Let S and S’ be the foci of an ellipse
Question:

Let $S$ and $S^{\prime}$ be the foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $\Delta \mathrm{S}^{\prime} \mathrm{BS}$ is a right angled triangle with right angle at $\mathrm{B}$ and area $\left(\Delta \mathrm{S}^{\prime} \mathrm{BS}\right)=8 \mathrm{sq}$. units, hen the length of a latus rectum of the ellipse is :

  1. (1) 4

  2. (2) $2 \sqrt{2}$

  3. (3) $4 \sqrt{2}$

  4. (4) 2


Correct Option: 1

Solution:

 $\because \triangle S B S$ is right angled triangle, then

$($ Slope of $B S) \times\left(\right.$ Slope of $\left.B S^{\prime}\right)=-1$

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

$b^{2}=a^{2} e^{2}$ …..(1)

Since, area of $\triangle S^{\prime} B S=8$

$\Rightarrow \quad \frac{1}{2} \cdot 2 a e \cdot b=8$

$b^{2}=8$ ….(2)

From eq $^{n}$ (1)

$a^{2} e^{2}=8$

Also,

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

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

$\Rightarrow 8=a^{2}-8$

$\Rightarrow a^{2}=16$

Hence, required length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2(8)}{4}$

$=4$ units

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