Intersection of Ellipse and Line

A line may meet the ellipse in one point or two distinct points or it may not meet the ellipse at all. If the line meets the ellipse at one point is called Tangent and If the line meets the ellipse at two points it is called a chord. In real life, we use tangents in the construction and navigation field to calculate distances, heights, and angles.

In this article, we will cover the concept of Line and the Ellipse. This category falls under the broader category of Coordinate Geometry, which is a crucial Chapter in class 11 Mathematics. It is not only essential for board exams but also for competitive exams like the Joint Entrance Examination(JEE Main) and other entrance exams such as SRMJEE, BITSAT, WBJEE, BCECE, and more. A total of twenty-five questions have been asked on JEE MAINS( 2013 to 2023) from this topic including one in 2014, one in 2021, two in 2022, and three in 2023.

Standard Equation of Ellipse

The standard form of the equation of an ellipse with centre $(0,0)$ and major axis on the $x$-axis is $\frac{x^2}{a^2}+\frac{y^2}{\mathbf{b}^2}=1 \quad$ where $b^2=a^2\left(1-e^2\right)$
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 $( \pm a, 0)$

1. $a > b$

2. the length of the major axis is $2a$

3. the length of the minor axis is $2b$

4. the coordinates of the vertices are $(±a, 0)$

Intersection of Line and the Ellipse

The equation of Ellipse and Line is
Ellipse : $\quad \frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Line: $
\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 m^2+b^2\right) x^2+2 m a^2 x+c^2 a^2-a^2 b^2=0
\end{aligned}
$
The above written equation is quadratic in $x$

Now depending on the value of the determinant of this equation we can have the following cases:

1. If $D>0$, then we have 2 real and distinct roots which means two distinct points of intersection of the line and the ellipse.
2. If $D=0$, then we have 1 real and repeated root which means one point of intersection of the line and the ellipse which means that the line is tangent to the ellipse. By putting values in $D=0$, we get $a^2 m^2+b^2=c^2$. This is also the condition of tangency for the line $y=m x+c$ to be tangent to the ellipse.
3. If $D<0$, then we do not have any real root, which means no point of intersection of the line and the ellipse.

Condition of Tangency

Using a²m²+b²=c² as the condition of tangency for the line y = mx + c to be tangent to the ellipse, the equation of tangent to the standard ellipse is $y=m x \pm \sqrt{a^2 m^2+b^2}$

Solved Examples Based on Intersection of Line and an Ellipse

Example 1: If the maximum distance of normal to the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1 b<2$ from the origin is 1, then the eccentricity of the ellipse is : [JEE MAINS 2023]

Solution

Normal to the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1 b<2$ at point $(\mathrm{a} \cos \theta, \mathrm{b} \sin \theta)$ is

$
\frac{a x}{\cos \theta}-\frac{b y}{\sin \theta}=a^2-b^2
$

Its distance from its origin is

$
\begin{aligned}
& \mathrm{d}=\frac{\left|\mathrm{a}^2-\mathrm{b}^2\right|}{\sqrt{\mathrm{a}^2 \sec ^2 \theta+\mathrm{b}^2 \operatorname{cosec}^2 \theta}} \\
& \mathrm{d}=\frac{\left|\mathrm{a}^2-\mathrm{b}^2\right|}{\sqrt{\mathrm{a}^2+\mathrm{b}^2+2 a b+(a \tan \theta-\mathrm{b} \cot \theta)^2}} \\
& \mathrm{~d} \frac{|(a-b)(a+b)|}{\sqrt{a^2+b^2+2 a b+(a \tan \theta-b \tan \theta)^2}} \\
& d_{\max }=\frac{|(a-b)(a+b)|}{a+b}=|a-b| \\
& \because d_{\max }=1 \\
& |2-\mathrm{b}|=1 \\
& 2-\mathrm{b}=1[\because \mathrm{b}<2] \\
& \mathrm{b}=1
\end{aligned}
$
Eccentricity $=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$

$
\Rightarrow \mathrm{e}=\frac{\sqrt{3}}{2}
$
Hence, the answer is $\frac{\sqrt{3}}{2}$

