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