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

• A. 20/3
  
• B. 15/3
  
• C. 40/3
  
• D. none of these

Answer 1

The correct option is A 20/3


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