Ellipse Ques 7
- In an ellipse, with centre at the origin, if the difference of the lengths of major axis and minor axis is 10 and one of the foci is at $(0,5 \sqrt{3})$, then the length of its latus rectum is
(2019 Main, 8 April I)
(a) 5
(b) 10
(c) 8
(d) 6
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Answer:
Correct Answer: 7.(a)
Solution:
Formula:
- One of the focus of ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ is on $Y$-axis $(0,5 \sqrt{3})$
$$ \therefore \quad b e=5 \sqrt{3} $$
[where $e$ is eccentricity of ellipse]
According to the question,
$$ \begin{aligned} & 2 b-2 a=10 \\ & b-a=5 \end{aligned} $$
On squaring Eq. (i) both sides, we get
$$ b^{2} e^{2}=75 $$
$\Rightarrow b^{2} 1-\frac{a^{2}}{b^{2}}=75$
$\because e^{2}=1-\frac{a^{2}}{b^{2}}$
$\Rightarrow \quad b^{2}-a^{2}=75$
$\Rightarrow(b+a)(b-a)=75$
$\Rightarrow \quad b+a=15 \quad$ [from Eq. (ii)]
On solving Eqs. (ii) and (iii), we get
$$ b=10 \text { and } a=5 $$
So, length of latusrectum is $\frac{2 a^{2}}{b}=\frac{2 \times 25}{10}=5$ units
BETA
