Question

If the length of the latus rectum of an ellipse is one-fourth of the major axis, then the eccentricity of the ellipse is

• A. \(\frac{\sqrt{3}}{2}\)
• B. \(\frac{\sqrt{3}}{4}\)
• C. \(\frac{\sqrt{5}}{4}\)
• D. \(\frac{\sqrt{5}}{6}\)
• E. \(\frac{2}{3}\)

Hint:

For ellipse questions, remember these two key formulas: major axis \(=2a\) and latus rectum \(=\frac{2b^2}{a}\). Most questions reduce to simple substitution from these.

Answer 1

The Correct Option is A

Step 1: Recall the standard ellipse and its important formulas.
For an ellipse in standard form:
\[ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1 \quad \text{with } a>b \] the length of the major axis is:
\[ 2a \] and the length of the latus rectum is:
\[ \frac{2b^2}{a} \]

Step 2: Use the condition given in the question.
It is given that the length of the latus rectum is one-fourth of the major axis. Therefore:
\[ \frac{2b^2}{a}=\frac{1}{4}(2a) \]

Step 3: Simplify the equation.
The right-hand side becomes:
\[ \frac{1}{4}(2a)=\frac{a}{2} \] So we have:
\[ \frac{2b^2}{a}=\frac{a}{2} \]

Step 4: Solve for the relation between \(a\) and \(b\).
Multiply both sides by \(2a\):
\[ 4b^2=a^2 \] Thus,
\[ b^2=\frac{a^2}{4} \]

Step 5: Recall the formula involving eccentricity.
For an ellipse, eccentricity \(e\) is given by:
\[ e=\sqrt{1-\frac{b^2}{a^2}} \] Now substitute \[ \frac{b^2}{a^2}=\frac{1}{4} \]

Step 6: Calculate the eccentricity.
\[ e=\sqrt{1-\frac{1}{4}} \] \[ e=\sqrt{\frac{3}{4}} \] \[ e=\frac{\sqrt{3}}{2} \]

Step 7: State the final answer.
Hence, the eccentricity of the ellipse is:
\[ \boxed{\frac{\sqrt{3}}{2}} \] which matches option \((1)\).