9y² = 45 …..(i)
and equation of line
y = x + 2 …..(ii)
From equation (i) and (ii),
5x² – 9(x + 2)² = 45
⇒ 5x² – 9(x² + 4x + 4) = 45
⇒ 5x² – 9y² – 36x – 36 – 45 = 0
⇒ -4x² – 36x – 81 = 0
⇒ 4x² + 36x + 81 = 0
On solving,

Question 7.
Prove that line lx + my = 1 will touch hyperbola

Solution:

Question 8.
Find the equation of tangents of hyperbola 4x² – 9y² = 1, which is parallel to line 4y = 5x + 7.
Solution:
Equation of hyperbola
4x² – 9y² = 1 ….(i)

Hence, required equation,
4y = 5x ⊥ 3/2

Question 9.
Prove that locus of foot of perpendicular drawn from focus at tangent of hyperbola is a circle.
Solution:
Let equation of hyperbola,

Eq. of line passing through (+c , 0) and perpendicular to line(ii)

Rajasthan Board RBSE Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

Question 1.
Radius of a circle
9x² +y² + 8x = 4(x² – y²) is :
(A) 1
(B) 2
(C) 4 / 5
(D) 5 / 4
Solution:
Equation of circle,
9x² + y² + 8x = 4(x² – y²)
⇒ 9x² + y² + 8x = 4x² – 4y²
⇒ 9x² – 4x² + 8x + y² + 4y² = 0
⇒ 5x² + 5y² + 8x = 0

Question 2.
Equation of a circle whose centre is in I quadrant as (α, β) and touches x-axis will be :
(A) x² + y² – 2αx – 2βy + α² = 0
(B) x² + y² + 2αx – 2βy + α² = 0
(C) x² + y² – 2αx + 2βy + α² = 0
(D) x² + y² + 2αx + 2βy + α² = 0
Solution:
Centre of circle,
(- g, -f) = α, β
g = – α

Question 3.
If liney = mx + c touches the circle x² + y² = 4y, then value of c is :

Solution:
Equation of line
y = mx + c …..(i)
and equation of circle
x² + y² = 4y
⇒ x² + y² – 4y = 0
⇒ x² + y² – 2.2.y + 4 = 4
⇒ x² + (y – 2)² = 4
⇒ x² + y² = 2²
where X = x, Y – y – 2, a = 2
According to condition of tangency

Thus option (C) is correct.

Question 4.
Line 3x + 4y = 25 touches the circle x² + y² = 25 at the point :
(A) (4, 3)
(B) (3, 4)
(C) (-3, -4)
(D) (3, -4)
Solution:
Equation of line,
3x + 4y = 25
and equation of circle

Question 5.
A conic section will be parabola if:
(A) e = 0
(B) e < 0
(C) e > 0
(D) e = 1
Solution:
Option (C) is correct.

Question 6.
Equation of directrix of parabola x² =- 8y is :
(A) y = -2
(B) y = 2
(C) x = 2
(D) x = – 2
Solution:
Equation of parabola
x² = – 8y
⇒ x² = -4 × 2 × y
Here a = 2
By equation of directrix y = a
y = 2
Thus, option (B) is correct.

Question 7.
Vertex of parabola x² + 4x + 2y = 0 is:
(A) (0, 0)
(B) (2, – 2)
(C) (- 2, -2)
(D) (-2, 2)
Solution:
Equation of parabola is :
x² + 4x + 2y = 0
⇒ x² + 4x + 4 = – 2y + 4
⇒ (x + 2)² = – 2(y – 2)
⇒ x² = – 2y
Here X = x + 2
and Y = y – 2
Vertex of parabola is (0, 0).
then x = x + 2 = 0
⇒ y = -2
y = y -2 = 0
⇒ y = 2
So vertex is (- 2, 2).
Thus, option (D) is correct.

Question 8.
If focus of any parabola is (-3,0) and directrix is x + 5 = 0, then its equation will be :
(A) y² = 4(x + 4)
(B) y² + 4x + 16 = 0
(C) y² + 4x = 16
(D) x² = 4(y + 4)
Solution:
Let P(x, y) be any point on parabola. By definition of parabola,
Distance between (x, y) and focus (- 3, 0)
= Length of ⊥ drawn from P(x, y) to directrix x + 5 = 0.

