Length of Sub-Tangent and Sub-Normal of an Ellipse

Ellipse

An ellipse is the locus of a point which moves such that its distance from a fixed point (focus) gives a constant. The standard form of the equation of an ellipse with centre $(0,0)$ and major axis on the x -axis is $\frac{\mathrm{x}^2}{\mathbf{a}^2}+\frac{\mathbf{y}^2}{\mathbf{b}^2}=1 \quad$ where $\mathrm{b}^2=\mathrm{a}^2\left(1-\mathrm{e}^2\right)$
1. $a>b$
2. the length of the major axis is $2 a$
3. the length of the minor axis is $2 b$
4. the coordinates of the vertices are $( \pm a, 0)$

Tangent of an Ellipse

Tangent of an ellipse is a line which touches the ellipse at only one point.

For any point $\left(x_1, y_1\right)$ on the ellipse, the equation of the tangent line at that point is given by:

$
\frac{x_1 x}{a^2}+\frac{y_1 y}{b^2}=1
$

Normal at a point of an Ellipse

Normal at a point of the ellipse is a line perpendicular to the tangent and passing through the point of contact. The equation of normal at $\left(x_1, y_1\right)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is

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

Length of sub-Tangent and sub-Normal of an Ellipse

The sub-Tangent and sub-Normal are particular portions of the tangent line and normal line respectively. The portion of the tangent along the $x$ axis from the point of contact of the tangent on the ellipse to the intersection of the tangent at the $x$-axis is called the sub-Tangent. Similarly, the portion of the normal along the $x$-axis from the point of contact of the tangent on the ellipse to the intersection of the normal at the $x$-axis is called the sub-Normal.

The tangent and normal of the ellipse at $\mathrm{P}\left(\mathrm{x}_1, \mathrm{y}_1\right)$ 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 $\mathrm{Eq}(\mathrm{i})$, 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 $P\left(x_1, y_1\right)$ to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ 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
$\Rightarrow \mathrm{x}=\mathrm{OR}$
$\therefore \mathrm{OR}=\mathrm{x}_1-\frac{\mathrm{b}^2}{\mathrm{a}^2} \mathrm{x}_1$
$\therefore$ Length of Subnormal $=$ RS $=$ OS - OR

$
\begin{aligned}
& =x_1-\left(x_1-\frac{b^2}{a^2} x_1\right) \\
& =\frac{b^2}{a^2} x_1=\left(1-e^2\right) x_1
\end{aligned}
$

Recommended Video Based on Length of sub-Tangent and sub-Normal of Ellipse

Solved Examples based on Length of sub-Tangent and sub-Normal of Ellipse

Example 1: If the normal to the ellipse $3 x^2+4 y^2=12$ at a point P on its parallel to the line, $2 x+y=4$ and the tangent to the ellipse at P passes through $Q(4,4)$ then PQ is equal to :
1)$
\frac{\sqrt{157}}{2}
$

2)$
\frac{\sqrt{221}}{2}
$

3)$
\frac{\sqrt{61}}{2}
$

4) $\frac{5 \sqrt{5}}{2}$

Solution

Equation of Norma to ellipse -
The equation of normal at $\left(\mathrm{x}_1, \mathrm{y}_1\right)$ to the ellipse, $\frac{\mathrm{A}}{\mathrm{a}^2}+\frac{y}{\mathrm{~b}^2}=1$ is $\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2$.

Distance formula -
The distance between the point $A\left(x_1, y_1\right)$ and $B\left(x_2, y_2\right)$
is $\sqrt{\left(x_1-x_2\right)^2+\left(y_1-y_2\right)^2}$
- wherein

$
\begin{aligned}
& 3 x^2+4 y^2=12 \\
& \frac{x^2}{4}+\frac{y^2}{3}=1
\end{aligned}
$

So, $x=2 \cos \theta \quad y=\sqrt{3} \sin \theta$
Let $\quad P(2 \cos \theta, \sqrt{3} \sin \theta)$
Equation of normal is $\frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2-b^2$

$
\begin{aligned}
& \frac{4 x}{2 \cos \theta}-\frac{3 y}{\sqrt{3} \sin \theta}=4-3 \\
& 2 x \sin \theta-\sqrt{3} y \cos \theta=\cos \theta \sin \theta \\
& \text { Slope }=\frac{2}{\sqrt{3}} \tan \theta=-2 \\
& \tan \theta=-\sqrt{3}
\end{aligned}
$

Equation of tangent is it passes through $(4,4)$
$12 \cos \theta+8 \sqrt{3} \sin \theta=6$

$
\begin{aligned}
& \frac{1}{2} \cos \theta+\frac{8 \sqrt{3}}{6} \sin \theta=1 \\
& \cos \theta=-\frac{1}{2}, \quad \sin \theta=\frac{\sqrt{3}}{2} \quad \therefore \theta=120^{\circ}
\end{aligned}
$

Hence point is $(2 \cos 120, \sqrt{3} \sin 120)$

$
\begin{aligned}
& P\left(-1, \frac{3}{2}\right), Q(4,4) \\
& P Q=\sqrt{(-1-4)^2+\left(\frac{3}{2}-4\right)^2}=\sqrt{25+\frac{25}{4}} \\
& =\frac{5 \sqrt{5}}{2}
\end{aligned}
$

