Ellipse Ques 1
- Let the length of the latus rectum of an ellipse with its major axis along $X$-axis and centre at the origin, be $8$ . If the distance between the foci of this ellipse is equal to the length of its minor axis, then which one of the following points lies on it?
(2019 Main, 11 Jan II)
(a) $(4 \sqrt{2}, 2 \sqrt{3})$
(b) $(4 \sqrt{3}, 2 \sqrt{2})$
(c) $(4 \sqrt{2}, 2 \sqrt{2})$
(d) $(4 \sqrt{3}, 2 \sqrt{3})$
Show Answer
Answer:
Correct Answer: 1.(b)
Solution: (b) Let the equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Then, according the problem, we have
$\frac{2 b^2}{a}=8 $ and $ 2 a e=2 b $
[Length of latusrectum $=\frac{2 b^2}{a} $ and length of minor axis $=2 b$]
$ \begin{array}{rlrl} \Rightarrow & b\left(\frac{b}{a}\right) =4 \text { and } \frac{b}{a}=e \\ \Rightarrow & b(e) =4 \\ \Rightarrow & b =4 \cdot \frac{1}{e} \quad ……..(i) \end{array} $
Also, we know that $b^2=a^2\left(1-e^2\right)$
$\Rightarrow \quad \frac{b^2}{a^2}=1-e^2 \Rightarrow \quad e^2=1-e^2$
$ \left[\because \frac{b}{a}=e\right] $
$\Rightarrow \quad$ $2 e^2=1$
$\Rightarrow \quad$ $ e=\frac{1}{\sqrt{2}} \quad ……..(ii)$
From Eqs. (i) and (ii), we get
Now, $ \begin{aligned} b & =4 \sqrt{2} \\ a^2 & =\frac{b^2}{1-e^2}=\frac{32}{1-\frac{1}{2}}=64 \end{aligned} $
$\therefore$ Equation of ellipse be $\frac{x^2}{64}+\frac{y^2}{32}=1$
Now, check all the options.
Only $(4 \sqrt{3}, 2 \sqrt{2})$, satisfy the above equation.
BETA
