Question

If (2, 4) and (10, 10) are the ends of a latus-rectum of an ellipse with eccentricity 1/2, then the length of semi-major axis is

विकल्प

• 20/3

• 15/3

• 40/3

• none of these

Answer 1

\[\frac{20}{3}\]
\[e = \frac{1}{2} \left(\text{ Given }\right)\]
\[\text{ Now, }e = \sqrt{1 - \frac{b^2}{a^2}}\]
\[ \Rightarrow \frac{1}{2} = \sqrt{1 - \frac{b^2}{a^2}}\]
On squaring both sides, we get:
\[\frac{1}{4} = 1 - \frac{b^2}{a^2}\]
\[ \Rightarrow \frac{1}{4} = \frac{a^2 - b^2}{a^2}\]
\[ \Rightarrow 4 a^2 - 4 b^2 = a^2 \]
\[ \Rightarrow 3 a^2 = 4 b^2 \]
\[ \Rightarrow \frac{b^2}{a^2} = \frac{3}{4}\]
\[ \Rightarrow a^2 = \frac{4 b^2}{3}or a = \frac{2b}{\sqrt{3}} . . . (1)\]
\[\text{ Latus rectum }=\frac{2 b^2}{a^2}\]
If (2, 4) and (10, 10) are the ends points of a latus rectum.
\[i . e . \sqrt{\left( 10 - 2 \right)^2 + \left( 10 - 4 \right)^2} = 2 b^2 \times \frac{\sqrt{3}}{2b}\]
\[ \Rightarrow \sqrt{64 + 36} = \sqrt{3}b\]
On squaring both sides, we get:
\[100 = 3 b^2 \]
\[ \Rightarrow b = \frac{10}{\sqrt{3}}\]
\[\text{ Now,} a = \frac{2b}{\sqrt{3}}\]
\[ \Rightarrow a = \frac{20}{\sqrt{3}} \times \frac{1}{\sqrt{3}}\]
\[ \Rightarrow a = \frac{20}{3}\]
\[\text{ So, the length of the semi major axis is }\frac{20}{3}.\]