Length of sub-Tangent and Sub-Normal of an Ellipse - Practice Questions & MCQ

The tangent and normal of the ellipse at P(x₁, y₁) meet the X-axis at Q and R respectively.

Then, the equation of the tangent at $P\left(x_1, y_1\right)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

$
\frac{\mathrm{xx}_1}{\mathrm{a}^2}+\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=1
$

$\because Q$ lies on X-axis, then put $\mathrm{y}=0$ in Eq (i), we get

$
\Rightarrow \mathrm{x}=\mathrm{OQ}
$

$\Rightarrow \mathrm{OQ}=\frac{\mathrm{a}^2}{\mathrm{x}_1}$ and $\mathrm{OS}=\mathrm{x}_1$

$
\text { length of subtangent }=S Q=O Q-O S=\frac{a^2}{x_1}-x_1
$
Equation of normal at $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ to the ellipse $\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ is

$
\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2
$

$\because R$ lies on X-axis, then put $\mathrm{y}=0$ in Eq (ii), we get

$
\Rightarrow \mathrm{x}=\mathrm{OR}
$

$\therefore \mathrm{OR}=\mathrm{x}_1-\frac{\mathrm{b}^2}{\mathrm{a}^2} \mathrm{x}_1$

\begin{aligned}
& \therefore \text { Length of Subnormal }=\mathrm{RS}=\mathrm{OS}-\mathrm{OR} \\
& =\mathrm{x}_1-\left(\mathrm{x}_1-\frac{\mathrm{b}^2}{\mathrm{a}^2} \mathrm{x}_1\right) \\
& =\frac{\mathrm{b}^2}{\mathrm{a}^2} \mathrm{x}_1=\left(1-\mathrm{e}^2\right) \mathrm{x}_1
\end{aligned}