SATHEE: Ellipse 1 Question 2

Ellipse 1 Question 2

2. 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

(d) 6

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Answer:

Correct Answer: 2. (a)

Solution:

One of the focus of ellipse $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ is on $Y$ -axis $(0, 5 \sqrt{3})$

$∴ b e = 5 \sqrt{3}$

[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

$b^{2} e^{2} = 75$

$\Rightarrow b^{2} 1 - \frac{a^{2}}{b^{2}} = 75$

$∵ e^{2} = 1 - \frac{a^{2}}{b^{2}}$

$\Rightarrow b^{2} - a^{2} = 75$

$\Rightarrow (b + a) (b - a) = 75$

$\Rightarrow b + a = 15$ [from Eq. (ii)]

On solving Eqs. (ii) and (iii), we get

$b = 10 and a = 5$

So, length of latusrectum is $\frac{2 a^{2}}{b} = \frac{2 \times 25}{10} = 5$ units

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Ellipse 1 Question 2

2. 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

Answer:

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Ellipse 1 Question 2

2. 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 (0,53), then the length of its latus rectum is

Answer:

इंजीनियरिंग की तैयारी

चिकित्सा तैयारी

एनसीईआरटी पुस्तकें और समाधान

महत्वपूर्ण संसाधन

अन्य संसाधन

पाठ्यक्रम

परीक्षा मंच

नमूना मॉक टेस्ट

सदस्यता

एनसीईआरटी व्यायाम वीडियो समाधान

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2. 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