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Ellipse question

2021 · 25 Jul · Shift 1 · Q37
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  5. /2021 · 25 Jul · Shift 1 · Q37

Ellipse question

2021 · 25 Jul · Shift 1 · Q37

JEE MainMathematicsEllipseMCQ+4 / −1
Let an ellipse E:x2a2+y2b2=1E:{{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1E:a2x2​+b2y2​=1, a2>b2{a^2} \gt {b^2}a2>b2, passes through (32,1)\left( {\sqrt {{3 \over 2}} ,1} \right)(23​​,1) and has eccentricity 13{1 \over {\sqrt 3 }}3​1​. If a circle, centered at focus F(α\alphaα, 0), α\alphaα> 0, of E and radius 23{2 \over {\sqrt 3 }}3​2​, intersects E at two points P and Q, then PQ2 is equal to :
  1. A
    83{8 \over 3}38​
  2. B
    43{4 \over 3}34​
  3. C
    163{{16} \over 3}316​
  4. D
    3
View written solutionFree

Correct answer: C

  1. Given ellipse and eccentricity

The ellipse is

E:x2a2+y2b2=1,a2>b2E:\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, \qquad a^2>b^2E:a2x2​+b2y2​=1,a2>b2

with eccentricity

e=13.e=\frac{1}{\sqrt{3}}.e=3​1​.

For an ellipse,

e=ca,c2=a2−b2.e=\frac{c}{a}, \qquad c^2=a^2-b^2.e=ac​,c2=a2−b2.

Thus

c=ae=a3.c=ae=\frac{a}{\sqrt{3}}.c=ae=3​a​.

So

c2=a23.c^2=\frac{a^2}{3}.c2=3a2​.

Using c2=a2−b2c^2=a^2-b^2c2=a2−b2,

a2−b2=a23a^2-b^2=\frac{a^2}{3}a2−b2=3a2​

which gives

b2=a2−a23=2a23.b^2=a^2-\frac{a^2}{3}=\frac{2a^2}{3}.b2=a2−3a2​=32a2​.
  1. Use the point on the ellipse

The ellipse passes through

(32,1).\left(\sqrt{\frac32},1\right).(23​​,1).

Substitute into the ellipse equation:

32a2+1b2=1.\frac{\frac32}{a^2}+\frac{1}{b^2}=1.a223​​+b21​=1.

Since b2=2a23b^2=\frac{2a^2}{3}b2=32a2​,

32a2+12a23=1.\frac{3}{2a^2}+\frac{1}{\frac{2a^2}{3}}=1.2a23​+32a2​1​=1.

Now

12a23=32a2,\frac{1}{\frac{2a^2}{3}}=\frac{3}{2a^2},32a2​1​=2a23​,

so

32a2+32a2=1\frac{3}{2a^2}+\frac{3}{2a^2}=12a23​+2a23​=1 3a2=1\frac{3}{a^2}=1a23​=1 a2=3.a^2=3.a2=3.

Hence

b2=2a23=2.b^2=\frac{2a^2}{3}=2.b2=32a2​=2.

So the ellipse is

x23+y22=1.\frac{x^2}{3}+\frac{y^2}{2}=1.3x2​+2y2​=1.
  1. Find the focus

We have

c=ae=3⋅13=1.c=ae=\sqrt{3}\cdot \frac{1}{\sqrt{3}}=1.c=ae=3​⋅3​1​=1.

So the right focus is

F(1,0).F(1,0).F(1,0).

Thus α=1\alpha=1α=1.

  1. Equation of the circle

The circle is centered at F(1,0)F(1,0)F(1,0) with radius 23\frac{2}{\sqrt{3}}3​2​, so its equation is

(x−1)2+y2=43.(x-1)^2+y^2=\frac{4}{3}.(x−1)2+y2=34​.
  1. Intersect the circle with the ellipse

From the ellipse,

x23+y22=1.\frac{x^2}{3}+\frac{y^2}{2}=1.3x2​+2y2​=1.

Multiply by 666:

2x2+3y2=6.(1)2x^2+3y^2=6. \tag{1}2x2+3y2=6.(1)

From the circle,

(x−1)2+y2=43(x-1)^2+y^2=\frac43(x−1)2+y2=34​ x2−2x+1+y2=43x^2-2x+1+y^2=\frac43x2−2x+1+y2=34​ x2+y2−2x+(1−43)=0x^2+y^2-2x+\left(1-\frac43\right)=0x2+y2−2x+(1−34​)=0 x2+y2−2x−13=0.x^2+y^2-2x-\frac13=0.x2+y2−2x−31​=0.

Multiply by 333:

3x2+3y2−6x−1=0.(2)3x^2+3y^2-6x-1=0. \tag{2}3x2+3y2−6x−1=0.(2)

From (1),

3y2=6−2x2.3y^2=6-2x^2.3y2=6−2x2.

Substitute into (2):

3x2+(6−2x2)−6x−1=03x^2+(6-2x^2)-6x-1=03x2+(6−2x2)−6x−1=0 x2−6x+5=0x^2-6x+5=0x2−6x+5=0 (x−1)(x−5)=0.(x-1)(x-5)=0.(x−1)(x−5)=0.

Since the ellipse has ∣x∣≤3|x|\le \sqrt3∣x∣≤3​, x=5x=5x=5 is impossible. Hence

x=1.x=1.x=1.

Then from the ellipse,

13+y22=1\frac{1}{3}+\frac{y^2}{2}=131​+2y2​=1 y22=23\frac{y^2}{2}=\frac{2}{3}2y2​=32​ y2=43.y^2=\frac{4}{3}.y2=34​.

So the intersection points are

P(1,23),Q(1,−23).P\left(1,\frac{2}{\sqrt3}\right), \qquad Q\left(1,-\frac{2}{\sqrt3}\right).P(1,3​2​),Q(1,−3​2​).
  1. Find PQ2PQ^2PQ2

The points have the same xxx-coordinate, so

PQ=23−(−23)=43.PQ=\frac{2}{\sqrt3}-\left(-\frac{2}{\sqrt3}\right)=\frac{4}{\sqrt3}.PQ=3​2​−(−3​2​)=3​4​.

Therefore

PQ2=(43)2=163.PQ^2=\left(\frac{4}{\sqrt3}\right)^2=\frac{16}{3}.PQ2=(3​4​)2=316​.
  1. Compare with options

Thus,

PQ2=163.PQ^2=\frac{16}{3}.PQ2=316​.

So the correct option is C.

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