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AP EAMCET · Maths · Ellipse

Let \(T_1\) be the tangent drawn at a point \(P(\sqrt{2}, \sqrt{3})\) on the ellipse \(\frac{x^2}{4}+\frac{y^2}{6}=1\). If \((\alpha, \beta)\) is the point where \(T_1\) intersects another tangent \(T_2\) to the ellipse perpendicularly, then \(\alpha^2+\beta^2=\)

  1. A 10
  2. B 52
  3. C 26
  4. D \(5 / 12\)
Verified Solution

Answer & Solution

Correct Answer

(A) 10

Step-by-step Solution

Detailed explanation

Given ellipse \(\frac{x^2}{4}+\frac{y^2}{6}=1 \Rightarrow a^2=4, b^2=6\) Equation of tangent in slope form is \(y=m x+\sqrt{a^2 m^2+b^2}\) \((\alpha, \beta)\) lies on it \(\Rightarrow \beta=m \alpha+\sqrt{4 m^2+6}\)…