Test: Parabola- 3 - JEE MCQ

# Test: Parabola- 3 - JEE MCQ

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## 30 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Parabola- 3

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Test: Parabola- 3 - Question 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

Detailed Solution for Test: Parabola- 3 - Question 1

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.

Test: Parabola- 3 - Question 2

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

Detailed Solution for Test: Parabola- 3 - Question 2

Solving x + y = k + 1 and y = x(1 – x) ⇒ x2 – 2x + k + 1 = 0  and Δ = 0

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Test: Parabola- 3 - Question 3

### 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)

Detailed Solution for Test: Parabola- 3 - Question 3

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

Test: Parabola- 3 - Question 4

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

Detailed Solution for Test: Parabola- 3 - Question 4

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.

Test: Parabola- 3 - Question 5

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

Detailed Solution for Test: Parabola- 3 - Question 5

It lies on y = 0.
∴

Test: Parabola- 3 - Question 6

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.

Detailed Solution for Test: Parabola- 3 - Question 6

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

Test: Parabola- 3 - Question 7

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

Detailed Solution for Test: Parabola- 3 - Question 7

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

Test: Parabola- 3 - Question 8

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

Detailed Solution for Test: Parabola- 3 - Question 8

Taking the new axes as  we see that the parabola can be  with the required condition |X| <  200

Test: Parabola- 3 - Question 9

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

Detailed Solution for Test: Parabola- 3 - Question 9

Center of such circle in the case of parabolas  y2 = 4ax, x2 = 4ay is

Test: Parabola- 3 - Question 10

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

Detailed Solution for Test: Parabola- 3 - Question 10

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

Test: Parabola- 3 - Question 11

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

Detailed Solution for Test: Parabola- 3 - Question 11

Let P be (t2, 2t) area of Δ PAB

it is maximum at t = 1/2.

Test: Parabola- 3 - Question 12

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

Detailed Solution for Test: Parabola- 3 - Question 12

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.

Test: Parabola- 3 - Question 13

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

Detailed Solution for Test: Parabola- 3 - Question 13

Test: Parabola- 3 - Question 14

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

Detailed Solution for Test: Parabola- 3 - Question 14

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

Test: Parabola- 3 - Question 15

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

Detailed Solution for Test: Parabola- 3 - Question 15

(y – 3)2 = –4 (x – 3) (given equation of parabola.
∴ equation of latus rectum is x = 2.

Test: Parabola- 3 - Question 16

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

Detailed Solution for Test: Parabola- 3 - Question 16

Test: Parabola- 3 - Question 17

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

Detailed Solution for Test: Parabola- 3 - Question 17

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

Test: Parabola- 3 - Question 18

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

Detailed Solution for Test: Parabola- 3 - Question 18

Test: Parabola- 3 - Question 19

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

Detailed Solution for Test: Parabola- 3 - Question 19

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,

Test: Parabola- 3 - Question 20

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

Detailed Solution for Test: Parabola- 3 - Question 20

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

Test: Parabola- 3 - Question 21

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

Detailed Solution for Test: Parabola- 3 - Question 21

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.

Test: Parabola- 3 - Question 22

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

Detailed Solution for Test: Parabola- 3 - Question 22

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

Test: Parabola- 3 - Question 23

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

Detailed Solution for Test: Parabola- 3 - Question 23

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

Test: Parabola- 3 - Question 24

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

Detailed Solution for Test: Parabola- 3 - Question 24

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

Test: Parabola- 3 - Question 25

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

Detailed Solution for Test: Parabola- 3 - Question 25

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

Test: Parabola- 3 - Question 26

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

Detailed Solution for Test: Parabola- 3 - Question 26

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).

Test: Parabola- 3 - Question 27

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

Detailed Solution for Test: Parabola- 3 - Question 27

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)

Test: Parabola- 3 - Question 28

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

Detailed Solution for Test: Parabola- 3 - Question 28

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) .

Test: Parabola- 3 - Question 29

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

Detailed Solution for Test: Parabola- 3 - Question 29

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)

Test: Parabola- 3 - Question 30

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

Detailed Solution for Test: Parabola- 3 - Question 30

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).

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