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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 (a)`1/(2x^2)+1/(4y^2)=1` (b) `1/(4x^2)+1/(2y^2)=1` (c)`(x^2)/2+y^2=1` (d) `(x^2)/4+(y^2)/2=1`

A

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

B

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

C

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

D

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

Text Solution

Verified by Experts

The correct Answer is:
A
Let the point P `( sqrt(2) cos theta , sin theta ) ` on `(x^(2))/(2)+(y^(2))/(1)=1.`

Equation of tangent is , `( x sqrt(2))/(2) cos theta + y sin =1`
whose intercept on coordinate axes are
`A ( sqrt(2) sec theta ,0) ` and B ( ` 0, cosec theta)`
`therefore ` Mid - point of its intercept between axes
`((sqrt(2))/(2)sex theta , (1)/(2) cos ec theta )=(h,k)`
`implies cos theta =(1)/(sqrt(2h))and sin theta =(1)/(2k)`
thus , focus of mid - point M is
`( cos ^(2) theta sin ^(2) theta )=(1)/(2h^(2))+(1)/(4k^(2))`
`implies (1)/(2x^(2))+(1)/(4y^(2))=1` is required locus .
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