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AP EAMCET · Maths · Ellipse

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

  1. A \(\frac{1+\sqrt{17}}{4}\)
  2. B \(\frac{-1+\sqrt{17}}{4}\)
  3. C \(\frac{-1+\sqrt{37}}{4}\)
  4. D \(\frac{1+\sqrt{37}}{4}\)
Verified Solution

Answer & Solution

Correct Answer

(A) \(\frac{1+\sqrt{17}}{4}\)

Step-by-step Solution

Detailed explanation

For normal of ellipse: \(\frac{x^2}{4}+\frac{y^2}{3}=1\) ...(i) \(y^{\prime}=-\frac{3}{4} \frac{x}{y}\) Slope of normal is \(m=-\frac{1}{y^{\prime}}=\frac{4}{3} \cdot \frac{y}{x}\) ...(ii) For tangent of hyperbola: \(\frac{x^2}{4}-\frac{y^2}{3}=1\) ...(iii)…