Parabola Test

Result

Parabola Test - 8

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Question 1
4 / -1

The combined equation to two parabolas, both have their axis along x-axis, is given by y⁴ - y² (4x + 4 - 2 sin²2α) + sin²2α (4x + 4x + sin² 2α) = 0. The locus of the point of intersection of tangents, one to each of the parabolas, when they include an angle of 90°
is

A
another parabola

B
circle

C
a line midway between the tangents at vertices

D
a line midway between their directrices

Solution

The parabolas are y² = 4sin² α(x + sin²α) and y² = 4cos² α(x + cos²α), hence the locus is x + cos²α + sin²α = 0 ⇒ x + 1 = 0
Assume the point as origin and line joining it to the centre as x-axis, the equation to the circle becomes x² + y² - 2x = 0, center is A₁ (1, 0) and the second circle has the equation x² + y² - 2√2 = 0 center A₀ (0, √2)
Similarly A₃(-2, 0), A₄ (0, -2√2) etc.
Question 2
4 / -1

The straight line x + y = k + 1 touches the parabola y = x(1 – x) if

A
k = -1

B
k = 0

C
k = 1

D
k takes any real value

Solution

Solving x + y = k + 1 and y = x(1 – x) ⇒ x² – 2x + k + 1 = 0 and Δ = 0
Question 3
4 / -1

Focus and vertex of the parabola that touches x-axis at (1, 0) and x = y at (1, 1) are (h, k) and (p, q) then the value of 25(p + q +h + k)

A
35

B
37

C
33

D
39

Solution

The x-axis touches at A(1, 0) and x = y touches at B(1, 1). Hence the equation to the curve through these points is given by y(y – x) + k(x – 1)² = 0. For this to represent a parabola, 4k = 1. The equation is x² – 4xy + 4y² – 2x + 1 = 0. Vertex focus
Question 4
4 / -1

Number of circles that touch a given parabola and one of its fixed focal chords at focus is

A
0

B
3

C
2

D
1

Solution

Let the parabola is y² = 4x. The equations (x - t²) + (y - 2t)² + 2λ (x - yt + t²) = 0 and (x - 1)² + y² + 2μ(mx - y - m) = 0 should represent the same circle touching mx – y – m = 0 at (1, 0) and the parabola at (t², 2t). Eliminating λ and μ, we get mt³ - 3t² + 3mt + 1 = 0 will give three values of t for any given m.
Question 5
4 / -1

The chord AB of the parabola y² = 4ax cuts the axis of the parabola at C. If and AC : AB = 1 : 3, then

A
t₂ = 2t₁

B
t₂ + 2t₁ = 0

C
t₁ + 2t₂ = 0

D
t₁ + t₂ = 0

Solution

It lies on y = 0.
∴
Question 6
4 / -1

In a knockout tournament 16 equally skilled players namely P₁, P₂, -------- P₁₆ are participating. In each round players are divided in pairs at random and winner from each pair moves in the next round. If P₂ reaches the semifinal, then the probability that P1 will win the tournament is.

A
3/64

B
1/16

C
1/20

D
1/15

Solution

Let E₁ = P₁ win the tournament, E² = P₂ reaches the semifinal since all players are equally skilled and there are 4 persons in the semifinal
both are in semifinal and P₁ wins in semifinal and final
Question 7
4 / -1

If the parametric equations of the parabola are given by x = 4t² - 2t + 1; y = 3t² + t + 1 and the vertex of the parabola also satisfies y - x = k/100, then the area of the circle x² + y² + 12x -10y + 2k = 0 in square units is

A
π

B
3π

C
4π

D
5π

Solution

Eliminating t, we get (3x - 4y + 2)² = 16x + 12y - 27. Vertex is
Hence k = 29 and the area is 3π.
Question 8
4 / -1

A parabola has focus at (0, 0) and passes through the points (4, 3) and (–4, –3). The number of lattice points (x, y) on the parabola such that |4x + 3y| < 1000 is

A
20

B
100

C
40

D
80

Solution

Taking the new axes as we see that the parabola can be with the required condition |X| < 200
Question 9
4 / -1

Center of the smallest circle that is drawn to touch the two parabolas given by y² + 2x + 2y + 3 = 0; x² + 2x + 2y + 3 = 0 is

A

B

C

D
(-1, -1)

Solution

Center of such circle in the case of parabolas y² = 4ax, x² = 4ay is
Question 10
4 / -1

The equation of directrix and latusrectum of a parabola are 3x – 4y + 27 = 0 and 3x – 4y + 2 = 0. Then the length of latusrectum is

A
5

B
10

C
15

D
20

Solution

where d is the distance between lines whose equations are ax+by+C₁ =0 & ax+by+C₂ = 0

= 5
d = 5
If the distance between vertex and latus rectum = distance of vertex from directrix = a then d = 2a = 5
⇒ Length of latus rectum = 4a = 10