Parabola Test

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Parabola Test - 10

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Weekly Quiz Competition

Question 1
4 / -1

If the tangents at two points (1, 2) and (3, 6) as a parabola intersect at the point (– 1, 1), then the slope of the directrix of the parabola is

A
√2

B
-2

C
-1

D
none of these

Solution

If the tangents at P and Q intersect at T, then axis of parabola is parallel to TR, where R is the mid point of P and Q. So, slope of the axis is 1.
∴ slope of the directrix = – 1.
Question 2
4 / -1

A variable chord PQ of the parabola y = 4x² substends a right angle at the vertex. Then the locus of points of intersection of the tangents at P and Q is

A
4y + 1= 16x²

B
y + 4 = 0

C
4y + 4 = 4x²

D
4y + 1 = 0

Solution

Slope of OP x slope of OQ = -1
⇒ 4t₁.4t₂ = -1
Eq of tangent at

Eq of tangent at

Let (x₁ , y₁) is the point of intersection
Question 3
4 / -1

Point on the curve y² = 4(x- 10) which is nearest to the line x + y = 4 may be

A
(11, 2)

B
(10, 0)

C
(11, -2)

D
None of these

Solution

P(x₀ , y₀) : pt on curve nearest to line.
Normal at P is perpendicular to the line Normal at P has slope
∴ y₀= 2 and x₀ = 11; P(11, -2)
Question 4
4 / -1

If (h, k) is a point on the axis of parabola 2(x -1)² + 2 (y -1)² = (x + y + 2)² from where three distinct normals can be drawn, then

A
h > 2

B
h < 4

C
h > 8

D
h < 8

Solution

which is of the following PM² = SP²
∴ Focus= (1,1) . Directrix is x+y+2 =0
Axis is x-y=0 and z= (-1,-1)
Vertex = (0,0). Parameter a = √2
The distance of the point from vertex and lie on axis from which 3 normals can be drawn must be greater 2a = 2√2. Hence point on axis at a distance 2√2 is (2,2), Hence h>2
Question 5
4 / -1

The equation of the common tangent touching the circle (x – 3)² + y² = 9 and the parabola y² = 4x below y² - 4x below the x–axis is

A

B

C

D

Solution

Equation of tangent at (x₁, y₁) on the parabola y² = 4x is
2x - y₁ + 2x₁ = 0
length of perpendicular from centre of the circle (x - 3)² + y² = 9 is

Hence required equation
Question 6
4 / -1

The line x + y = 6 is a normal to the parabola y² = 8x at the point

A
(18, - 12)

B
(4, 2)

C
(2, 4)

D
(3, 3)

Solution

Slope of the normal is given to be -1. We know that, foot of the normal is (am², -2am). Here a = 2, m = -1. Hence the required point is (2, 4).
Question 7
4 / -1

The tangent and normal at the point P(4, 4) to the parabola, y² = 4x intersect the x–axis at the points Q and R respectively. Then the cirucm centre of the ΔPQR is

A
(2, 0)

B
(2, 1)

C
(1, 0)

D
(1, 2)

Solution

Eq. of tangent 2y = x + 4
∴ Q ≡ (-4,0)
Eq. of normal is y - 4 = -2 (x - 4)
⇒ y + 2x = 12
Clearly QR is diameter of the required circle.
⇒ (x + 4) (x – 6) + y² = 0
⇒ x² + y² – 2x – 24 = 0
centre (1, 0)
Question 8
4 / -1

All chords of the parabola y² = 4x which subtend right angle at the origin are concurrent at the point:

A
(1, 0)

B
(2, 0)

C
(3, 0)

D
(4, 0)

Solution

Let y = mx+ c be such chord with extremities A and B .
∴ The combined equation of the pair of lines OA and OB is
∴ Coeff of x² + Coeff of y² = 0

∴ c = -4 m
∴ The chord equation is y = m (x- 4) .
Question 9
4 / -1

The mirror image of the parabola y² = 4x in the tangent to the parabola to the point (1,2) is

A
(x - 1)² = 4(y + 1)

B
(x + 1)² = 4(y + 1)

C
(x + 1)² = 4(y - 1)

D
(x - 1)² = 4(y - 1)

Solution

Any point on the given parabola is (t², 2t). The equation of the tangent at (1,2) is x-y +1 = 0. The image (h,k) of the point (t²,2t) in x-y + 1 = 0 is given by

∴ h = t² - t² + 2t - 1 = 2t - 1
Eliminating t from h = 2t – 1 and k = t²+1
we get, (h+1)² = 4(k-1)
The required equation of reflection is (x+1)² = 4(y-1)
Question 10
4 / -1

The locus of mid–point of family of chords λx + y - 5 = 0 (parameter) of the parabola x² = 20y is

A
y² = 10(x - 5)

B
x² = 10(y - 5)

C
x² + y² = 25

D
none of these

Solution

λx + y - 5 = 0 is the focal chord
Since it passes through (0, 5)
Let h, k be the mid point of all such chord h = 5(t₁ + t₂),

On eliminating t₁, t₂
x² = 10 (y - 5).