Ellipse Test

Result

Ellipse Test - 6

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Question 1
4 / -1

Q is a point on the auxiliary circle corresponding to the point P of the ellipse If T is the foot of the perpendicular dropped from the focus S on to the tangent to the auxiliary circle at Q then the ΔSPT is :

A
isosceles

B
equilateral

C
right angled

D
right isosceles

Solution

Tangent at Q is x cos θ + y sin θ = a
ST = |a e cosθ - a| = a(1 - e cos θ)
Also SP = e PM

ST = SP ⇒ isosceles.
Question 2
4 / -1

Let x and y satisfy the relation x² + 9y² - 4x + 6y + 4 = 0, then maximum value of the expression (4x - 9y),

A
15

B
11

C
16

D
21

Solution

Given equation is

Let x - 2 = cos θ
Then 4x - 9y = 11+ 4 cosθ - 3sinθ
Question 3
4 / -1

A rectangle ABCD has area 200 square units. An ellipse with area 200π passes through A and C and has foci at B and D, then

A
perimeter of the rectangle is 80 units

B
dimensions of the rectangle are 4 and 20

C
dimensions of the rectangle are 4 and 10

D
perimeter of the rectangle is 40 units
Question 4
4 / -1

A tangent to the ellipse at any point P meet the line x = 0 at a point Q. Let R be the image of Q in the line y = x, then circle whose extremities of a diameter are Q and R passes through a fixed point. The fixed point is

A
(3, 0)

B
(5, 0)

C
(0, 0)

D
(4, 0)

Solution

Equation of the tangent to the ellipse at P (5 cos θ , 4 sin θ) is
It meets the line x = 0 at Q (0, 4 cosec θ)
Image of Q in the line y = x is R (4 cosec θ , 0)
∴ Equation of the circle is
X (x – 4 cosec θ) + y(y – 4 cosec θ) = 0
i.e. x² + y² – 4 (x + y) cosec θ = 0
∴ each member of the family passes through the intersection of x² + y² = 0 and x + y = 0
i.e. the point (0, 0).
Question 5
4 / -1

If the curve x² + 3y² = 9 subtends an obtuse angle at the point (2α, α), then a possible value of α² is

A
1

B
2

C
3

D
4

Solution

The given curve is whose director circle is x² + y² = 12. For the required condition (2α, α) should lie inside the circle and outside the ellipse i.e.,
Question 6
4 / -1

Let S = 0 be the equation of reflection of about the line x – y – 2 = 0. Then the locus of point of intersection of perpendicular tangents of S is

A
x² + y² + 10x - y - 4 = 0

B
x² +y² = 25

C
x² + y² -10x - 4 y + 4 = 0

D
x² + y² -10x - 4 y = 0

Solution

(x - 5)² + (y - 2)² = 16 + 9
Question 7
4 / -1

Any ordinate MP of a ellipse meets the auxiliary circle in Q, then locus of point of intersection of normals at P and Q to the respective curves, is

A
x² + y² = 8

B
x² + y² = 34

C
x² + y² = 64

D
x² + y² = 15

Solution

Equation of normal to the ellipse at ‘P’ is
5x sec θ – 3y cosec θ = 16 …(1)
Equation of normal to the circle x² + y² = 25 at point Q is –

y = x tan θ
Eliminating θ from (1) & (2). We get x² + y² = 64.
Question 8
4 / -1

The radius of the largest circle whose centre at (-3,0) and is inscribed in the ellipse 16x² + 25y² = 400 is

A
2

B
3

C
4

D
1
Question 9
4 / -1

The line passing through the extremely A of the major axis and extremity B of the minor axis of the ellipse x² + 9y² = 9 meets its auxiliary circle at the point M . Then the area of the triangle with vertices at A,M and the origin O is

A
31/10

B
29/10

C
21/10

D
27/10

Solution

Equation of line AM is
x +3y -3 = 0
Perpendicular distance of line from the origin
Length of AM
Question 10
4 / -1

Coordinate of the vertices B and C of triangle ABC are respectively (2, 0) and (8, 0). The vertex ‘A’ is varying in such a way that Then the locus of ‘A’ is an ellipse whose major axis is of length.

A
6

B
8

C
15

D
10

Solution

Let BC = a, CA = b & AB = c

∴ b + c = 2s – a = 10.
Locus of A is an ellipse with major axis of length 10