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Ellipse

JEE Mathematics · 90 questions · Page 1 of 9 · Click an option or "Show Solution" to reveal answer

Q1
Let x = 4 be a directrix to an ellipse whose centre is at the origin and its eccentricity is 12{1 \over 2}. If P(1, β\beta ), β\beta > 0 is a point on this ellipse, then the equation of the normal to it at P is :
A 4x – 3y = 2
B 8x – 2y = 5
C 7x – 4y = 1
D 4x – 2y = 1
Correct Answer
Option D
Solution
e=12e = {1 \over 2}
x=ae=4x = {a \over e} = 4

⇒\Rightarrow a = 2

e2=1−b2a2{e^2} = 1 - {{{b^2}} \over {{a^2}}}
⇒14=1−b24\Rightarrow {1 \over 4} = 1 - {{{b^2}} \over 4}
b24=34⇒b2=3{{{b^2}} \over 4} = {3 \over 4} \Rightarrow {b^2} = 3

∴\therefore Ellipse

x24+y23=1{{{x^2}} \over 4} + {{{y^2}} \over 3} = 1

P(1, β\beta) is on the ellipse

14+β23=1{1 \over 4} + {{{\beta ^2}} \over 3} = 1
β23=34⇒β=32{{{\beta ^2}} \over 3} = {3 \over 4} \Rightarrow \beta = {3 \over 2}
⇒P(1,32)\Rightarrow P\left( {1,{3 \over 2}} \right)

Equation of normal

a2xx1−b2yy1=a2−b2{{{a^2}x} \over {{x_1}}} - {{{b^2}y} \over {{y_1}}} = {a^2} - {b^2}

⇒\Rightarrow

4x1−3y32=4−3{{4x} \over 1} - {{3y} \over {{3 \over 2}}} = 4 - 3

⇒\Rightarrow

4x−2y=14x - 2y = 1
Q2
If the co-ordinates of two points A and B are (7,0)\left( {\sqrt 7 ,0} \right) and (−7,0)\left( { - \sqrt 7 ,0} \right) respectively and P is any point on the conic, 9x2 + 16y2 = 144, then PA + PB is equal to :
A 8
B 9
C 16
D 6
Correct Answer
Option A
Solution

9x2 + 16y2 = 144 ⇒\Rightarrow

x216+y29=1{{{x^2}} \over {16}} + {{{y^2}} \over 9} = 1

∴\therefore a = 4; b = 3; Now e =

1−916=74\sqrt {1 - {9 \over {16}}} = {{\sqrt 7 } \over 4}

A and B are foci PA + PB = 2a = 2 × 4 = 8

Q3
Which of the following points lies on the locus of the foot of perpedicular drawn upon any tangent to the ellipse, x24+y22=1{{{x^2}} \over 4} + {{{y^2}} \over 2} = 1 from any of its foci?
A (−1,3)\left( { - 1,\sqrt 3 } \right)
B (−2,3)\left( { - 2,\sqrt 3 } \right)
C (−1,2)\left( { - 1,\sqrt 2 } \right)
D (1,2)\left( {1,2 } \right)
Correct Answer
Option A
Solution

Let foot of perpendicular is (h, k) Given

x24+y22=1{{{x^2}} \over 4} + {{{y^2}} \over 2} = 1

∴\therefore a = 2, b =

2\sqrt 2

and e =

1−24\sqrt {1 - {2 \over 4}}

=

12{1 \over {\sqrt 2 }}

∴\therefore Focus(ae, 0) =

(2,0)\left( {\sqrt 2 ,0} \right)

Equation of tangent y = mx +

a2m2+b2\sqrt {{a^2}{m^2} + {b^2}}

⇒\Rightarrow y = mx +

4m2+2\sqrt {4{m^2} + 2}

Passes through (h, k) (k – mh)2 = 4m2 + 2 .....(1) Line perpendicular to tangent will have slope

−1m- {1 \over m}

. y - 0 =

−1m(x−2)- {1 \over m}\left( {x - \sqrt 2 } \right)

my = -x +

2{\sqrt 2 }

∴\therefore (h + mk)2 = 2 .....(

2) Add equation (1) and (2) k2(1 + m2) + h2 (1 + m2) = 4 (1 + m2) h2 + k2 = 4 x2 + y2 = 4 (Auxiliary circle) ∴\therefore

(−1,3)\left( { - 1,\sqrt 3 } \right)

lies on the locus.

