Wait, why is the major axis vertical? How do I know that?

Look at the coordinates of the foci: $(2, 5)$ and $(2, -3)$. The x-coordinates are both $2$. This means the line connecting the foci is a vertical line. Since the major axis of an ellipse always passes through the foci, it must be vertical.

Can I use $b^2 = a^2(e^2 - 1)$?

No, that is the relation for a hyperbola! For an ellipse, the eccentricity $e$ is less than $1$, so $1 - e^2$ is positive. Always remember: Ellipse is $b^2 = a^2(1 - e^2)$, and Hyperbola is $b^2 = a^2(e^2 - 1)$.

Is the latus rectum length always $\frac{2b^2}{a}$?

Yes, that is the standard definition for the length of the latus rectum for any ellipse, regardless of its orientation. It is the length of the chord passing through the focus perpendicular to the major axis.

JEE Main 2025 April

LEVELJEE Main

Animated Solution for Mathematics - Conic Sections: The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, - 3)$ and eccentricity is $\frac{4}{5}$ , is

Select Answer:

Visualized Solution

Identify the Foci

Given Foci: $F_{1} (2, 5)$ and $F_{2} (2, - 3)$
Eccentricity $e = \frac{4}{5}$

Determine Orientation

Notice the x-coordinates of both foci are $2$ .
Since x-coordinates are identical, the major axis is vertical.

Distance Between Foci Formula

For any ellipse, the distance between its foci is $2 a e$ .

Calculate Distance

$2 a e = (2 - 2)^{2} + (5 - (- 3))^{2}$
$2 a e = 0^{2} + 8^{2} = 8$
Therefore, $a e = 4$

Substitute Eccentricity

We know $e = \frac{4}{5}$
Substitute into $a e = 4$ :
$a (\frac{4}{5}) = 4$

Calculate Semi-major Axis $a$

$a (\frac{4}{5}) = 4 ⟹ a = 5$
The semi-major axis $a$ is $5$ .

Relation Between $a, b, e$

The fundamental relation for an ellipse is:
$b^{2} = a^{2} (1 - e^{2})$

Substitute Values for $b^{2}$

Substitute $a = 5$ and $e = \frac{4}{5}$ :
$b^{2} = 5^{2} (1 - (\frac{4}{5})^{2})$

Calculate $b^{2}$

$b^{2} = 25 (1 - \frac{16}{25})$
$b^{2} = 25 (\frac{9}{25}) = 9$

Latus Rectum Formula

The length of the Latus Rectum is given by $\frac{2 b ^{2}}{a}$

Substitute for Latus Rectum

Substitute $b^{2} = 9$ and $a = 5$ into the formula:
$Length = \frac{2 ( 9 )}{5}$

Final Answer

$Length = \frac{18}{5}$
The length of the latus-rectum is $\frac{18}{5}$ .

00:00 / 00:00

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Solution Diagram

Analyzing the Setup

Imagine you are standing on a coordinate plane, looking at two points:

F_{1} (2, 5)

and

F_{2} (2, - 3)

. These are the two foci that define the shape of the ellipse.

Notice that the

x

-coordinates are identical. This indicates that the ellipse is a vertical ellipse, with its major axis parallel to the

y

-axis.

The Geometry of Distance

The distance between the two foci of any ellipse is defined as

2 a e

. By calculating the difference in the

y

-coordinates of our given points, we find:

2 a e = 5 - (- 3) = 8

This simplifies to the first pillar of our solution:

a e = 4

Finding the Semi-major Axis

The problem provides the eccentricity

e = \frac{4}{5}

. We can now solve for the semi-major axis

a

by substituting this value into our previous equation:

a (\frac{4}{5}) = 4

By canceling the fours, we find the value of the semi-major axis:

a = 5

The Final Piece

The Latus Rectum

To find the length of the latus rectum, defined by the formula

\frac{2 b ^{2}}{a}

, we must first determine

b^{2}

. We use the fundamental relationship for an ellipse:

b^{2} = a^{2} (1 - e^{2})

Substituting our known values into this equation yields:

b^{2} = 5^{2} (1 - (\frac{4}{5})^{2})

b^{2} = 25 (1 - \frac{16}{25}) = 25 (\frac{9}{25}) = 9

Finally, we calculate the length of the latus rectum:

Length = \frac{2 b ^{2}}{a} = \frac{2 ( 9 )}{5} = \frac{18}{5}

The final result is $\frac{18}{5}$ .

