Question

Tangents are drawn to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at points where it is intersected by the line $\ell x+my+n=0$. Find the point of intersection of tangents at these points.

• A. $\frac{-a^2}{n},\frac{-b^{2}m}{n}$
• B. $\frac{-a^2\ell }{n},\frac{-b^{2}m}{n}$
• C. $\frac{-a^2\ell }{n},\frac{-b^2}{n}$
• D. None of these

Answer 1

The correct option is B $\frac{-a^2\ell }{n},\frac{-b^{2}m}{n}$

Let $P(x_1,y_1)$ be the point of intersection of the
Line $lx+my+n=0$ and the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Then the equation of tangent at $P$ is
$\frac{{xx}_1}{a^2}+\frac{{yy}_1}{b^2}=1\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(i\right)$
Since $(x_1,y_1)$ is the point of intersection of the line
$lx+my+n=0\cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \cdot \left(ii\right)$
Clearly $\left(i\right)$ and $\left(ii\right)$ represent the same line. Therefore,
$\therefore \frac{x_1}{a^{2}l}=\frac{y_1}{b^{2}m}=\frac{1}{-n}$
$x_1=\frac{{-a}^{2}l}{n},y_1=\frac{-b^{2}m}{n}$
Therefore, the point of intersection of given line and the given ellipse is
$(-\frac{a^{2}l}{n},\frac{b^{2}m}{n})$
Hence, option 'B' is correct.