Ellipse 2 Question 3
3. If the line $x-2 y=12$ is tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at the point $3, \frac{-9}{2}$, then the length of the latusrectum of the ellipse is
(2019 Main, 10 April I)
(a) $8 \sqrt{3}$
(b) 9
(c) 5
(d) $12 \sqrt{2}$
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Answer:
Correct Answer: 3. (b)
Solution:
Key Idea Write equation of the tangent to the ellipse at any point and use formula for latusrectum of ellipse.
Equation of given ellipse is
$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 $
Now, equation of tangent at the point $3,-\frac{9}{2}$ on the ellipse (i) is
$\Rightarrow \quad \frac{3 x}{a^{2}}-\frac{9 y}{2 b^{2}}=1$
$\left[\because\right.$ the equation of the tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ at the point $\left(x _1, y _1\right)$ is $\left.\frac{x x _1}{a^{2}}+\frac{y y _1}{b^{2}}=1\right]$
$\because$ Tangent (ii) represent the line $x-2 y=12$, so
$ \begin{aligned} & \frac{1}{\frac{3}{a^{2}}}=\frac{2}{\frac{9}{2 b^{2}}}=\frac{12}{1} \\ & \Rightarrow \quad a^{2}=36 \text { and } b^{2}=27 \end{aligned} $
Now, Length of latusrectum $=\frac{2 b^{2}}{a}=\frac{2 \times 27}{6}=9$ units
