Position of a point with respect to Ellipse

An ellipse is the locus of a point which moves such that its distance from a fixed point is constant. Each fixed point is called a focus (plural: foci). The position of point concerning the ellipse shows whether the give n point lies on, inside or outside the ellipse.

This article is about the position of the point concerning the ellipse which falls under the broader category of two dimensional analytical Geometry. This concept has applications in various fields like calculus, physics etc. This is one of the impootant 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.

Ellipse

An ellipse is the locus of a point which moves such that its distance from a fixed point is constant. Each fixed point is called a focus (plural: foci). An ellipse is formed when a cone is intersected by a plane at an angle with respect to its base.

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 $\frac{\mathrm{x}^2}{\mathbf{a}^2}+\frac{\mathbf{y}^2}{\mathbf{b}^2}=1 \quad$ 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 $( \pm a, 0)$

Position of a point concerning Ellipse
The position of a point with respect to ellipse shows whether the given point lies on, inside or outside of the ellipse. The positon of the point with respect to the elliipse can be determined by using the standard equation of the ellipse, $\frac{\mathrm{x}^2}{\mathbf{a}^2}+\frac{\mathbf{y}^2}{\mathbf{b}^2}=1 \quad$.

Let $P\left(x_1, y_1\right)$ is any point in the plane
Then,
(a) P lies outside of the ellipse, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-1>0$
(b) P lies on of the ellipse, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-1=0$
(c) P lies inside of the ellipse, then $\frac{x^2}{a^2}+\frac{y^2}{b^2}-1<0$

Solved Examples Based on the Position of a Point Concerning Ellipse
Example 1: Where does the point $(5,5)$ lie w.r.t the ellipse $\frac{x^2}{5}+\frac{y^2}{10}=2$ ?
1) Outside
2) On the ellipse
3) Inside
4) None of these

Solution
Ellipse equation can be re-written in standard form as

$
\begin{aligned}
& \frac{x^2}{10}+\frac{y^2}{20}=1 \\
& \frac{x^2}{10}+\frac{y^2}{20}-1=0
\end{aligned}
$
Checking the given point

$
S_1=\frac{5^2}{10}+\frac{5^2}{20}-1=2.75>0
$

Thus, it lies outside the ellipse.
Hence, the answer is the option 1.

Example 2: The number of real tangents that can be drawn to the ellipse $3 x^2+5 y^2=32$ passing through $(3,5)$ is.
1) 2
2) 8
3) 7
4) 5

Solution
For $(3,5)$

$
S_1: 3(9)+5(25)-32>0
$

So, the given Point lies outside the ellipse and hence two tangents can be drawn from the given point to the ellipse.
Hence, the answer is (2).

Example 3: The number of real tangents that can be drawn to the ellipse $3 x^2+5 y^2=32$ passing through $(3,5)$ is
1) 0
2) 1
3) 2
4) infinite

Solution
Since $\qquad$ So, the given point lies outside the ellipse.
So, two real tangents can be drawn from the point to the ellipse.
Hence, the answer is the option (3).

Example 4:There are exactly two points on the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ whose distance from its centre is same and is equal to $\sqrt{\frac{a^2+2 b^2}{2}}$. Find the eccentricity of the ellipse.
1) $\frac{1}{2}$
2) $\frac{1}{\sqrt{2}}$
3) $\frac{1}{\sqrt{3}}$
4) $\sqrt{\frac{2}{3}}$

Solution

Since there are exactly two points on the ellipse $\frac{\mathrm{x}^2}{\mathrm{a}^2}+\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1$ whose distance from the centre is the same, the points would be either endpoints of the major axis or of the minor axis.
But $\sqrt{\frac{a^2+2 b^2}{2}}>b$ so the points are the vertices of the major axis.
Hence $a=\sqrt{\frac{a^2+2 b^2}{2}} \Rightarrow a^2=2 b^2$

Therefore

$
\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\frac{1}{\sqrt{2}}
$

Hence, the answer is the option (2).

Example 5: If a point $\mathrm{P}(\mathrm{x}, \mathrm{y})$ moves along the ellipse $\frac{\mathrm{x}^2}{25}+\frac{\mathrm{y}^2}{16}=1$ and if C is the center of the ellipse, then,
1) 39
2) 37
3) 40
4) 41

Solution

Vertex $:(5,0) \operatorname{SayP}(x, y)=(5,0)$ AndcentreC $=(0,0) \operatorname{So} 4 \max \{C P\}+5 \min \{C P\} \Rightarrow 4 \times 5+5 \times 4 \Rightarrow 20+20 \Rightarrow 40$

Hence, the answer is the option (3).

Summary

The position of a point with respect to ellipse shows whether the given point lies on, inside or outside of the ellipse. It is determined using the equation of the ellipse. Understanding the concepts related toe ellipse would give an idea to solve more complex problems in calculus, physics, engineering etc.