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Test: Ellipse- 2 - JEE MCQ


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20 Questions MCQ Test Mathematics (Maths) for JEE Main & Advanced - Test: Ellipse- 2

Test: Ellipse- 2 for JEE 2024 is part of Mathematics (Maths) for JEE Main & Advanced preparation. The Test: Ellipse- 2 questions and answers have been prepared according to the JEE exam syllabus.The Test: Ellipse- 2 MCQs are made for JEE 2024 Exam. Find important definitions, questions, notes, meanings, examples, exercises, MCQs and online tests for Test: Ellipse- 2 below.
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Test: Ellipse- 2 - Question 1
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If distance between the directrices be thrice the distance between the foci, then eccentricity of ellipse is

Detailed Solution for Test: Ellipse- 2 - Question 1

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

Test: Ellipse- 2 - Question 2
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If the eccentricity of an ellipse be 5/8 and the distance between its foci be 10, then its latus rectum is

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Test: Ellipse- 2 - Question 3
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The curve represented by x = 3(cost + sint), y = 4(cost – sint), is

Test: Ellipse- 2 - Question 4
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If the distance of a point on the ellipse +  = 1 from the centre is 2, then the eccentric angle is
Test: Ellipse- 2 - Question 5
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An ellipse having foci at (3, 3) and (–4, 4) and passing through the origin has eccentricity equal to
Test: Ellipse- 2 - Question 6
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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
Detailed Solution for Test: Ellipse- 2 - Question 6

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

Test: Ellipse- 2 - Question 7
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The equation to the locus of the middle point of the portion of the tangent to the ellipse  +  = 1 included between the co-ordinate axes is the curve
Test: Ellipse- 2 - Question 8
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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
Detailed Solution for Test: Ellipse- 2 - Question 8

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.

Test: Ellipse- 2 - Question 9
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Which of the following is the common tangent to the ellipses  +  = 1 & + = 1 ?
Detailed Solution for Test: Ellipse- 2 - Question 9

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}

Test: Ellipse- 2 - Question 10
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Angle between the tangents drawn from point (4, 5) to the ellipse  +  = 1 is
Test: Ellipse- 2 - Question 11
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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
Test: Ellipse- 2 - Question 12
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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
Detailed Solution for Test: Ellipse- 2 - Question 12

Test: Ellipse- 2 - Question 13
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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
Detailed Solution for Test: Ellipse- 2 - Question 13

P = (acosθ, bsinθ)
Q = (acosθ, asinθ)
equation of tangent at Q 

Test: Ellipse- 2 - Question 14
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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
Detailed Solution for Test: Ellipse- 2 - Question 14

Q = (acosθ, asinθ) 
 P = (acosθ, bsinθ) 
 ∆ SPT is an isosceles triangle.

Test: Ellipse- 2 - Question 15
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The equation of the normal to the ellipse  +  = 1 at the positive end of latus rectum is
Detailed Solution for Test: Ellipse- 2 - Question 15

 The positive end of latus rectum is

Test: Ellipse- 2 - Question 16
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The eccentric angle of the point where the line, 5x - 3y = 8√2 is a normal to the ellipse  +  = 1 is
Test: Ellipse- 2 - Question 17
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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
Detailed Solution for Test: Ellipse- 2 - Question 17

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.

Test: Ellipse- 2 - Question 18
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Which of the following is the eccentricity for ellipse?
Detailed Solution for Test: Ellipse- 2 - Question 18

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.

Test: Ellipse- 2 - Question 19
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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
Detailed Solution for Test: Ellipse- 2 - Question 19

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

Test: Ellipse- 2 - Question 20
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If tan q1. tan q2 = – then the chord joining two points q1 & q2 on the ellipse +  = 1 will subtend a right angle at
Detailed Solution for Test: Ellipse- 2 - Question 20


Clearly, they subtend right angle at centre.

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