Animated Solution for Mathematics - Conic Sections: The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, - 3)$ and eccentricity is $\frac{4}{5}$ , is

Select Answer:

Visualized Solution

Identify the Foci

Determine Orientation

Distance Between Foci Formula

Calculate Distance

Substitute Eccentricity

Calculate Semi-major Axis $a$

Relation Between $a, b, e$

Substitute Values for $b^{2}$

Calculate $b^{2}$

Latus Rectum Formula

Substitute for Latus Rectum

Final Answer

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Geometry of Distance

Finding the Semi-major Axis

The Final Piece

Similar Questions

An ellipse has its center at $(1, - 2)$ , one focus at $(3, - 2)$ and one vertex at $(5, - 2)$ . Then the length of its latus rectum is :

In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at $(0, 53)$ , then the length of its latus rectum is:

For some $θ \in (0, \frac{π}{2})$ , if the eccentricity of the hyperbola, $x^{2} - y^{2} sec^{2} θ = 10$ is $5$ times the eccentricity of the ellipse, $x^{2} sec^{2} θ + y^{2} = 5$ , then the length of the latus rectum of the ellipse, is :

If the length of the latus rectum of an ellipse is 4 units and the distance between a focus and its nearest vertex on the major axis is $\frac{3}{2}$ units, then its eccentricity is :

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$ . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is

Let $P$ be a parabola with vertex $(2, 3)$ and directrix $2 x + y = 6$ . Let an ellipse $E : \frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ of eccentricity $\frac{1}{2}$ pass through the focus of the parabola $P$ . Then the square of the length of the latus rectum of $E$ , is

Let $S$ and $S^{'}$ be the foci of the ellipse and $B$ be any one of the extremities of its minor axis. If $S^{'} B S$ is a right angled triangle with right angle at $B$ and area $(S^{'} B S) = 8$ sq. units, then the length of a latus rectum of the ellipse is :

Let the foci and length of the latus rectum of an ellipse $\frac{x ^{2}}{a ^{2}} + \frac{y ^{2}}{b ^{2}} = 1, a > b$ be $(\pm 5, 0)$ and $50$ , respectively. Then, the square of the eccentricity of the hyperbola $\frac{x ^{2}}{b ^{2}} - \frac{y ^{2}}{a ^{2} b ^{2}} = 1$ equals

Animated Solution for Mathematics - Conic Sections: .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } The length of the latus-rectum of the ellipse, whose foci are (2,5) and (2,−3) and eccentricity is 54​, is

Select Answer:

Visualized Solution

Identify the Foci

Determine Orientation

Distance Between Foci Formula

Calculate Distance

Substitute Eccentricity

Calculate Semi-major Axis .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a

Relation Between .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } a,b,e

Substitute Values for .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Calculate .math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } b2

Latus Rectum Formula

Substitute for Latus Rectum

Final Answer

The Sigma Insight: Standard Equations of Parabola, Ellipse, and Hyperbola

Analyzing the Setup

The Geometry of Distance

Finding the Semi-major Axis

The Final Piece

Similar Questions

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } An ellipse has its center at (1,−2), one focus at (3,−2) and one vertex at (5,−2). Then the length of its latus rectum is :

.math-text-wrapper .katex { font-size: 1.05em !important; } .math-text-wrapper .katex-display { margin: 1rem 0 !important; } Let the foci of a hyperbola coincide with the foci of the ellipse 36x2​+16y2​=1. If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :

If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is

Animated Solution for Mathematics - Conic Sections: The length of the latus-rectum of the ellipse, whose foci are $(2, 5)$ and $(2, - 3)$ and eccentricity is $\frac{4}{5}$ , is

Calculate Semi-major Axis $a$

Relation Between $a, b, e$

Substitute Values for $b^{2}$

Calculate $b^{2}$

An ellipse has its center at $(1, - 2)$ , one focus at $(3, - 2)$ and one vertex at $(5, - 2)$ . Then the length of its latus rectum is :

Let the foci of a hyperbola coincide with the foci of the ellipse $\frac{x ^{2}}{36} + \frac{y ^{2}}{16} = 1$ . If the eccentricity of the hyperbola is 5, then the length of its latus rectum is :