Question

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 5 (b) 25 (c) 16 (d) none of these

Answer 1

Equation of Normal to a hyperbola is:- `axcos(theta)+bycot(theta)=a^2+ b^2` By given hyperbola,`a=2, b=1`
So,`2xcos(theta)+ycot(theta)=5` As the normal makes equal intercepts on positve x & y-axis, So, `slope=-1=m`
Therefore,` (-2cos(theta)/(cot(theta)))=-1`
`Sin(theta)=1/2`
`theta=pi/6`
For y-intercept of the normal(x=0):-
y-intercept=`5/cot(theta)=5/sqrt3=c`
As it touches the ellipse `(x/a)^2+ (y/b)^2=1`
So, `c^2=a^2m^2+ b^2` Hence, `a^2(-1)^2 + b^2= (5/sqrt3)^2` `a^2+b^2=25/3`