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  • If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is : Option: 1 <img alt="2\sqrt{3}" src="https://learn.careers360.com/latex-image/?2%5Csqrt%

If the distance between the foci of an ellipse is 6 and the distance between its directrices is 12, then the length of its latus rectum is :
Option: 1 2\sqrt{3}      
Option: 2 \sqrt{3}      
Option: 3 \frac{3}{\sqrt{2}}      
Option: 4 3\sqrt{2}      
 

Answers (1)

best_answer

 

 

Length of Latusrectum -

Length of Latusrectum

\\ {\text { Let Latusrectum } \mathrm{LL}^{\prime}=2 \alpha} \\ {\mathrm{S}(\mathrm{ae}, 0) \text { is focus, then } \mathrm{LS}=\mathrm{SL}^{\prime}=\alpha} \\ {\text { coordinates of } \mathrm{L} \text { and } \mathrm{L}^{\prime} \text { become }(\mathrm{ae}, \alpha) \text { and }(\mathrm{ae},-\alpha) \text { respectively }} \\ {\text { Eq of ellipse, } \frac{\mathrm{x}^{2}}{\mathrm{a}^{2}}+\frac{\mathrm{y}^{2}}{\mathrm{b}^{2}}=1} \\ {\therefore \frac{(\mathrm{ae})^{2}}{\mathrm{a}^{2}}+\frac{\alpha^{2}}{\mathrm{a}^{2}}=1 \Rightarrow \alpha^{2}=\mathrm{b}^{2}\left(1-\mathrm{e}^{2}\right)} \\ {\alpha^{2}=\mathrm{b}^{2}\left(\frac{\mathrm{b}^{2}}{\mathrm{a}^{2}}\right) \quad\left[\mathrm{b}^{2}=\mathrm{a}^{2}\left(1-\mathrm{e}^{2}\right)\right]} \\ {\alpha=\frac{\mathrm{b}^{2}}{\mathrm{a}}} \\ {\Rightarrow 2 \alpha=\mathrm{LL}^{\prime}=\frac{2 \mathrm{b}^{2}}{\mathrm{a}}}

End Point of Latus rectum:

\mathrm{L}=\left(\mathrm{ae}, \frac{\mathrm{b}^{2}}{\mathrm{a}}\right) \text { and } \mathrm{L}^{\prime}=\left(\mathrm{ae},-\frac{\mathrm{b}^{2}}{\mathrm{a}}\right)

 

Focal Distance of a Point:

The sum of the focal distance of any point on the ellipse is equal to the major axis.

Let P(x, y) be any point on the ellipse.

 

\\ {\text { Here, \;\;\;\;\;\;\;\;\;} \quad S P=e P M=e\left(\frac{a}{e}-x\right)=a-e x} \\\\ {\quad \quad\quad\quad\quad\quad\; S^{\prime} P=e P M'=e\left(\frac{a}{e}+x\right)=a+e x}

Now, SP + S’P = a – ex + a + ex = 2a =AA' =  constant. 

Thus the sum of the focal distances of a point on the ellipse is constant.

-

 

 

 

given

2ae = 6 \\ \therefore ae = 3.......(1)

Also,

 \frac{2a}{e}=126 \\ \therefore \frac{a}{e}=63.......(2)

from (1) and (2) 

e=\frac{1}{\sqrt{2}}, a = 3\sqrt{2}

since , b^2= a^2(1-e^2) \\

Substitute the values of 'e' and 'a' in above equation .

\Rightarrow b^2= 9 \\ \therefore b= \pm 3

length of latus rectum = \frac{2b^2}{a} =\frac{ 2 \times 9}{3\sqrt{2}}=3\sqrt{2}

Correct Option (4)

Posted by

Ritika Jonwal

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