- Home
- Standard 11
- Mathematics
10-2. Parabola, Ellipse, Hyperbola
hard
Let the tangent and normal at the point $(3 \sqrt{3}, 1)$ on the ellipse $\frac{x^2}{36}+\frac{y^2}{4}=1$ meet the $y$-axis at the points $A$ and $B$ respectively. Let the circle $C$ be drawn taking $A B$ as a diameter and the line $x =2 \sqrt{5}$ intersect $C$ at the points $P$ and $Q$. If the tangents at the points $P$ and $Q$ on the circle intersect at the point $(\alpha, \beta)$, then $\alpha^2-\beta^2$ is equal to
A
$\frac{314}{5}$
B
$\frac{304}{5}$
C
$60$
D
$61$
(JEE MAIN-2023)
Solution
Given ellipse $\frac{x^2}{36}+\frac{y^2}{4}=1$
$\frac{x}{4 \sqrt{3}}+\frac{y}{4}=1$
$y=4$
$\frac{x}{4}-\frac{4}{4 \sqrt{3}}=\frac{2}{\sqrt{3}}$
$y=-8$
$x^2+y^2+4 y-32=0$
$h x+k y+2(y+k)-32=0$
$k=-2$
$h x+2 k-32=0$
$h x=36$
$\alpha=h=\frac{36}{2 \sqrt{5}}$
$\beta=k=-2$
$\alpha^2-\beta^2=\frac{304}{5}$
Standard 11
Mathematics