A normal to the hyperbola $\frac{x^2}{4} - \frac{y^2}{1} = 1$ has equal intercepts on positive $x$ and $y$-axes. If this normal touches the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, then find $[a^2 + b^2]$. ($[|$ represents greatest integer function$)$
Step-by-Step Solution
Key Concept: Use the normal form of hyperbola $ rac{x^2}{a^2} - rac{y^2}{b^2} = 1$ as $y = mx + rac{a^2m^2 + b^2}{m}$ and apply the condition of equal intercepts (slope = -1) to find the specific normal, then use the tangency condition $c^2 = a^2m^2 + b^2$ for the ellipse to determine $a^2 + b^2$.
The normal to the hyperbola is $2\cos\theta + \cot\theta = 5$. With equal intercepts, the normal has slope $-1$, giving $\sin\theta = \frac{1}{2}$. The normal equation becomes $x + y = \frac{5}{\sqrt{3}}$. For this to touch the ellipse $c^2 = a^2m^2 + b^2$, we get $a^2 + b^2 = \frac{25}{3}$.
Correct Answer: 8