Ellipse 2 Question 5
5. 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 to
(a) $\frac{128}{17}$
(b) $\frac{64}{17}$
(c) $\frac{4}{17}$
(d) $\frac{2}{17}$
(2019 Main, 8 April I)
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Answer:
Correct Answer: 5. (d)
Solution:
- Equation of given ellipse is
$\Rightarrow \quad \begin{aligned} & 4 x^{2}+y^{2}=8 \ & \frac{x^{2}}{2}+\frac{y^{2}}{8}=1 \Rightarrow \frac{x^{2}}{(\sqrt{2})^{2}}+\frac{y^{2}}{(2 \sqrt{2})^{2}}=1\end{aligned}$
Now, equation of tangent at point $(1,2)$ is
$ 2 x+y=4 $
$\left[\because\right.$ equation of tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at $\left(x _1, y _1\right)$
is $\left.\frac{x x _1}{a^{2}}+\frac{y y _1}{b^{2}}=1\right]$
and equation of another tangent at point $(a, b)$ is
$ 4 a x+b y=8 $
Since, lines (ii) and (iii) are perpendicular to each other.
$\therefore \quad-\frac{2}{1} \times-\frac{4 a}{b}=-1$
[if lines $a _1 x+b _1 y+c _1=0$ and $a _2 x+b _2 y+c _2=0$
are perpendicular, then $-\frac{a _1}{b _1} \quad-\frac{a _2}{b _2}=-1$ ]
$\Rightarrow \quad b=-8 a$
Also, the point $(a, b)$ lies on the ellipse (i), so
$ 4 a^{2}+b^{2}=8 $
$ \begin{aligned} & \Rightarrow & 4 a^{2}+64 a^{2} & =8 \\ & \Rightarrow & 68 a^{2} & =8 \Rightarrow a^{2}=\frac{8}{68} \\ & \Rightarrow & a^{2} & =\frac{2}{17} \end{aligned} $
[from Eq.(iv)]
