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 :

Question

The area bounded by $ y - 1 = |x| $ and $ y + 1 = |x| $ is: (a) $ \frac{1}{2} $

Answer 1

To solve the problem, we need to use the properties of the ellipse. Given the conditions in the problem, we can derive the eccentricity of the ellipse step-by-step.

Understanding the Latus Rectum: The length of the latus rectum ($L\_r$) of an ellipse is given by the formula:

$L\_r = \frac{2b^2}{a}$

According to the problem, $L\_r = 4$ units.

Distance from Focus to Vertex: The distance from a focus to its nearest vertex on the major axis is given as $\frac{3}{2}$ units, which is the same as the distance denoted by $ae$ (where $e$ is the eccentricity).

Thus, we have:

$ae = \frac{3}{2}$

We can relate $a$, $b$, and $e$ using the formula:

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

Substitute the known values and solve for $e$.

From $ae = \frac{3}{2}$, isolate $a$:

$a = \frac{3}{2e}$

Using the formula for the latus rectum:

$\frac{2b^2}{a} = 4 \Rightarrow b^2 = 2a$

Substitute $a = \frac{3}{2e}$ into the equation:

$b^2 = 2 \times \frac{3}{2e} = \frac{3}{e}$

Substituting $b = a\sqrt{1-e^2}$, we have:

$b^2 = a^2 (1-e^2)$

Equating this with $\frac{3}{e}$, we get:

$\left(\frac{3}{2e}\right)^2 (1-e^2) = \frac{3}{e}$

After rearranging and solving for $e$, you will find:

$e = \frac{1}{3}$

Conclusion: The eccentricity of the ellipse is $\frac{1}{3}$. Option $\frac{1}{3}$ is correct.

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 :

The Correct Option is B

Solution and Explanation

Top Questions on Conic sections

Questions Asked in JEE Main exam