Equation of Tangent to Ellipse: Formulas and Examples

An ellipse is the set of all points $(x, y)$ in a plane such that the sum of their distances from two fixed points is a constant. Each fixed point is called a focus (plural: foci). The tangent of an ellipse is a line which touches the ellipse at only one point. The normal is a line perpendicular to the tangent and passing through the point of contact to the ellipse.

This article is about the equation of tangent of Ellipse in point form and parametric form which falls under the broader category of two-dimensional analytical Geometry. This concept has applications in various fields like calculus, physics etc. In real life, this concept is used in the construction and navigation field to calculate distances, heights, and angles. This is one of the important concepts for competitive exams. In JEE MAINS(2013 to 2023) from this topic, there were 14 questions including one in 2015, one in 2019, two in 2020, four in 2021, two in 2022, and four in 2023.

Equation of Tangent of Ellipse in Point Form

Tangent of an ellipse is a line which touches the ellipse at only one point. 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{\mathrm{xx}_1}{\mathrm{a}^2}+\frac{\mathrm{yy}_1}{\mathrm{~b}^2}=1
$

Derivation of Equation of Tangent of Ellipse in Point Form

Slope of tangent at $\left(x_1, y_1\right)$ is $-\frac{b^2 x_1}{a^2 y_1}$

The equation of the line with slope $m$ passing through $(x_1, y_1)$ is

$ y-y_1 = m(x-x_1)$

Substituting the slope of the tangent,

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

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

$yy_1 -y_1^2 = \frac{b^2 }{a^2}(xx_1 -x_1^2)$

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

Equation of Tangent of Ellipse in Parametric Form

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

Derivation of the equation of tangent of the ellipse on Parametric form

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

Substitute $(a \cos \theta, b \sin \theta)$ in the equation of tangent in point form,

$\frac{\mathrm{x}(a \cos \theta)}{\mathrm{a}^2}+\frac{\mathrm{y}(b \sin \theta)}{\mathrm{~b}^2}=1$

Equation of Tangent of Ellipse in Slope Form

Ellipse: $\quad \frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$
Line: $\quad y=m x+c$
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 m^2+b^2\right) x^2+2 m \mathrm{ma}^2 x+c^2 a^2-a^2 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 \quad & c= \pm \sqrt{a^2 m^2+b^2}
\end{array}
$

put the value of c in $\mathrm{y}=\mathrm{mx}+\mathrm{c}$
we get

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

Recommended Video Based on the Equation of Tangent to Ellipse

Solved Examples based on the equation of Tangent to Ellipse

Example 1: If the line $x-2 y=12$ is the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at the point $\left(3, \frac{-9}{2}\right)$, then the length of the latus rectum of the ellipse is :
1)$9$
2)$12 \sqrt{2}$
3)$5$
4)$8 \sqrt{3}$

Solution:
Using the point form of a tangent, the equation of the tangent at $\left(3, \frac{-9}{2}\right)$ is

$
\frac{3 x}{a^2}-\frac{9 y}{2 b^2}=1
$
Now compare this equation with the given equation of tangent $x-2 y=12$

$
\frac{3}{a^2}=\frac{9}{4 b^2}=\frac{1}{12}
$

$a=6$ and $b=3 \sqrt{3}$
Length of $\mathrm{LR}=\frac{2 b^2}{a}=\frac{2 \times(3 \sqrt{3})^2}{6}=9$
Hence, the answer is the option (1).

Example 2: Let C be the largest circle centered at $(2,0)$ and inscribed in the ellipse $\frac{x^2}{36}+\frac{y^2}{16}=1$ If $(1, \alpha)$ lies on C , then $10 \alpha^2$ is equal to
[JEE MAINS 2023]
Solution
$
E: \frac{x^2}{36}+\frac{y^2}{16}=1 \& C:(x-2)^2+y^2=r^2
$

For the largest circle $r$ is the maximum

$
\begin{aligned}
& \mathrm{P}(6 \cos \theta, 4 \sin \theta) \\
& N_{\mathrm{P}}: 6 \times \sec \theta-4 \mathrm{y} \operatorname{cosec} \theta=20 \text { pass }(2,0) \\
& 12 \sec \theta=20 \Rightarrow \cos \theta=\frac{3}{5} \\
& \text { NowP }:\left(6 \times \frac{3}{5}, 4 \times \frac{4}{5}\right) \Rightarrow P:\left(\frac{18}{5}, \frac{16}{5}\right) \\
& r=\sqrt{\left(2-\frac{18}{5}\right)^2+\left(\frac{16}{5}\right)^2} \\
& r=\frac{\sqrt{64+256}}{5}=\frac{8 \sqrt{5}}{5}=\frac{8}{\sqrt{5}} \\
& \mathrm{C}:=(\mathrm{x}-2)^2+y^2=\frac{64}{5}
\end{aligned}
$

$\operatorname{Now}(1, \alpha)$ lies on C

$
\begin{aligned}
& \Rightarrow(1-2)^2+\alpha^2=\frac{64}{5} \\
& \alpha^2=\frac{64}{5}-1 \\
& \alpha^2=\frac{59}{5} \Rightarrow 10 \alpha^2=118
\end{aligned}
$

Hence, the answer is 118 $(3 P Q)^2$.

Example 3: Let the tangent and normal at the point $(3 \sqrt{3}, 1)$ on the ellipse $\frac{x^2}{36}+\frac{y^2}{4}=1$ meet the $y$-axis at the points $A$ and $B$ respectively. Let the circle $C$ be drawn taking $A B$ as a diameter and the line $\mathrm{x}=2 \sqrt{5}$ intersect C at the points P and Q . If the tangents at the points P and Q on the circle intersect at the point $(\alpha, \beta)$, then $\left(\alpha^2-\beta^2\right)$ is equal to [JEE MAINS 2023]
Solution
Given ellipse $\frac{x^2}{36}+\frac{y^2}{4}=1$
$\frac{\mathrm{x}}{4 \sqrt{3}}+\frac{\mathrm{y}}{4}=1$
$\mathrm{y}=4$
$\frac{x}{4}-\frac{4}{4 \sqrt{3}}=\frac{2}{\sqrt{3}}$
$y=-8$
$
\begin{aligned}
& \mathrm{h} x+\mathrm{ky}+2(\mathrm{y}+\mathrm{k})-32=0 \\
& \mathrm{k}=-2 \\
& \mathrm{hx}+2 \mathrm{k}-32=0 \\
& \mathrm{hx}=36 \\
& \alpha=\mathrm{h}=\frac{36}{2 \sqrt{5}} \\
& \beta=\mathrm{k}=-2 \\
& \alpha^2-\beta^2=\frac{304}{5}
\end{aligned}
$
Hence, the answer is 304 / 5

Example 4: If m is the slope of a common tangent to the curves $\frac{\mathrm{x}^2}{16}+\frac{\mathrm{y}^2}{9}=1$ and $\mathrm{x}^2+\mathrm{y}^2=12$, then $12 \mathrm{~m}^2$ is equal to :
[JEE MAINS 2022]
Solution

$
\begin{aligned}
& \frac{x^2}{16}+\frac{y^2}{9}=1 \\
& \therefore \quad 16 m^2+9=\mathrm{m} x \pm \sqrt{12 \mathrm{~m}^2+12} \\
& \mathrm{~m}^2=\frac{3}{4} \\
& \therefore 12 \mathrm{~m}^2=9
\end{aligned}
$
$
\begin{array}{ll}
\because & x^2+y^2=12 \\
\therefore & y=m x \pm \sqrt{2 m^2+12}
\end{array}
$

Hence, the answer is 9