# If the tangents on the ellipse

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

1. (1) $\frac{128}{17}$

2. (2) $\frac{64}{17}$

3. (3) $\frac{4}{17}$

4. (4) $\frac{2}{17}$

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:

$\Rightarrow a=\pm \frac{2}{\sqrt{34}} \Rightarrow a^{2}=\frac{2}{17}$