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 \(\alpha^2+\beta^2=\)
- A 10
- B 52
- C 26
- D \(5 / 12\)
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 \(\Rightarrow \beta=m \alpha+\sqrt{4 m^2+6}\)…
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