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Question

If the foci of the ellipse x29+y2=1 subtend right angle at a point P. Then, the locus of P is


A

x2+y2=1

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B

x2+y2=2

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C

x2+y2=4

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D

x2+y2=8

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Solution

The correct option is D

x2+y2=8


Explanation for the correct option:

Step 1: Solve for the foci of the ellipse

Given that equation of the ellipse is x29+y2=1

We know that the general equation of ellipse is x2a2+y2b2=1; where aand bare the lengths of semi-major and semi-minor axis respectively,

Here, a=3 and b=1

Eccentricity of the ellipse e=1-ba2

=1-132=83e=223

∴ Two foci are given by ±ae,0=±3×223,0=±22,0

Step 2: Solve for the locus of point P

Let the co-ordinates of point P be h,k

Given that foci subtend right angle at P⇒product of slopes=-1

Let m1,m2 be the slopes of the lines joining the points 22,0 and -22,0 with P respectively

We know that slope of a line joining two points x1,y1 and x2,y2 is m=y2-y1x2-x1

∴m1=k-0h-22=kh-22m2=k-0h+22=kh+22

Now, according to the given condition m1×m2=-1

⇒kh-22×kh+22=-1⇒k2h2-8=-1⇒k2=-h2+8⇒h2+k2=8

To obtain the locus, replace h,kwith x,y⇒x2+y2=8

Hence, the correct answer is option (D) i.e. x2+y2=8.


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