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

Explanation

To solve the problem, we need to find the value of \( a^2 + b^2 \) given that a normal to the hyperbola \( \frac{x^2}{4} - \frac{y^2}{1} = 1 \) has equal intercepts on the positive x- and y-axis and touches the ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).

### Step 1: Identify the hyperbola and its properties
The given hyperbola is:
\[ \frac{x^2}{4} - \frac{y^2}{1} = 1 \]
This is a hyperbola centered at the origin with \( a^2 = 4 \) and \( b^2 = 1 \).
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