Horizontal and Vertical Ellipse: Definition and Examples

What is Ellipse?

The locus of a point moves in a plane such that the ratio of the distance from a fixed point (focus) to the distance from a fixed line (directrix) is constant. The constant is known as eccentricity e and for ellipse 0 < e < 1.

Horizontal Ellipse

When the major axis is along the X-axis and the minor axis is along the Y-axis, i.e. a > b, it is called a horizontal ellipse.

the length of the major axis is 2a

the length of the minor axis is 2b

The foci are S (0, ae) and S’(0, -ae)

Vertical Ellipse

When the major axis is along the Y-axis and the minor axis is along the X-axis, i.e. b > a, it is called a vertical ellipse.

Then, AA’ = 2a and BB’ = 2b

the length of the major axis is 2b

the length of the minor axis is 2a

The foci are S (0, be) and S’(0, -be)

The equation of directrix MZ and M’Z’ are $
y=\frac{b}{e} \text { and } y=-\frac{b}{e}
$

Recommended Video Based on Horizontal and Vertical Ellipse

Solved Examples Based on Horizontal and Vertical Ellipse

Example 1: Let $E_1: \frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>\mathrm{b}$. Let $E_2$ be another ellipse such that it touches the endpoints of the major axis of $E_1$ and the foci of $E_2$ are the endpoints of the minor axis of $\mathrm{E}_1$. If $\mathrm{E}_1$ and $\mathrm{E}_2$ have the same eccentricities, then its value is : [JEE MAINS 2021]

Solution

$
\begin{aligned}
&\text { As per the question, }\\
&\begin{aligned}
& b=c e_2 \text { and } e_1=e_2 \\
\Rightarrow & b=c e_2 \text { and } e_1^2=e_2^2 \\
\Rightarrow & b=c e_2 \text { and } 1-\frac{b^2}{a^2}=1-\frac{a^2}{c^2} \\
\Rightarrow & b=c e_2 \quad \text { and } \quad \frac{b^2}{a^2}=\frac{a^2}{c^2} \\
\Rightarrow & b=c e_2 \quad \text { and } \quad c^2=\frac{a^4}{b^2} \\
\Rightarrow & c=\frac{b}{e_2} \quad \text { and } \quad c=\frac{a^2}{b} \\
\Rightarrow & \frac{b}{e_2}=\frac{a^2}{b} \\
\Rightarrow & e_2=\frac{b^2}{a^2} \\
\Rightarrow & e_2=1-e_1^2 \\
\Rightarrow & e_1=1-e_1^2 \\
\Rightarrow & e_1^2+e_1-1=0 \\
\Rightarrow & e_1=\frac{\sqrt{5}-1}{2}
\end{aligned}
\end{aligned}
$

Hence, the correct answer is $\frac{-1+\sqrt{5}}{2}$
Example 2: Let $\mathrm{S}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{N} \times \mathbb{N}: 9(\mathrm{x}-3)^2+16(\mathrm{y}-4)^2 \leq 144\right\}$ and $\mathrm{T}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathbb{R} \times \mathbb{R}:(\mathrm{x}-7)^2+(\mathrm{y}-4)^2 \leq 36\right\}$. Then $\mathrm{n}(\mathrm{S} \cap \mathrm{T})$ is equal to $\qquad$
[JEE MAINS 2022]
Solution

For $\mathrm{x}=1 \Rightarrow(1,4)$
For $\mathrm{x}=2 \Rightarrow(2,4),(2,5),(2,3),(2,6),(2,2)$
For $\mathrm{x}=3 \Rightarrow(3,4),(3,5),(3,3),(3,6),(3,2)(3,7),(3,1)$
For $\mathrm{x}=4 \Rightarrow(4,4),(4,5),(4,3),(4,6),(4,2)$
For $\mathrm{x}=5 \Rightarrow(5,4),(5,5),(5,3),(5,2),(5,6)$
For $\mathrm{x}=6 \Rightarrow(6,4),(6,3),(6,5)$
For $\mathrm{x}=7 \Rightarrow(7,4)$
$\therefore$ Total 27 common points
Ans:27

Example 3: If $x^2+9 y^2-4 x+3=0, x, y \in \mathbb{R}$, then x and $y$ respectively lie in the intervals :
[JEE MAINS 2021]
Solution: To find the Range of x , make quadratic in y

$
\begin{aligned}
& 9 y^2+x^2-4 x+3=0 \\
& D \geq 0 \\
& \Rightarrow 0-4 \times 9 \times\left(x^2-4 x+3\right) \geq 0 \\
& \Rightarrow(x-1)(x-3) \leq 0 \Rightarrow x \epsilon[1,3]
\end{aligned}
$
To Find the Range of y , make quadratic in x

$
\begin{aligned}
& x^2-4 x+9 y^2+3=0 \\
& D \geq 0 \\
& \Rightarrow 4^2-4 \times\left(9 y^2+3\right) \geq 0 \Rightarrow 9 y^2-1 \leq 0 \\
& \Rightarrow(3 y-1)(3 y+1) \leq 0 \Rightarrow y \epsilon\left[\frac{-1}{3}, \frac{1}{3}\right]
\end{aligned}
$
Hence, the answer is $[1,3]$ and $\left[-\frac{1}{3}, \frac{1}{3}\right]$

Example 4: What are the coordinates of the midpoints of the ellipse $x^2+4 y^2-6 x+40 y=0$
Solution: We know that the center of the Ellipse is the mid-point of the two foci of the ellipse.

$
\begin{aligned}
& x^2-6 x+4\left(y^2+10 y\right)=0 \\
& x^2-6 x+9+4\left(y^2+10 y+25\right)=100+9 \\
& (x-3)^2+4(y+5)^2=109
\end{aligned}
$
Centre is $(3,-5)$
Hence, the answer is $(3,-5)$

Example 5: Consider an ellipse, whose center is at the origin and its major axis is along the $x$-axis. If its eccentricity is $\frac{3}{5}$ and the distance between its foci is 6 , then the area (in sq. units) of the quadrilateral inscribed in the ellipse, with the vertices as the endpoints of major and minor axes of the ellipse, is

Solution

Given

$
e=\frac{3}{5}
$

and

$
2 a e=6 \Rightarrow a e=3
$
Hence $a=5$

$
b^2=a^2\left(1-e^2\right) \Rightarrow b=4
$
Area of quadrilateral $\mathrm{ABCD}=4 \operatorname{Ar}(\Delta A O B)$

$
\begin{aligned}
& =4 \times \frac{1}{2} \times a \times b \\
& =4 \times \frac{1}{2} \times 5 \times 4 \\
& =40
\end{aligned}
$
Hence, the answer is 40