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`

Options

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`

Answer

A

Explanation

we have ,
`x^(2)+2y^(2)=2or ,(x^(2))/((sqrt(2))^(2))+(y^(2))/(1^(2))=1`
the equation of any tangent to this ellipse in parametric form at point `(sqrt(2) cos theta , sin theta )` is
`(x)/(sqrt(2))cos theta + ysin theta =1`
this cuts the coordinate axes at point `A((sqrt(2))/(cos theta ),0) and b(0,(1)/(sin theta ))` .LEt P(h,K) be the mid -point of intercept AB. then ,
`h=(sqrt(2) cos theta)/(2) and K =(1)/(2 sin theta)`
`implies cos theta =(1)/(sqrt(2h))and sin theta =(1)/(2k)`
`therefore cos^(2) + sin^(2) theta theta =1 implies (1)/(2h^(2))+(1)/(4k^(2))=1`
hence the locus of (h,k) is ` (1)/(2x^(2))+(1)/(4y^(2))=1`