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Q)

Find the equation of the ellipse referred to its centre whose minor axis is equal to the distance between the foci and whose latus is 10?

$\begin{array}{1 1}(a)\;\large\frac{x^2}{100}+\frac{y^2}{50}\normalsize=1\\(b)\;\large\frac{x^2}{10}+\frac{y^2}{5}\normalsize=1\\(c)\;\large\frac{x^2}{200}+\frac{y^2}{100}\normalsize=1\\(d)\;\text{None of these}\end{array}$

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A)
Let the equation of the ellipse is $\large\frac{x^2}{a^2}+\frac{y^2}{b^2}$$=1\qquad a > b$
Then the foci are $S(ae,0)$ and $S'(-ae,0)$
Length of minor axes =2b
Length of lactus rectum =$\large\frac{2b^2}{a}$
According to the question
$BB'=SS'$
$2b=2ae$
$b=ae$
$\large\frac{2b^2}{a}$$=10$
$b^2=5a$
$b^2=a^2(1-e^2)$
$a^2e^2=a^2(1-e^2)$
$e=\large\frac{1}{\sqrt 2}$
$b=\large\frac{a}{\sqrt 2}$
$b^2=\large\frac{a^2}{2}$
$5a=\large\frac{a^2}{2}$
$a=10$
$b^2=5\times 10=50$
Hence equation of ellipse
$\large\frac{x^2}{100}+\frac{y^2}{50}$$=1$
Hence (a) is the correct answer.
 
Comment
A)
nate Geometry Next similar questionQ) Find the equation of the ellipse referred to its centre whose minor axis is equal to the distance between the foci and whose latus is 10? (a)x2100+y250=1(b)x210+y25=1(c)x2200+y2100=1(d)None of these Your answer Email me (dattawakde7507@gmail.com) if my answer is selected or commented on Homework help Revision notes Donate A) Let the equation of the ellipse is x2a2+y2b2=1a>b Then the foci are S(ae,0) and S′(−ae,0) Length of minor axes =2b Length of lactus rectum =2b2a According to the question BB′=SS′ 2b=2ae b=ae 2b2a=10 b2=5a b2=a2(1−e2) a2e2=a2(1−e2) e=12–√ b=a2–√ b2=a22 5a=a22 a=10 b2=5×10=50 Hence equation of ellipse x2100+y250=1 Hence (a) is the
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