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)/(x^(2))+(1)/(2y^(2)) =1`
B. `(1)/(4x^(2))+(1)/(2y^(2))=1`
C. `(1)/(2x^(2))+(1)/(4y^(2))=1`
D. `(1)/(2x^(2)) +(1)/(y^(2))=1`

Answer

C

Explanation

Ellipse`(x^(2))/(2)+(y^(2))/(1)=1`
Let the equation of tangent
`(x)/(sqrt(2)) " cos "0 +(y)/(1) " sin "0 =1`
If (h, k) be the mid point of the portion of the tangent intercepted between the axes then

`2h =sqrt(2)/("cos"0), 2K =(1)/("sin"0)`
`rArr " cos "0 =(1)/sqrt(2h) and`
squring and adding , we get `rArr (1)/(2h^(2)) +(1)/(4K^(2)) =1`
`:. " locus " (1)/(2x^(2)) +(1)/(4y^(2)) =1`