Q.If the eccentricity of an ellipse is $\dfrac{5}{8}$ and the distance between its foci is 10, then find latus rectum of the ellipse.

The key idea here involves understanding the fundamental properties of an ellipse, specifically the definitions of eccentricity, the distance between foci, and the latus rectum.

Given the eccentricity $e = \frac{5}{8}$ and the distance between its foci is 10.

The distance between the foci of an ellipse is $2ae$.

So, $2ae = 10$, which implies $ae = 5$.

Substituting the value of $e$: $a \left(\frac{5}{8}\right) = 5$.

This gives $a = 8$.

For an ellipse, the relationship between $a$, $b$, and $e$ is $b^2 = a^2(1 - e^2)$.

Substitute $a=8$ and $e=\frac{5}{8}$:

$b^2 = 8^2 \left(1 - \left(\frac{5}{8}\right)^2\right)$

$b^2 = 64 \left(1 - \frac{25}{64}\right)$

$b^2 = 64 \left(\frac{64 - 25}{64}\right)$ …

To find the latus rectum of an ellipse, we need to understand a few key properties: its eccentricity, the location of its foci, and the lengths of its semi-major and semi-minor axes. These properties are interconnected, and the problem provides us with enough information to determine them.

An ellipse is essentially a "stretched" circle. Its shape is defined by its eccentricity, $e$.

• Eccentricity ($e$): This value tells us how "flat" or "circular" an ellipse is. For an ellipse, $0 < e < 1$. A value closer to 0 means it's more circular, and a value closer to 1 means it's more elongated.
• Foci ($F_1, F_2$): These are two fixed points inside the ellipse. The sum of the distances from any point on the ellipse to the two foci is constant.
• Semi-major axis ($a$): Half the length of the longest diameter of the ellipse.
• Semi-minor axis ($b$): Half the length of the shortest diameter of the ellipse.
• Distance between foci: If the foci are at $(\pm ae, 0)$ for a standard ellipse centered at the origin, the distance between them is $2ae$.
• Latus Rectum (LR): This is a chord passing through a focus and perpendicular to the major axis. Its length is given by the formula $\frac{2b^2}{a}$.

The core idea here is to use the given eccentricity and the distance between the foci to first find the value of $a$ (the semi-major axis). Once we have $a$ and $e$, we can find $b^2$ (the square of the semi-minor axis) using the fundamental relationship between $a$, $b$, and $e$ for an ellipse. Finally, with $a$ and $b^2$, we can directly calculate the latus rectum.

Here's how we proceed:

1. Identify the given information.

   We are given:

   • Eccentricity, $e = \frac{5}{8}$.
   • Distance between foci, $2ae = 10$.

2. Use the distance between foci to find the semi-major axis ($a$).

   We know that the distance between the foci of an ellipse is $2ae$.

   We have the equation:

   $2ae = 10$

   Substitute the given value of $e$:

   $2a \left(\frac{5}{8}\right) = 10$

   Simplify the left side:

   $a \left(\frac{5}{4}\right) = 10$

   Now, solve for $a$:

   $a = 10 \times \frac{4}{5}$

   $a = 2 \times 4$

   $a = 8$

   So, the length of the semi-major axis is 8 units. …