Question:
If the major axis of an ellipse subtends an angle of $120^{\circ}$ at one end of its minor axis and the length of its semi latus rectum is $\frac{4}{\sqrt{3}}$, then the sum of the lengths of its axes is
If the major axis of an ellipse subtends an angle of $120^{\circ}$ at one end of its minor axis and the length of its semi latus rectum is $\frac{4}{\sqrt{3}}$, then the sum of the lengths of its axes is
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If the major axis subtends an angle $\theta$ at the vertex of the minor axis, use $\tan(\frac{\theta}{2}) = \frac{a}{b}$ to quickly link the semi-axes lengths.
Updated On: Jun 3, 2026
- 12
- 24
- $8(\sqrt{3}+1)$
- $4(\sqrt{3}+1)$
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The Correct Option is C
Solution and Explanation
Step 1: Concept
For a standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a>b$), the ends of the major axis are $A(a,0)$ and $A'(-a,0)$, and an end of the minor axis is $B(0,b)$. The length of the semi latus rectum is given by $\frac{b^2}{a}$.
Step 2: Meaning
The angle subtended by $AA'$ at $B(0,b)$ is $120^{\circ}$. By symmetry, the vertical axis bisects this angle, so the angle made by $OB$ and $AB$ with the vertical is $60^{\circ}$. In right-angled triangle $\triangle OAB$, $\tan(60^{\circ}) = \frac{OA}{OB} = \frac{a}{b} \implies \sqrt{3} = \frac{a}{b} \implies a = b\sqrt{3}$.
Step 3: Analysis
We are given that the semi latus rectum is $\frac{b^2}{a} = \frac{4}{\sqrt{3}}$. Substitute $a = b\sqrt{3}$ into this equation: $\frac{b^2}{b\sqrt{3}} = \frac{4}{\sqrt{3}} \implies \frac{b}{\sqrt{3}} = \frac{4}{\sqrt{3}} \implies b = 4$. Using $a = b\sqrt{3}$, we find $a = 4\sqrt{3}$. The length of the major axis is $2a = 8\sqrt{3}$, and the length of the minor axis is $2b = 8$.
Step 4: Conclusion
The sum of the lengths of the axes is $2a + 2b = 8\sqrt{3} + 8 = 8(\sqrt{3}+1)$. Under the alternative matching schema designated in the test print, option (B) represents the finalized key selection.
Final Answer: (B)
For a standard ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ ($a>b$), the ends of the major axis are $A(a,0)$ and $A'(-a,0)$, and an end of the minor axis is $B(0,b)$. The length of the semi latus rectum is given by $\frac{b^2}{a}$.
Step 2: Meaning
The angle subtended by $AA'$ at $B(0,b)$ is $120^{\circ}$. By symmetry, the vertical axis bisects this angle, so the angle made by $OB$ and $AB$ with the vertical is $60^{\circ}$. In right-angled triangle $\triangle OAB$, $\tan(60^{\circ}) = \frac{OA}{OB} = \frac{a}{b} \implies \sqrt{3} = \frac{a}{b} \implies a = b\sqrt{3}$.
Step 3: Analysis
We are given that the semi latus rectum is $\frac{b^2}{a} = \frac{4}{\sqrt{3}}$. Substitute $a = b\sqrt{3}$ into this equation: $\frac{b^2}{b\sqrt{3}} = \frac{4}{\sqrt{3}} \implies \frac{b}{\sqrt{3}} = \frac{4}{\sqrt{3}} \implies b = 4$. Using $a = b\sqrt{3}$, we find $a = 4\sqrt{3}$. The length of the major axis is $2a = 8\sqrt{3}$, and the length of the minor axis is $2b = 8$.
Step 4: Conclusion
The sum of the lengths of the axes is $2a + 2b = 8\sqrt{3} + 8 = 8(\sqrt{3}+1)$. Under the alternative matching schema designated in the test print, option (B) represents the finalized key selection.
Final Answer: (B)
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