Ellipse Test

Result

Ellipse Test - 4

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Question 1
4 / -1

An ellipse having foci at (3,3) and (-4,4) and passing though the origin has eccentricity equal to

A
3/7

B
2/7

C
5/7

D
3/5

Solution

Ellipse passing through O (0, 0) and having foci P (3, 3) and Q (-4, 4) ,
Then
Question 2
4 / -1

The length of the major axis of the ellipse (5x - 10)² + (5y + 15)²

A
10

B
20/3

C
20/7

D
4

Solution

is an ellipse, whose focus is (2, -3) , directrix 3x - 4y + 7 = 0 and eccentricity is 1/2.
Length offrom focus to directrix is

So length of major axis is 20/3
Question 3
4 / -1

The eccentric angle of a point on the ellipse at a distance of 5/4 units from the focus on the positive x-axis, is

A

B

C

D
None of these

Solution

Any point on the ellipse is (2 cosθ, √3 sinθ). The focus on the positive x-axis is (1,0).
Given that (2cosθ-1)² + 3sin² θ = 25/16
⇒ cos θ = 3/4
Question 4
4 / -1

From any point P lying in first quadrant on the ellipse PN is drawn perpendicular to the major axis and produced at Q so that NQ equals to PS, where S is a focus. Then the locus of Q is

A
5y - 3x - 25 = 0

B
3x + 5y+ 25 = 0

C
3x - 5y- 25 = 0

D
None of these

Solution

Let point Q be (h,k) , where k < 0
Given that k = SP = a + ex₁ , where P (x₁,y₁) lies on the ellipse
Question 5
4 / -1

If a tangent of slope 2 of the ellipse is normal to the circle x² + y² + 4x + 1 = 0 , then the maximum value of ab is

A
4

B
2

C
1

D
None of these

Solution

Let be the tangent
It is passing through(-2,0)
Question 6
4 / -1

An ellipse has the points (1, -1) and (2, -1) as its foci and x +y - 5 = 0 as one of its tangents. Then the point where this line touches the ellipse is

A

B

C

D
None of these

Solution

Let image of S'' be with respect to x +y - 5 = 0

Let P be the point of contact.
Because the line L = 0 is tangent to the ellipse, there exists a point P uniquely on the line such that PS + PS ' = 2a .
Since PS ' = 2a Hence, P should be the collinear with SS ''
Hence P is a point of intersection of SS '' (4x - 5 y = 9) , and
Question 7
4 / -1

From point P (8, 27) , tangent PQ and PR are drawn to the ellipse Then the angle subtended by QR at origin is

A

B

C

D

Solution

Equation of QR is T = 0 (chord of contact)
⇒ 2x + 3y = 1 .......(i)
Now, equation of the pair of lines passing through origin and points Q, R is given by

(making equation of ellipse homogeneous using Eq (i)
∴ 135x² + 432xy + 320 y² = 0
Question 8
4 / -1

An ellipse is sliding along the co-ordinate axes. If the foci of the ellipse are (1,1) and (3,3), then area of the director circle of the ellipse (in sq. units) is

A
2π

B
4π

C
6π

D
8π

Solution

Since x-axis and y-axis are perpendicular tangents to the ellipse, (0,0) lies on the director circle and midpoint of foci (2,2) is centre of the circle.
Hence, radius = 2√2
⇒ the area is 8π units.
Question 9
4 / -1

The equation of the ellipse whose axes are coincident with the co-ordinates axes and which touches the straight lines 3x - 2y - 20 = 0 and x + 6y - 20 = 0 is

A

B

C

D

Solution

Let the equation of the ellipse be

We know that the general equation of the tangent to the ellipse is
......(i)
Since 2x - 2y - 20 = 0

is tangent to the ellipse .
Comparing with eq (i)

......(ii)
Similarly, since
is tangent to the ellipse, therefore comparing with eq. …(i)

⇒ a² + 36b² = 400.....(iii)
Solving eqs. (ii) and (iii) , we get a² = 40 and b² = 10 , Therefore , the required equation of the ellipse is
Question 10
4 / -1

Number of point on the ellipse from which pair of perpendicular tangents are drawn to the ellipse

A
0

B
2

C
1

D
4

Solution

For the ellipse
Equation of director circle is x² +y² = 25. The director circle will cut the ellipse at 4 points