Question:
If the tangents on the ellipse $4 x^{2}+y^{2}=8$ at the points $(1,2)$ and $(a, b)$ are perpendicular to each other, then $a^{2}$ is equal
Correct Option: , 4
Solution:
Since $(a, b)$ touches the given ellipse $4 x^{2}+y^{2}=8$
$\therefore 4 a^{2}+b^{2}=8$ $\ldots(1)$
Equation of tangent on the ellipse at the point $A(1,2)$ is:
$4 x+2 y=8 \Rightarrow 2 x+y=4 \Rightarrow y=-2 x+4$
But, also equation of tangent at $P(a, b)$ is:
$4 a x+b y=8 \Rightarrow y=\frac{-4 a}{b}+\frac{8}{b}$
Since, tangents are perpendicular to each other.
$\Rightarrow \frac{-4 a}{b}=\frac{-1}{2}$
$\Rightarrow b=8 a$ $\ldots(2)$
from (1) \& (2) we get:
$\Rightarrow a=\pm \frac{2}{\sqrt{34}} \Rightarrow a^{2}=\frac{2}{17}$