Q4
If the normal at an end of a latus rectum of an ellipse passes through an extremity of the minor axis, then the eccentricity e of the ellipse satisfies :
A e4 + 2e2 – 1 = 0
B e4 + e2 – 1 = 0
C e2 + 2e – 1 = 0
D e2 + e – 1 = 0
Correct Answer
Option B
Solution

Equation of normal at

(ae,b2a)\left( {ae,{{{b^2}} \over a}} \right)
a2xae−b2yb2a=a2−b2{{{a^2}x} \over {ae}} - {{{b^2}y} \over {{{{b^2}} \over a}}} = {a^2} - {b^2}

It passes through (0,–b) ∴\therefore

0−b2(−b)b2a=a2−b20 - {{{b^2}\left( { - b} \right)} \over {{{{b^2}} \over a}}} = {a^2} - {b^2}

⇒\Rightarrow

aa

b =

a2−b2{a^2} - {b^2}

⇒\Rightarrow

aa

b =

a2e2{a^2}{e^2}

[as b2 =

a2(1−e2){a^2}\left( {1 - {e^2}} \right)

] ⇒\Rightarrow

aa

2b2 =

a4e4{a^4}{e^4}

⇒\Rightarrow

b2a2{{{{b^2}} \over {{a^2}}}}

= e4 ⇒\Rightarrow

1−e2{1 - {e^2}}

= e4 ⇒\Rightarrow e4 + e2 – 1 = 0

Q5
If the curve x2 + 2y2 = 2 intersects the line x + y = 1 at two points P and Q, then the angle subtended by the line segment PQ at the origin is :
A π2−tan⁡−1(14){\pi \over 2} - {\tan ^{ - 1}}\left( {{1 \over 4}} \right)
B π2+tan⁡−1(13){\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 3}} \right)
C π2−tan⁡−1(13){\pi \over 2} - {\tan ^{ - 1}}\left( {{1 \over 3}} \right)
D π2+tan⁡−1(14){\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)
Correct Answer
Option D
Solution

Ellipse :

x2+y1=1{x \over 2} + {y \over 1} = 1

Line :

x+y=1x + y = 1

Using homogenisation

x2+2y2=2(1)2{x^2} + 2{y^2} = 2{(1)^2}
x2+2y2=2(x+y)2{x^2} + 2{y^2} = 2{(x + y)^2}
x2+2y2=2x2+2y2+4xy{x^2} + 2{y^2} = 2{x^2} + 2{y^2} + 4xy
x2+4xy=0{x^2} + 4xy = 0

for

ax2+2hxy+by2=0a{x^2} + 2hxy + b{y^2} = 0
tan⁡θ=∣2h2−aba+b∣\tan \theta = \left| {{{2\sqrt {{h^2} - ab} } \over {a + b}}} \right|
tan⁡θ=∣2(2)2−01+0∣\tan \theta = \left| {{{2\sqrt {{{(2)}^2} - 0} } \over {1 + 0}}} \right|
tan⁡θ=−4\tan \theta = - 4
cot⁡θ=−14\cot \theta = - {1 \over 4}
θ=cot⁡−1(−14)\theta = {\cot ^{ - 1}}\left( { - {1 \over 4}} \right)
θ=π−cot⁡−1(14)\theta = \pi - {\cot ^{ - 1}}\left( {{1 \over 4}} \right)
θ=π−(π2−tan⁡−1(14))\theta = \pi - \left( {{\pi \over 2} - {{\tan }^{ - 1}}\left( {{1 \over 4}} \right)} \right)
θ=π2+tan⁡−1(14)\theta = {\pi \over 2} + {\tan ^{ - 1}}\left( {{1 \over 4}} \right)
Q6
If the points of intersections of the ellipse x216+y2b2=1{{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1 and the circle x2 + y2 = 4b, b > 4 lie on the curve y2 = 3x2, then b is equal to :
A 12
B 10
C 6
D 5
Correct Answer
Option A
Solution
x216+y2b2=1{{{x^2}} \over {16}} + {{{y^2}} \over {{b^2}}} = 1

... (1)

x2+y2=4b{x^2} + {y^2} = 4b

.... (2)

y2=3x2{y^2} = 3{x^2}

.... (3) From eq (2) and (3) x2 = b and y2 = 3b From equation (1)

b16+3bb2=1{b \over {16}} + {{3b} \over {{b^2}}} = 1
⇒b2+48=16b\Rightarrow {b^2} + 48 = 16b
⇒b=12\Rightarrow b = 12
Q7
Let a tangent be drawn to the ellipse x227+y2=1{{{x^2}} \over {27}} + {y^2} = 1 at (33cos⁡θ,sin⁡θ)(3\sqrt 3 \cos \theta ,\sin \theta ) where 0∈(0,π2)0 \in \left( {0,{\pi \over 2}} \right). Then the value of θ\theta such that the sum of intercepts on axes made by this tangent is minimum is equal to :
A π6{{\pi \over 6}}
B π3{{\pi \over 3}}
C π8{{\pi \over 8}}
D π4{{\pi \over 4}}
Correct Answer
Option A
Solution

Tangent =

x33cos⁡θ+ysin⁡θ=1{x \over {3\sqrt 3 }}\cos \theta + y\sin \theta = 1

x-intercept =

33{3\sqrt 3 }

secθ\theta y-intercept = cosecθ\theta sum =

33{3\sqrt 3 }

secθ\theta + cosecθ\theta = f(θ\theta) θ\theta

∈\in
(0,π2)\left( {0,{\pi \over 2}} \right)

⇒\Rightarrow f'(θ\theta) =

33{3\sqrt 3 }

secθ\thetatanθ\theta −- cosecθ\theta cotθ\theta = 0 ⇒\Rightarrow

33sin⁡θcos⁡2θ=cos⁡θsin⁡θ{{3\sqrt 3 \sin \theta } \over {{{\cos }^2}\theta }} = {{\cos \theta } \over {\sin \theta }}
⇒tan⁡3θ=(13)3\Rightarrow {\tan ^3}\theta = {\left( {{1 \over {\sqrt 3 }}} \right)^3}
⇒tan⁡θ=13\Rightarrow \tan \theta = {1 \over {\sqrt 3 }}

⇒\Rightarrow θ\theta =

π6{{\pi \over 6}}

also f'(θ\theta) changes sign −- to + hence minimum.

