Parabola Test

Result

Parabola Test - 9

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Weekly Quiz Competition

Question 1
4 / -1

Let A ≡ (9, 6), B(4, -4) be two points on parabola y² = 4x and P(t², 2t), t∈[-2, 3] be a variable point on it such that area of DPAB is maximum, then point P will be

A
(4, 4)

B

C
(4, 1)

D

Solution

Let P be (t², 2t) area of Δ PAB

it is maximum at t = 1/2.
Question 2
4 / -1

Two tangents to the parabola y² = 4x, one drawn at a point P and another drawn at the point where the normal at the image of P in the axis of the parabola meets the curve again, include an angle

A
There will be only one such point P on the parabola for any such θ

B
There will be three such points P on the parabola for any such θ

C
No such point P is possible for any given θ

D
P must be one extremity of the latus rectum

Solution

If P(t) is the point and Q(T) is another point in the question, we have and

can have only one real root, there will be only one such point P.
Question 3
4 / -1

If the chord joining the points t₁ and t₂ on the parabola y² = 4ax subtends a right angle at its vertex then t₂ =

A

B

C

D

Solution
Question 4
4 / -1

The equation of the latusrectum of the parabola y² – 6y + 4x – 3 = 0 is

A
x = –2

B
x = 2

C
x = –3/2

D
x = 3/2

Solution

(y - 3)² = -4 (x - 3) (given equation of parabola.
∴ equation of latus rectum is x = 2.
Question 5
4 / -1

The equation of the latusrectum of the parabola y² – 6y + 4x – 3 = 0 is

A
x = –2

B
x = 2

C
x = –3/2

D
x = 3/2

Solution

(y – 3)² = –4 (x – 3) (given equation of parabola.
∴ equation of latus rectum is x = 2.
Question 6
4 / -1

If the line x – 1 = 0 is the directrix of the parabola y² – kx + 8 = 0, then one of the value of k is

A
1/8

B
8

C
4

D
1/4

Solution
Question 7
4 / -1

The curve described parametrically by x = t² + t + 1, y = t² - t + 1 represents

A
a circle

B
a parabola

C
an ellipse

D
a hyperbola

Solution

x = t² +t +1andy = t² -t +1
⇒ x + y - 2 = 2t² and x - y = 2t
⇒ 2(x + y -2) = (x - y)² ⇒ x² + y² -2xy -2x -2y + 4 = 0
Comparing with the equation
ax² + 2hxy + by² + 2gx + 2fy + c = 0,
∴ abc + 2fgh - af² - bg² - ch² ≠ and h² = ab
Question 8
4 / -1

The length of the latus rectum of the parabola 169{(x - 1)² + (y - 3)²} = (5x - 12y + 17)²} is

A
14/13

B
28/13

C
12/13

D
13/28
Question 9
4 / -1

The angle between the tangents drawn from the point (3, 4) to the parabola y² - 2y + 4x = 0 is

A

B

C

D

Solution

The equation of the parabola is

The equation of any tangent to this parabola is
If it passes through (3, 4), then
⇒ 11m² -12m - 4 = 0
Let m₁ & m₂ be the roots of this equation. Then,

Let θ be the angle between the tangents. Then,
Question 10
4 / -1

Equation of common tangent of parabola y² = 8x and x² + y = 0 is

A
y = 2x + 1

B
x = y + 1

C
2x - y + 1 = 0

D
x + 2y + 1 = 0

Solution

if it is common tangent, then m³ = 8 m = 2.