Question 9.
If vertex and focus of any parabola are (2,0) and (5, 0) respectively, then its equation will be :
(A) y² = 12x + 24
(B) y² = 12x – 24
(C) y² = – 12x – 24
(D) y² = – 12x + 24
Solution:
Vertex = (2, 0),
Focus = (5, 0)
Distance between vertex and focus

∴ X = x – 2
and Y = y
Thus equation of parabola
Y² = 4aX
⇒ y² = 4 × 3 × (x – 2)
⇒ y² = 12x – 24
Thus, optional (B) is correct.

Question 10.
Focus of parabola x² = – 8y is :
(A) (2, 0)
(B) (0, 2)
(C) (-2, 0)
(D) (0, – 2)
Solution:
Equation of parabola,
x² = – 8y
⇒ x² = -4 × 2 × y
Coordinates of focus = (0, – a) – (0, – 2)
Thus, optional (D) is correct.

Question 11.
Equation of any tangent at parabola y² = x is :
(A) y = mx + 1/m
(B) y = mx + 1 /4m
(C) y = mx + 4/m
(D) y = mx + 4m
Solution:
Equation of parabola,
y² = x

Thus, optional (B) is correct.

Question 12.
If line 2y – x = 2 touches the parabola y² = 2x then tangent point is :
(A) (4, 3)
(B) (-4, 1)
(C) (2, 2)
(D) (1, 4)
Solution:
Equation of line,
2y – x = 2
x = 2y – 2 ….(i)
Equation of parabola,
y² = 2x
From equation (i) and (ii),
y² = 2(2y – 2)
⇒ y² = 4y – 4
⇒ y² – 4y + 4 = 0
⇒ (y – 2)² = 0
⇒ y – 2 = 0
⇒ y = 2
Put value ofy in equation (i),
x = 2 × 2 – 2 = 2
Thus tangent point, (x, y) = (2, 2)
Thus, optional (C) is correct.

Question 13.
Tangent equation of parabola x² = 8y parallel to line x + 2y + 1 = 0 is :
(A) x + 2y + 1 = 0
(B) x – 2y + 1 = 0
(C) x + 2y – 1 = 0
(D) x – 2y – 1 = 0
Solution:
Equation of parabola,
x² = 8y
⇒ x² = 4 × 2 × y
here a = 2
Equation of line,
x + 2y + 1 = 0
Equation of its parallel line.
x + 2y + λ = 0
According to condition of tandency

Thus, equation of tangent line x + 2y – 4 = 0
Thus, optional (A) is correct.

Question 14.
A normal of parabola y² = 4x is :
(A) y = x + 4
(B) y + x = 3
(C) y + x = 2
(D) y + x = 1
Solution:
Equation of parabola
y² = 4x
y² = 4 × 1 × x
Here a = 1
To find normal, a point should be given which is not given here. So Question is incomplete.

Question 15.
Length of semi latus rectum of ellipse x² + 4y² = 12 will be :

Solution:
Education of ellipse
3x² + 4y² =12

Thus option (A) is correct.

Question 16.
Eccentricity of ellipse 3x² + 4y² = 12 will be :
(A) -2
(B) 1/2
(C) 1
(D) 2
Solution:

Question 17.
If line y = mx + c touches the ellipse

then value of c will be:

Solution:
Option (C) is correct.

Question 18.
Coordinates of focus of ellipse

(A) (± ae, 0)
(B) (± be, 0)
(C) (0, ± ae)
(D) (0, ± be)
Solution:
Option (D) is correct.

Question 19.
Eccentricity of rectangular hyperbola will be:
(A) 0
(B) 1
(C) √2
(D) 2
Solution:
Option (C) is correct.

Question 20.
Eccentricity of hyperbola 9×2 – 16y2 = 144 will be :
(A) 1
(B) 0
(C) 5/16
(D) 5 / 4
Solution:
Equation of hyperbola,
9x² – 16y² = 144


Thus Option (D) is correct.

Question 21.
Write the equation of circle whose centre is (a cos α, a sin α) and radius is a.
Solution:
Centre of circle
(- g, -f) – (a cos α, a sin α)
or (g,f) = (-a cos α, – a sin α)

Then equation of circle,
x² + y² + 2gx + 2fy + c = 0
⇒ x² + y² – 2a cos α . x – 2a sin α. y = 0
This is required equation.