Q8
Let E1:x2a2+y2b2=1,a>b{E_1}:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1,a > b. Let E2 be another ellipse such that it touches the end points of major axis of E1 and the foci of E2 are the end points of minor axis of E1. If E1 and E2 have same eccentricities, then its value is :
A −1+52{{ - 1 + \sqrt 5 } \over 2}
B −1+82{{ - 1 + \sqrt 8 } \over 2}
C −1+32{{ - 1 + \sqrt 3 } \over 2}
D −1+62{{ - 1 + \sqrt 6 } \over 2}
Correct Answer
Option A
Solution
e2=1−b2a2{e^2} = 1 - {{{b^2}} \over {{a^2}}}
e2=1−a2c2{e^2} = 1 - {{{a^2}} \over {{c^2}}}
⇒b2a2=a2c2\Rightarrow {{{b^2}} \over {{a^2}}} = {{{a^2}} \over {{c^2}}}
⇒c2=a4b2⇒c=a2b\Rightarrow {c^2} = {{{a^4}} \over {{b^2}}} \Rightarrow c = {{{a^2}} \over b}

Also b = ce

⇒c=be\Rightarrow c = {b \over e}
be=a2b{b \over e} = {{{a^2}} \over b}
⇒e=b2a2=1−e2\Rightarrow e = {{{b^2}} \over {{a^2}}} = 1 - {e^2}
⇒e2+e−1=0\Rightarrow {e^2} + e - 1 = 0
⇒e=−1+52\Rightarrow e = {{ - 1 + \sqrt 5 } \over 2}
Q9
Let an ellipse E:x2a2+y2b2=1E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1, a2>b2{a^2} > {b^2}, passes through (32,1)\left( {\sqrt {{3 \over 2}} ,1} \right) and has eccentricity 13{1 \over {\sqrt 3 }}. If a circle, centered at focus F(α\alpha, 0), α\alpha > 0, of E and radius 23{2 \over {\sqrt 3 }}, intersects E at two points P and Q, then PQ2 is equal to :
A 83{8 \over 3}
B 43{4 \over 3}
C 163{{16} \over 3}
D 3
Correct Answer
Option C
Solution
32a2+1b2=1{3 \over {2{a^2}}} + {1 \over {{b^2}}} = 1

and

1−b2a2=131 - {{{b^2}} \over {{a^2}}} = {1 \over 3}
⇒a2=3b2=3\Rightarrow {a^2} = 3{b^2} = 3
⇒x23+y22=1\Rightarrow {{{x^2}} \over 3} + {{{y^2}} \over 2} = 1

...... (i) Its focus is (1, 0) Now, equation of circle is

(x−1)2+y2=43{(x - 1)^2} + {y^2} = {4 \over 3}

..... (ii) Solving (i) and (ii) we get

y=±23,x=1y = \pm {2 \over {\sqrt 3 }},x = 1
⇒PQ2=(43)2=163\Rightarrow P{Q^2} = {\left( {{4 \over {\sqrt 3 }}} \right)^2} = {{16} \over 3}
Q10
If a tangent to the ellipse x2 + 4y2 = 4 meets the tangents at the extremities of it major axis at B and C, then the circle with BC as diameter passes through the point :
A (3,0)(\sqrt 3 ,0)
B (2,0)(\sqrt 2 ,0)
C (1, 1)
D (−-1, 1)
Correct Answer
Option A
Solution
x24+y21=1{{{x^2}} \over 4} + {{{y^2}} \over 1} = 1

Equation of tangent i (cosθ\theta)x + 2sinθ\thetay = 2

B(−2,1+cos⁡θsin⁡θ),C(2,1−cos⁡θsin⁡θ)B\left( { - 2,{{1 + \cos \theta } \over {\sin \theta }}} \right),C\left( {2,{{1 - \cos \theta } \over {\sin \theta }}} \right)
B(−2,cot⁡θ2)B\left( { - 2,\cot {\theta \over 2}} \right)
C(2,tan⁡θ2)C\left( {2,\tan {\theta \over 2}} \right)

Equation of circle is

(x+2)(x−2)+(y−cot⁡θ2)(y−tan⁡θ2)=0(x + 2)(x - 2) + \left( {y - \cot {\theta \over 2}} \right)\left( {y - \tan {\theta \over 2}} \right) = 0
x2−4+y2−(tan⁡θ2+cot⁡θ2)y+1=0{x^2} - 4 + {y^2} - \left( {\tan {\theta \over 2} + \cot {\theta \over 2}} \right)y + 1 = 0

so,

(3,0)\left( {\sqrt 3 ,0} \right)

satisfying option (1)

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