Let an ellipse E: x^2 a^2+ y^2 b^2=1,a^2>b^2, passes through (√ 3 2,1) and has eccentricity 1 √3. If a circle, centered at focus F(α ,0),α >0, of E…
JEE Main 2021 — Mathematics Coordinate Geometry
2021mcqmedium
Let an ellipse E:a2x2+b2y2=1,a2>b2, passes through (23,1) and has eccentricity 31. If a circle, centered at focus F(α,0),α>0, of E and radius 32, intersects E at two points P and Q, then PQ2 is equal to :
Official previous-year question
Held on 25 Jul 2021 · Verified 6 Jul 2026.
Options
A
38
B
34
C
316
D
3
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Solution
The ellipse a2x2+b2y2=1 passes through the point (23,1)
So, 2a23+b21=1...(1) and 1−a2b2=31...(2)
On solving equations (1) and (2), we get
\Rightarrow {a}^{2}=3&{b}^{2}=2
⇒3x2+2y2=1...(3)
We know that, focii of the ellipse a2x2+b2y2=1 is (±ae,0)
Here, a=\sqrt{3}&e=\frac{1}{\sqrt{3}}
∴ Its focus is (1,0)(∵α>0)
Now, equation of circle is
(x−1)2+y2=34...(4)
Solving (3) and (4) we get
y=±32,x=1
⇒PQ2=(1−1)2+(32+32)2
=(34)2=316
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