Question

If a normal drawn to the ellipse \[ \frac{x^2}{4} + \frac{y^2}{3} = 1 \] touches the hyperbola \[ \frac{x^2}{4} - \frac{y^2}{3} = 1, \] then the square of the slope of that normal is:

• A. \( \frac{1 + \sqrt{17}}{4} \)
• B. \( \frac{-1 + \sqrt{17}}{4} \)
• C. \( \frac{-1 + \sqrt{37}}{4} \)
• D. \( \frac{1 + \sqrt{37}}{4} \)

Hint:

Use slope form of normals and discriminant-zero condition for tangents to conics.

Answer 1

The Correct Option is B

Let slope of normal be \( m \), use parametric form of ellipse and derive normal equation. If this normal is tangent to hyperbola, it satisfies discriminant = 0 condition. After solving the resulting quadratic, the slope square reduces to: \[ m^2 = \frac{-1 + \sqrt{17}}{4} \]