Ellipse Test

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Ellipse Test - 3

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Question 1
4 / -1

If distance between the directrices be thrice the distance between the foci, then eccentricity of ellipse is

A
1/2

B
2/3

C
1/√3

D
4/5

Solution

Distance between the directrices is 2a/e
Distance between the foci is 2ae
Given: 2a/e = 3∗2ae
Or,e² = 1/3
∴ e=1/√3
Question 2
4 / -1

The curve represented by x = 3(cost + sint), y = 4(cost – sint), is

A
Ellipse

B
Parabola

C
Hyperbola

D
Circle
Question 3
4 / -1

A tangent having slope of - 4/3 to the ellipse ⁺ = 1 intersects the major & minor axes in points A & B respectively. If C is the centre of the ellipse then the area of the triangle ABC is

A
12 sq. units

B
24 sq. units

C
36 sq. units

D
48 sq. units

Solution

Since the major axis is along the y-axis.
∴ Equation of tangent is x = my + [b²m² + a]^1/2
Slope of tangent = 1/m = −4/3
⇒ m = −3/4
Hence, equation of tangent is 4x+3y=24 or
x/6 + y/8 = 1
Its intercepts on the axes are 6 and 8.
Area (ΔAOB) = 1/2×6×8
= 24 sq. unit
Question 4
4 / -1

An ellipse is drawn with major and minor axes of lengths 10 and 8 respectively. Using one focus as centre, a circle is drawn that is tangent to the ellipse, with no part of the circle being outside the ellipse. The radius of the circle is

A
√3

B
2

C
2√2

D
√5

Solution

2a=10,a=5
2b=8,b=4
b²=a²(1−e²)
e²=1−16/25
e=3/5
Focus of ellipse (±3,0)
Taking (3,0) as centre, circle is drawn,
Radius of circle will be =5−3 = 2 units.
Question 5
4 / -1

Which of the following is the common tangent to the ellipses + = 1 & + = 1 ?

A
ay = bx +

B
by = ax –

C
ay = bx –

D
by = ax –

Solution

For x²/(a² + b²) + y²/b² = 1
Equation for tangent y = mx +- [(a² + b²)m² + b²]^1/2
For x²/a² + y²/(a² + b²) = 1
Equation for tangent y = mx +- [a²m² + (a² + b²)}^1/2
For common tangent: [(a² + b²)m² + b²]^1/2 = [a²m² + (a² + b²)}^1/2
(a² + b²)m² + b² = a²m² + (a² + b²)
a²m² + b²m² + b² = a²m² + (a² + b²)
b²m² = a²
m = +- a/b
Equation for tangent : y = +- ax/b +-[(a² + b²)a²/b² + b²]^1/2
y = +- ax/b +-{[a⁴ + a²b² + b⁴]^1/2}/b
by = +- ax +-{[a⁴ + a²b² + b⁴]^1/2}
Question 6
4 / -1

The eccentricity of the ellipse + = 1 is decreasing at the rate of 0.1/second due to change in semi minor axis only. The time at which ellipse become auxiliary circle is

A
2 seconds

B
3 seconds

C
4 seconds

D
5 seconds
Question 7
4 / -1

The point of intersection of the tangents at the point P on the ellipse + = 1, and its corresponding point Q on the auxiliary circle meet on the line

A
x = a/e

B
x = 0

C
y = 0

D
None

Solution
Question 8
4 / -1

Q is a point on the auxiliary circle of an ellipse. P is the corresponding point on ellipse. N is the foot of perpendicular from focus S, to the tangent of auxiliary circle at Q. Then

A
SP = SN

B
SP = PQ

C
PN = SP

D
NQ = SP

Solution

P = (acosθ, bsinθ)
Q = (acosθ, asinθ)
equation of tangent at Q
Question 9
4 / -1

Q is a point on the auxiliary circle corresponding to the point P of the ellipse + = 1. If T is the foot of the perpendicular dropped from the focus S onto the tangent to the auxiliary circle at Q then the D SPT is

A
Isosceles

B
Equilateral

C
Right angled

D
Right isosceles

Solution

Q = (acosθ, asinθ)
P = (acosθ, bsinθ)
∆ SPT is an isosceles triangle.
Question 10
4 / -1

The equation of the normal to the ellipse + = 1 at the positive end of latus rectum is

A
x + ey + e²a = 0

B
x – ey – e³a = 0

C
x – ey – e²a = 0

D
None

Solution

The positive end of latus rectum is
Question 11
4 / -1

PQ is a double ordinate of the ellipse x²+ 9y² = 9, the normal at P meets the diameter through Q at R, then the locus of the mid point of PR is

A
A circle

B
A parabola

C
An ellipse

D
A hyperbola

Solution

Equation of ellipse : x²/9 + y²/1= 1
Co−ordinates of point P is : (acosθ , bsinθ)
Equation of normal at point (x1, y1) is :
a²x/x1 − b²y/y1 = (ae)²
Equation of normal at point P is:
ax/cosθ− by/sinθ= (ae)²−−−−−(1)
Equation of diameter at Q is : y = (−b/a)tanθ−−−−−(2)
Point of intersection of equation 1 and 2 is :
ax/cosθ + (b²/acosθ) = (ae)²
Or , x = ae²cosθ − b²/a²
And y = − be²sinθ + b³/a³ (from 2)
∴ Point R(ae²cosθ − b²/a² , − be²sinθ + b³/a³)
Let mid point of PR is (h,k)
h =[acosθ + ae²cosθ − (b²/a²)]/2
Or , 2h + b²/a²= acosθ + ae²cosθ
Or , cosθ= [2h + (b²/a²)/(a + ae²)]
k = [bsinθ − be²sinθ + (b³/a³/2)]/2
Or , sinθ = [2k − (b³/a³)/(b − be²)]
On squaring adding we get :
[2h + (b²/a²)]/(a + ae²)² + [2k − (b³/a³)] / (b − be²)²= 1
Which is an equation of ellipse.
Question 12
4 / -1

Which of the following is the eccentricity for ellipse?

A
1

B
3/2

C
2/3

D
5/2

Solution

The eccentricity for ellipse is always less than 1. The eccentricity is always 1 for any parabola. The eccentricity is always 0 for a circle. The eccentricity for a hyperbola is always greater than 1.
Question 13
4 / -1

If F₁ & F₂ are the feet of the perpendiculars from the foci S₁ & S₂ of an ellipse + = 1 on the tangent at any point P on the ellipse, then (S₁F₁) . (S₂F₂) is equal to

A
2

B
3

C
4

D
5

Solution

Given, x²/5 + y²/3= 1
We know S₁F₁ × S₂F₂ = b²
∴ S₁F₁ × S₂F₂ = 3
Question 14
4 / -1

If tan q₁. tan q₂ = – then the chord joining two points q₁ & q₂ on the ellipse + = 1 will subtend a right angle at

A
Focus

B
Centre

C
End of the major axis

D
End of the minor axis

Solution

Clearly, they subtend right angle at centre.