Example 2: Let the tangents at the points P and Q on the ellipse $\frac{\mathrm{x}^2}{2}+\frac{\mathrm{y}^2}{4}=1$ meet at the point $\mathrm{R}(\sqrt{2}, 2 \sqrt{2}-2)$. If S is the focus of the ellipse on its negative major axis, then $\mathrm{SP}^2+\mathrm{SQ}^2$ is equal to $\qquad$
MAINS 2022]
Solution
Equation of chard of contact is $\mathrm{T}=0$

$
\begin{aligned}
& \Rightarrow \quad \frac{\sqrt{2} \mathrm{x}}{2}+\frac{(2 \sqrt{2}-2) \mathrm{y}}{4}=1 \\
& \Rightarrow \quad 2 \sqrt{2} \mathrm{x}+(2 \sqrt{2}-2) \mathrm{y}=4 \\
& \Rightarrow \quad \mathrm{x}=\frac{4-(2 \sqrt{2}-2) \mathrm{y}}{2 \sqrt{2}}
\end{aligned}
$
Finding point of intersection with ellipse

$
\begin{aligned}
& \frac{(4-(2 \sqrt{2}-2) \mathrm{y})^2}{8 \times 2}+\frac{y^2}{4}=1 \\
& \Rightarrow \frac{(2-(\sqrt{2}-1) \mathrm{y})^2}{4}+\frac{\mathrm{y}^2}{4}=1 \\
& \Rightarrow \quad 4+(\sqrt{2}-1)^2 \mathrm{y}^2-4(\sqrt{2}-1) \mathrm{y}+\mathrm{y}^2=4 \\
& \Rightarrow \quad(3-2 \sqrt{2}+1) \mathrm{y}^2-4(\sqrt{2}-1) \mathrm{y}=0 \\
& \Rightarrow \quad 2 \sqrt{2}(\sqrt{2}-1) \mathrm{y}^2-4(\sqrt{2}-1) \mathrm{y}=0 \\
& \Rightarrow 2 \sqrt{2}(\sqrt{2}-1) \mathrm{y}(\mathrm{y}-\sqrt{2})=0 \\
& \Rightarrow \mathrm{y}=0, \sqrt{2} \\
& \Rightarrow \mathrm{x}=\sqrt{2}, 1
\end{aligned}
$

$
\begin{aligned}
& \frac{a^2 x}{x_1}-\frac{b^2 y}{y_1}=a^2 e^2 \\
& \frac{a^2 x}{a e}-\frac{b^2 y}{b^2} \cdot a=a^2 e^2 \\
& \frac{a x}{e}-a y=a^2 e^2 \Rightarrow \frac{x}{e}-y=a e^2
\end{aligned}
$

passes through $(0, b)$

$
\begin{aligned}
-b=a e^2 & \Rightarrow b^2=a^2 e^4 \\
a^2\left(1-e^2\right) & =a^2 e^4 \Rightarrow e^4+e^2=1
\end{aligned}
$

$
\begin{aligned}
& \Rightarrow \mathrm{x}=\sqrt{2}, 1 \\
& \therefore \mathrm{P}(\sqrt{2}, 0), \mathrm{Q}(1, \sqrt{2})
\end{aligned}
$
S is $(0,-\mathrm{be})$

$
\begin{aligned}
& \mathrm{e}=\sqrt{1-\frac{2}{4}}=\frac{1}{\sqrt{2}} \\
& \therefore \mathrm{S} \text { is }(0,-\sqrt{2}) \\
& \therefore \mathrm{PS}^2+\mathrm{SQ}^2=4+9=13
\end{aligned}
$

Hence, the answer is 13 .

Example 3: If the normal at the point $\mathrm{P}(\theta)$ to the ellipse $\frac{\mathrm{x}^2}{14}+\frac{\mathrm{y}^2}{5}=1$ intersects it again at the point $\mathrm{Q}(2 \theta)$, then $\cos \theta$ is equal to
1) $2 / 3$
2) $-2 / 3$
3) $3 / 4$
4) None of these

Solution

$
\begin{aligned}
& \frac{\sqrt{14} \mathrm{x}}{\cos \theta}-\frac{\sqrt{5} \mathrm{y}}{\sin \theta}=14-5 ; \text { as it passes through }(\sqrt{14} \cos 2 \theta, \sqrt{5} \sin 2 \theta) \\
& \text { so, } \frac{14\left(2 \cos ^2 \theta-1\right)}{\cos \theta}-\frac{5 \times 2 \sin \theta \cos \theta}{\sin \theta}=9 \\
& \Rightarrow 28 \cos \theta-\frac{14}{\cos \theta}-10 \cos \theta=9 \\
& \Rightarrow 18 \cos ^2 \theta-9 \cos \theta-14=0 \\
& \Rightarrow(3 \cos \theta+2)(6 \cos \theta-7)=0 \Rightarrow \cos \theta=-\frac{2}{3}
\end{aligned}
$
Hence, the answer is the option (2).

Example 4: 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 5: 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