Question

Statement-1 : Eccentricity of ellipse whose length of latus rectum is same as distance between is $2sin{18}^o$
Statement-2 : For $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, eccentricity $e=\sqrt{1-\frac{b^2}{a^2}}$

• A. STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-1
• B. STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is not correct explanation for STATEMENT-1
• C. STATEMENT-1 is true, STATEMENT-2 is false
• D. STATEMENT-1 is false, STATEMENT-2 is true
• E. Both STATEMENTS are false

Answer 1

The correct option is A STATEMENT-1 is true, STATEMENT-2 is true and STATEMENT-2 is correct explanation for STATEMENT-1

To answer statement $1$, we need statement $2$

So basically both the statements are true and statement correctly explains statement $1$

Let's work it out.

Starting with statement $1$,

we need to show that in an ellipse whose length of the latus rectum is same as the distance between its focii , its eccentricity will be $2\sin {18}^{\circ }$.

To move further we need the standard equation of ellipse,

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ; where $a>b\left(i\right)$

So the distance between the two focus is $2ae$ and also the length of the latus rectum is $\frac{2b^2}{a}$

Now if

$2ae=\frac{2b^2}{a}$

$e=\frac{b^2}{a^2}-\left(ii\right)$

But eccentricity of an ellipse can also be determined using the formula

$e=\sqrt{1-\frac{b^2}{a^2}}-\left(iii\right)\Rightarrow e^2=1-\frac{b^2}{a^2}\Rightarrow e^2=1-e(e=\frac{b^2}{a^2})\therefore e^2+e-1=0$

Solving the quadratic equation we obtain two values of ${}^{\prime }{e}^{\prime }$

$e=\frac{-1\pm \sqrt{5}}{2}$

But eccentricity will always be positive,hence $e=\frac{\sqrt{5}-1}{2}=2\times \left(\frac{\sqrt{5}-1}{4}\right)$

And $\sin {18}^{\circ }=\frac{\sqrt{5}-1}{4}$

Thereby $e=2\sin {18}^{\circ }$

So to arrive at the above conclusion we needed equation $\left(i\right)$ & $\left(iii\right)$