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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

Any tangent to ellipse
`(x^(2))/(4)+(y^(2))/(1)=1` is given by
`(x cos theta)/(sqrt(2))+y sin theta=1`

Let it meet axes at A and B
`:. A-=(sqrt(2)sec theta, 0) and B-=(0, "coses" theta)`
Let p(h,k) be the mid point of AB.
Hence, `2h=sqrt(2)sec theta` and `2k= "cosec"` which is locus of P
`((1)/(sqrt(2)h))^(2)+((1)/(2k))^(2)=1`

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