Question

Let P $(x_1,y_1)$ & Q $(x_2,y_2)$ are points lying on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ . If the tangents drawn at the points P & Q are perpendicular, then the value of $\frac{x_1{x}_2}{y_1{y}_2}$

• A. $-\frac{a^2}{b^2}$
• B. $\frac{a^2}{b^2}$
• C. $-\frac{a^4}{b^4}$
• D. $\frac{a^4}{b^4}$

Answer 1

The correct option is C $-\frac{a^4}{b^4}$

Slope of tangent at any point on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
is, $m=-\frac{b^{2}x}{a^{2}y}$
Thus slope of tangent at $P(x_1,y_1)$ is $m_1=-\frac{b^2{x}_1}{a^2{y}_1}$
and slope of tangent at $Q(x_2,y_2)$ is $m_2=-\frac{b^2{x}_2}{a^2{y}_2}$
Now, $\because$ both tangents are perpendicular, $m_1.m_2=-1$
$\Rightarrow \frac{b^2{x}_1}{a^2{y}_1}.\frac{b^2{x}_2}{a^2{y}_2}=-1$
$\Rightarrow \frac{x_1{x}_2}{y_1{y}_2}=-\frac{a^4}{b^4}$