Question 22.
If tangents at points (x₁, y₁) and (x₂, y₂) of circle x² + y² + 2gx + 2fy + c = 0 are perpendicular to each other than prove that
x₁x₂ + y₁y₂ + g(x₁ + x₂) + f(y₁ + y₂) + g² + f² = 0
Solution:
Given equation of circle
x² + y² + 2gx + 2fy + c = 0 …..(i)
Equation of tangent at point (x₁, y₁)
xx₁ + yy₁ + g(x + x₁) +f(y + y₁) + c = o
⇒ x₆x + y₆y + g(x₁ + x) +f(y₁ + y) + c = 0 …..(ii)

Question 23.
Find the equation of circle with radius r, whose centre lies in 1st quadrant and touches y-axis at a distance of h from the origin. Find the equation of other tangent which passes through origin.
Solution:
Centre of circle = (r, h)
Radius of circle = r
Thus, equation of circle
(x – r)² + (y – h)² = r²
Let tangent OB touches the circle at B. Tangent passes through origin. Let tangent touches the circle at point (x₁,y₁).
∴ Equation of tangent at point of circle,
xx₁ + yy₁ + g(x + x₁) + f(y + y₁) + c = 0
xx₁ + yy₁ – r(x + x₁) + h(y + y₁) + h² = 0
Since the line passes through origin.
∴ x × 0 + y × 0 – r(x + 0) – h(y + 0) + h² = 0
⇒ – rx – hy + h² = 0
⇒ rx + hy – h² = 0
This is required equation.

Question 24.
Tangents drawn at point (α, β) of circle x² + y² = a² meets the axis at points A and B respectively. Prove that area of ∆OAB will be,

where O is origin.
Solution:
Tangent equation of circle
x² + y² = a²
Equation of tangent at point (α, β)
xx₁ + yy₁ = a²
At x-axis x × a + y × β = a²

Question 25.
Find the equation of tangent at circle x² + y² = a² which makes a triangle of area a² with axis.
Solution:
Equation of circle,
x² + y² = a²
Area of ∆ABO = a²

Question 26.
Write the coordinates of the focus of parabola x² – 4x – 8y = 4.
Solution:
Equation of parabola,
x² – 4x – 8y = 4
⇒ x² – 2.2x + (2)² = 8y + 4 + 2²
⇒ (x – 2)² = 8y + 8
= 8(y + 1)
(x-2)² = 4.2.(y + 1)
X² = 4.2.Y
Where x = x – 2 and Y = y + 1
a = 2
Coordinates of focus = (0, a)
x = x- 2 = 0
⇒ x = 2
Y = y + 1 = 2
⇒ y = 1
Thus coordinates of focus = (2, 1)

Question 27.
Write the eccentricity of parabola
x² – 4x – 4y + 4 = 0.
Solution:
Eccentricity of parabola is 1.

Question 28.
Write the condition for which line lx + my + n = 0 touches the parabola y² = 4ax
Solution:
Required condition: In = am²

Question 29.
Write the equation of parabola whose vertex is (0, 0) and focus is (0, – a).
Solution:
Vertex (0, 0) and focus is at (0, – a) which lies at y-axis.
Thus axis of parabola is -ve y-axis.
Equation of parabola is of the form x² = – 4ay
Thus equation of parabola,
x² = 4(-a)y
⇒ x² = – 4ay
This is required equation.

Question 30.
Write the equation of axis of parabola
9y² – 16x – 12y – 57 = 0.
Solution:
Equation of parabola,

Question 31.
Write the coordinates of centre of ellipse

Solution:
Equation of ellipse

Question 32.
Write the condition for which line x cos α + y sin α = p touches the ellipse

Solution:
Putting value ofy form line x cos a + y sin α =p in the ellipse.

or x²(α² cos² α + b² sin² α) – 2a² px cos α
+ (α²p² – a²b² sin² α) = 0 …(i)
Given line will touch ellipse if eqⁿ. (i) has equal roots.
(- 2a² p cos α)² – 4(α² cos² α + b² sin² α)
(α²p² – α²b² sin² α) = 0
or 4α²b² sin² α[α² cos² α – p² + b² sinz α) = 0
Thus p² = a² cos² α + b² sin² α

Question 33.
Write the equation of hyperbola whose trans¬verse axis and conjugate axis are 4 and 5 respectively.
Solution:
Equation of hyperbola,

Question 34.
Write the coordinates of centre of hyperbola

Solution:
Equation of hyperbola

Comparing eqⁿ. (1) by standard equation of hyperbola, coordinates of centre.
X = x – 1 = 0
⇒ x = 1
Y = y + 2 = 0
⇒ y = – 2
Thus, coordinates of centre = (1, – 2).

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