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

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