SATHEE: Ellipse 1 Question 3

Ellipse 1 Question 3

3. Let $S$ and $S^{'}$ be the foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $Δ S^{'} B S$ is a right angled triangle with right angle at $B$ and area $(Δ S^{'} B S) = 8 s q$ units, then the length of a latus rectum of the ellipse is

(a) $2 \sqrt{2}$

(b) $4 \sqrt{2}$

(d) 4

(2019 Main, 12 Jan II)

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Answer:

Correct Answer: 3. (d)

Solution:

Let the ellipse be $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1$ .

Then, according to given information, we have the following figure.

Clearly, slope of line $S B = \frac{b}{- a e}$ and slope of line $S^{'} B = \frac{b}{a e}$

$∵$ Lines $S B$ and $S^{'} B$ are perpendicular, so

$\frac{b}{- a e} \cdot \frac{b}{a e} = - 1$

$[∵$ product of slopes of two

$\Rightarrow b^{2} = a^{2} e^{2}$

Also, it is given that area of $Δ S^{'} B S = 8$

$∴ \frac{1}{2} a^{2} = 8$

$[∵ S^{'} B = S B = a because S^{'} B + S B = 2 a and S^{'} B = S B]$

$\Rightarrow a^{2} = 16 \Rightarrow a = 4$

$∵ e^{2} = 1 - \frac{b^{2}}{a^{2}} = 1 - e^{2}$

[from Eq. (i)]

$\Rightarrow 2 e^{2} = 1$

$\Rightarrow e^{2} = \frac{1}{2}$

From Eqs. (i) and (iii), we get

$\begin{array}{cc} b^{2} = a^{2} \frac{1}{2} = 16 \frac{1}{2} [using Eq. (ii)] \\ \Rightarrow & b^{2} = 8 \end{array}$

Now, length of latus rectum $= \frac{2 b^{2}}{a} = \frac{2 \times 8}{4} = 4$ units

4 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 $= 2 b]$

$\begin{aligned} \Rightarrow b \frac{b}{a} = 4 and \frac{b}{a} = e \\ \Rightarrow b (e) = 4 \\ \Rightarrow b = 4 \cdot \frac{1}{e} \end{aligned}$

Also, we know that $b^{2} = a^{2} (1 - e^{2})$

$\begin{aligned} \Rightarrow & \frac{b^{2}}{a^{2}} = 1 & - e^{2} & \Rightarrow e^{2} = 1 - e^{2} \\ \Rightarrow & 2 e^{2} & = 1 \\ \Rightarrow & e & = \frac{1}{\sqrt{2}} \end{aligned}$

From Eqs. (i) and (ii), we get

Now, $a^{2} = \frac{b^{2}}{1 - e^{2}} = \frac{32}{1 - \frac{1}{2}} = 64$

$∴$ 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.

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Ellipse 1 Question 3

3. Let $S$ and $S^{'}$ be the foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $Δ S^{'} B S$ is a right angled triangle with right angle at $B$ and area $(Δ S^{'} B S) = 8 s q$ units, then the length of a latus rectum of the ellipse is

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Ellipse 1 Question 3

3. Let S and S′ be the foci of an ellipse and B be any one of the extremities of its minor axis. If ΔS′BS is a right angled triangle with right angle at B and area (ΔS′BS)=8sq units, then the length of a latus rectum of the ellipse is

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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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3. Let $S$ and $S^{'}$ be the foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $Δ S^{'} B S$ is a right angled triangle with right angle at $B$ and area $(Δ S^{'} B S) = 8 s q$ units, then the length of a latus rectum of the ellipse is