Ellipse 2 Question 19
19. Prove that, in an ellipse, the perpendicular from a focus upon any tangent and the line joining the centre of the ellipse of the point of contact meet on the corresponding directrix.
(2002, 5M)
Show Answer
Answer:
Correct Answer: 19. (d)
Solution:
- Any point on the ellipse
$$ \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1 \text { be } P(a \cos \theta, b \sin \theta) $$
The equation of tangent at point $P$ is given by
$$ \frac{x \cos \theta}{a}+\frac{y \sin \theta}{b}=1 $$
The equation of line perpendicular to tangent is
$$ \frac{x \sin \theta}{b}-\frac{y \cos \theta}{a}=\lambda $$
Since, it passes through the focus ( $a e, 0)$, then
$$ \begin{array}{cc} & \frac{a e \sin \theta}{b}-0=\lambda \\ \Rightarrow \quad \lambda=\frac{a e \sin \theta}{b} \\ \therefore \quad & \text { Equation is } \frac{x \sin \theta}{b}-\frac{y \cos \theta}{a}=\frac{a e \sin \theta}{b} \end{array} $$
Equation of line joining centre and point of contact $P(a \cos \theta, b \sin \theta)$ is
$$ y=\frac{b}{a}(\tan \theta) x $$
Point of intersection $Q$ of Eqs. (i) and (ii) has $x$ coordinate, $\frac{a}{e}$. Hence, $Q$ lies on the corresponding directrix $x=\frac{a}{e}$.
