RBSE Solutions for Class 11 Maths Chapter 12 Conic Section Miscellaneous Exercise

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) \(\frac { 1 }{ 2 } \)
(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) \(\sqrt { 2 }\)
(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).

RBSE Solutions for Class 11 Maths