Equation of Tangent to Ellipse - Practice Questions & MCQ

Equation of Tangent of Ellipse in Point Form:

The equation of tangent to the ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at point $\left(x_1, y_1\right)$ is $\frac{x^{\prime}}{a_1}+\frac{y_{y_1}}{\mathbf{b}^2}=1$

Differentiating $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ w.r.t. $x$, we have

$
\begin{array}{ll} 
& \frac{2 x}{a^2}+\frac{2 y}{b^2} \frac{d y}{d x}=0 \\
\Rightarrow & \frac{d y}{d x}=-\frac{b^2 x}{a^2 y} \\
\Rightarrow & \left(\frac{d y}{d x}\right)_{(x, y)}=-\frac{b^2 x_1}{a^2 y_1}
\end{array}
$
Hence, equation of the tangent is $y-y_1=-\frac{b^2 x_1}{a^2 y_1}\left(x-x_1\right)$ or

$
\frac{\mathrm{xx}_1}{\mathrm{a}^2}+\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=\frac{\mathrm{x}_1^2}{\mathrm{a}^2}+\frac{\mathrm{y}_1^2}{\mathrm{~b}^2}
$
But $\left(x_1, y_1\right)$ lies on the ellipse $\Rightarrow \frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}=1$

Hence, equation of the tangent is

$
\begin{aligned}
& \frac{x x_1}{a^2}+\frac{y y_1}{b^2}=1 \\
& \text { or } \quad \frac{x x_1}{a^2}+\frac{y y_1}{b^2}-1=0 \text { or } T=0 \\
& \text { where } \quad T=\frac{x x_1}{a^2}+\frac{y y_1}{b^2}-1
\end{aligned}
$

Note:

The equation T = 0 can also be used to find the equation of tangent at point (x₁, y₁) lying on any general ellipse.

Equation of Tangent of Ellipse in Parametric Form and Slope Form

Parametric Form:

The equation of tangent to the ellipse, $\frac{\mathrm{x}^{-}}{\mathrm{a}^2}+\frac{\mathrm{y}^{-2}}{\mathrm{~b}^2}=1$ at $(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)$ is

$
\frac{\mathrm{x}}{\mathrm{a}} \cos \theta+\frac{\mathrm{y}}{\mathrm{~b}} \sin \theta=1
$

(This can be easily derived using the point form of tangent to an ellipse)

Slope Form:

Ellipse: $\quad \frac{\lambda}{\mathrm{a}^2}+\frac{y}{\mathrm{~b}^2}=1$
Line: $\quad \mathrm{y}=\mathrm{mx}+\mathrm{c}$
After solving Eq. (i) and Eq. (ii)

$
\begin{aligned}
& \frac{x^2}{a^2}+\frac{(m x+c)^2}{b^2}=1 \\
\Rightarrow \quad & \left(a^2 \mathrm{~m}^2+\mathrm{b}^2\right) x^2+2 \mathrm{mca}^2 \mathrm{x}+\mathrm{c}^2 \mathrm{a}^2-\mathrm{a}^2 \mathrm{~b}^2=0
\end{aligned}
$
For tangent, $\mathrm{D}=0$

$
\begin{array}{ll} 
& 4 a^4 m^2 c^2-4\left(a^2 m^2+b^2\right) a^2\left(c^2-b^2\right)=0 \\
\Rightarrow & c^2=a^2 m^2+b^2 \\
\therefore & c= \pm \sqrt{a^2 m^2+b^2}
\end{array}
$

put the value of $c$ in $y=m x+c$
we get

$
\mathrm{y}=\mathrm{mx} \pm \sqrt{\left(\mathrm{a}^2 \mathrm{~m}^2+\mathrm{b}^2\right)}
$

The equation of tangent of slope m to the ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ are $=m x \pm \sqrt{a^2 m^2+b^2}$