Example 2: Let an ellipse with centre $(1,0)$ and latus rectum of length $\frac{1}{2}$ have its major axis along the $x$-axis. If its minor axis subtends an angle of $60^{\circ}$ at the foci, then the square of the sum of the lengths of its minor and major axes is equal to [JEE MAINS 2023]

Solution

$
\begin{aligned}
& \text { L.R. }=\frac{2 b^2}{a}=\frac{1}{2} \\
& 4 b^2=a \\
& \text { Elipse } \frac{(x-1)^2}{a^2}+\frac{y^2}{b^2}=1 \\
& m_{b_1 k_1}=\frac{1}{\sqrt{3}} \\
& \frac{b}{a}=\frac{1}{\sqrt{3}} \\
& 3 b^2=a^2 e^2=a^2-b^2 \\
& 4 b^2=a^2
\end{aligned}
$
From (i) and (ii)

$
\begin{aligned}
& a=a^2 \\
& \therefore a=1 \\
& b^2=\frac{1}{4} \\
& ((2 a)+(2 b))^2=9
\end{aligned}
$

Hence, the answer is 9

Example 3: The line $\mathrm{y}=\mathrm{x}+1$ meets the ellipse $\frac{\mathrm{x}^2}{4}+\frac{\mathrm{y}^2}{2}=1$ at two points P and Q . If r is the radius of the circle with PQ as diameter then $(3 r)^2$ is equal to : [JEE MAINS 2022]

Solution

Let P be the point $\left(-1+\frac{\mathrm{p}}{\sqrt{2}}, \frac{\mathrm{p}}{\sqrt{2}}\right)$
Since it lies o the point $\frac{x^2}{4}+\frac{y^2}{2}=1$ we get

$
\begin{aligned}
& \frac{\left(-1+\frac{p}{\sqrt{2}}\right)^2}{4}+\frac{\left(\frac{p}{\sqrt{2}}\right)^2}{2}=1 \Rightarrow \frac{3 p^2}{8}-\frac{p}{2 \sqrt{2}}-\frac{3}{4}=0 \\
& \quad\left(P_1-P_2\right)^2=\left(\frac{2 \sqrt{2}}{3}\right)^2-4(-2)=\frac{8}{9}+8=\frac{80}{9}
\end{aligned}
$

We know, $\left(\mathrm{P}_1-\mathrm{P}_2\right)^2=\mathrm{PQ}^2$

$
\begin{aligned}
& \Rightarrow \mathrm{PQ}^2=\frac{80}{9} \\
& \text { Now } r=\frac{\mathrm{PQ}}{2} \Rightarrow 9 \mathrm{r}^2=9 \frac{(\mathrm{PQ})^2}{4}=\frac{9}{4} \times \frac{80}{9}=20
\end{aligned}
$

Hence the correct answer is 20

Example 4: Let E be an ellipse whose axes are parallel to the co-ordinates axes, having its centre at $(3,-4)$, one focus at $(4,-4)$ and one vertex at $(5,-4)$. If $m x-y=4, m>0$ is a tangent to the ellipse $E$ then the value of $5 \mathrm{~m}^2$ is equal to [JEE MAINS 2021]

Solution

$
\begin{aligned}
& a=5-3=2 \\
& a e=4-3=1 \\
\Rightarrow & e=\frac{1}{2}
\end{aligned}
$

$
\begin{aligned}
& e^2=1-\frac{b^2}{a^2} \Rightarrow \frac{1}{4}=1-\frac{b^2}{4} \\
& \Rightarrow 1=4-b^2 \Rightarrow b^2=3
\end{aligned}
$
Any tangent to this ellipse is

$
\begin{aligned}
& (y+4)=m(x-3) \pm \sqrt{4 m^2+3} \\
& y=m x-3 m-4 \pm \sqrt{4 m^2+3}
\end{aligned}
$

Comparing it to the given line $y=m x-4$

$
\begin{aligned}
& \Rightarrow-3 m-4 \pm \sqrt{4 m^2+3}=-4 \\
& \Rightarrow \pm \sqrt{4 m^2+3}=3 m \\
& \Rightarrow 4 m^2+3=9 m^2 \\
& \Rightarrow 5 m^2=3
\end{aligned}
$
Hence the correct answer is 3