A normal to the hyperbola `(x^2)/4-(y^2)/1=1` has equal intercepts on the positive x- and y-axis. If this normal touches the ellipse `(x^2)/(a^2)+(y^2)/(b^2)=1` , then `a^2+b^2` is equal to (a)5 (b) 25 (c) 16 (d) none of these

Options

A. 5
B. 25
C. 16
D. none of these

Answer

D

Explanation

The equation of the normal to the hyperbola
`(x^(2))/(4)-(y^(2))/(1)=1`
at `(2 sec theta, tan theta)` is `2x cos theta+y cot theta=5`.
The slope of the normal is
`-2 sin theta=-1`
`"or "sin theta=(1)/(2)or theta=(pi)/(6)`
Y-intercept of the normal `=(5)/(cot theta)=(5)/(sqrt3)`
As it touches the ellipse
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
As it touches the ellipse
`(x^(2))/(a^(2))+(y^(2))/(b^(2))=1`
we have `a^(2)+b^(2)=(25)/(3)`