Question

The chord joining two points ${\theta }_1\text{ and }{\theta }_2$on the ellipse ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ such that $\tan {\theta }_1\tan {\theta }_2=-{\displaystyle \frac{a^2}{b^2}}$ will subtend a right angle at
$$\begin{gathered} (a)\text{ focus} \\ (b)\text{ center} \\ (c)\text{ end of the major axis} \\ (d)\text{ end of the minor axis} \end{gathered}$$

Answer 1

Hint: In this question suppose two points ${\theta }_1$ and ${\theta }_2$ such that ${\theta }_1=\left(a\cos {\theta }_1,b\sin {\theta }_1\right)$ and ${\theta }_2=\left(a\cos {\theta }_2,b\sin {\theta }_2\right)$through which the chord passes. Then use the concept of slope of line passing through two given points to find the slope of $O{\theta }_1\text{ and O}{\theta }_2$ where O is the origin. Use the concept that if two lines are perpendicular then their slopes are related as $m_1\times m_2=-1$.

Complete step-by-step answer:

The chord joining two points $\left({\theta }_1,{\theta }_2\right)$ on the ellipse ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ is shown above.
As we know that the ellipse is having a center (O) = (0, 0) is also shown in the figure.
Let us suppose the point ${\theta }_1=\left(a\cos {\theta }_1,b\sin {\theta }_1\right)$ and ${\theta }_2=\left(a\cos {\theta }_2,b\sin {\theta }_2\right)$ is also shown in the figure.
Now as we know that the slope between two points $(x_1,y_1)$ and $(x_2,y_2)$ is given as
Slope (m) = ${\displaystyle \frac{y_2-y_1}{x_2-x_1}}$
So find out the slopes of $\left(O{\theta }_1\right)$ and $\left(O{\theta }_2\right)$.
Let O = $(x_1,y_1)$ = (0, 0)
${\theta }_1=\left(x_2,y_2\right)=\left(a\cos {\theta }_1,b\sin {\theta }_1\right)$
${\theta }_2=\left(x_3,y_3\right)=\left(a\cos {\theta }_2,b\sin {\theta }_2\right)$
So let the slope of $\left(0{\theta }_1\right)$ be m1.
$\Rightarrow m_1={\displaystyle \frac{y_2-y_1}{x_2-x_1}}={\displaystyle \frac{b\sin {\theta }_1-0}{a\cos {\theta }_1-0}}={\displaystyle \frac{b}{a}}\tan {\theta }_1$
Now let the slope of $\left(O{\theta }_2\right)$ be m2.
$\Rightarrow m_1={\displaystyle \frac{y_3-y_1}{x_3-x_1}}={\displaystyle \frac{b\sin {\theta }_2-0}{a\cos {\theta }_2-0}}={\displaystyle \frac{b}{a}}\tan {\theta }_2$
Now multiply the slopes we have
$\Rightarrow m_1\times m_2={\displaystyle \frac{b}{a}}\tan {\theta }_1\times {\displaystyle \frac{b}{a}}\tan {\theta }_2={\displaystyle \frac{b^2}{a^2}}\tan {\theta }_1\tan {\theta }_2$........................ (1)
Now it is given that
$\tan {\theta }_1\tan {\theta }_2=-{\displaystyle \frac{a^2}{b^2}}$
Now substitute this value in equation (1) we have,
$\Rightarrow m_1\times m_2={\displaystyle \frac{b^2}{a^2}}\tan {\theta }_1\tan {\theta }_2={\displaystyle \frac{b^2}{a^2}}\times {\displaystyle \frac{-a^2}{b^2}}=-1$
So multiplication of slopes is (-1) which is the condition of the right angle.
Therefore the chord joining two points $\left({\theta }_1,{\theta }_2\right)$ on the ellipse ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ will subtend a right angle at origin or center.
Hence option (B) is correct.

Note: The center of the given ellipse that is ${\displaystyle \frac{x^2}{a^2}}+{\displaystyle \frac{y^2}{b^2}}=1$ is (0, 0) that is the origin that’s why option (c) is correct. The equation of shifted ellipse or the ellipse whose center is not at origin is given by ${\displaystyle \frac{{\left(x-p\right)}^2}{a^2}}+{\displaystyle \frac{{\left(y-q\right)}^2}{b^2}}=1$ here the center is at (p, q).