Let $T_1$ be the tangent drawn at a point $P(\sqrt{2}, \sqrt{3})$ on the 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 $\alpha^2+\beta^2=$
10
52
26
$5 / 12$
Solution
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 $\Rightarrow \beta=m \alpha+\sqrt{4 m^2+6}$
$\begin{aligned}
& \Rightarrow \beta^2+m^2 \alpha^2-2 m \alpha \beta=4 m^2+6 \\
& \Rightarrow m^2\left(\alpha^2-4\right)-2 m \alpha \beta+\beta^2-6=0
\end{aligned}$ Tangents are perpendicular $\Rightarrow m_1 m_2=-1$
$\Rightarrow \frac{\beta^2-6}{\alpha^2-4}=-1 \Rightarrow \alpha^2+\beta^2=10$