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If tangents are drawn to the ellipse x^2...

If tangents are drawn to the ellipse `x^2+2y^2=2,` then the locus of the midpoint of the intercept made by the tangents between the coordinate axes is `1/(2x^2)+1/(4y^2)=1` (b) `1/(4x^2)+1/(2y^2)=1` `(x^2)/2+y^2=1` (d) `(x^2)/4+(y^2)/2=1`

A

`x^2/4+y^2/2=1`

B

`1/(4x^2)+1/(2y^2)=1`

C

`1/(2x^2)+1/(4y^2)=1`

D

`x^2/2+y^2/4=1`

Text Solution

Verified by Experts

The correct Answer is:
C
`x^2/2+y^2/1=1`
Let the point of contact be (`sqrt2 cos theta , sin theta`)
`therefore` Equation of tangent is `x/sqrt2 cos theta + y sin theta =1`
So, `A-=(sqrt2/(cos theta), 0)` and `B-=(0,1/(sintheta))`
Let the midpoint of AB is (h,k)
`therefore h=(sqrt2/(costheta)+0)/2rArr cos theta =1/sqrt(2h)`..(i)
and `k=(0+1/(sintheta))/2rArr sintheta=1/(2k)` ...(ii)
squaring and adding (i) and (ii) `1/(2x^2)+1/(4y^2)=1`
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