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  • The equation of the ellipse referred to its centre whose minor axis is equal to the distance between the foci and whose latus rectum is 10 is <img alt="\mathrm{m x^2+n y^2=100}" src="https://entran

The equation of the ellipse referred to its centre whose minor axis is equal to the distance between the foci and whose latus rectum is 10 is \mathrm{m x^2+n y^2=100}, when \mathrm{m, n \in N}, then the value of \mathrm{10^m+20^n} must be

Option: 1

400


Option: 2

410


Option: 3

310


Option: 4

210


Answers (1)

best_answer

Let the equation of the ellipse is

\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1(\text { let } a>b) }

Then, the foci are \mathrm{S(a e, 0) \, \, and \, \, S^{\prime}(-a e, 0)} length of minor

axis \mathrm{ B B^{\prime}=2 b} and length of latus rectum \mathrm{ =\frac{2 b^2}{a}}

\mathrm{\therefore \quad \text { According to the question } B B^{\prime}=S S^{\prime}}

\mathrm{\begin{aligned} & \Rightarrow \quad 2 b=2 a e \\ & \Rightarrow \quad b=a e\, \, \, \, \, \, \, \, \, \, \, ....(i) \\ & \quad \text { and } \frac{2 b^2}{a}=10 \\ & \Rightarrow \quad b^2=5 a \end{aligned}}

\mathrm{Also, we \, \, have \, \, b^2=a^2\left(1-e^2\right)\\ }                 ....(ii)

putting the value of b from equation (i) in equation (ii) we have

\mathrm{\begin{aligned} & a^2 e^2=a^2\left(1-e^2\right) \\\\ & \Rightarrow \quad e^2=1-e^2 \\\\ & \Rightarrow \quad 2 e^2=1 \\\\ & \therefore \quad e=\frac{1}{\sqrt{2}} \end{aligned}}

From equation (i) we have 

\mathrm{\begin{aligned} &\begin{aligned} & b=\frac{a}{\sqrt{2}} \\ \therefore & b^2=\frac{a^2}{2} \\ \Rightarrow & 5 a=\frac{a^2}{2} \\ \Rightarrow & a=10 \\ \therefore & b^2=5 \times 10=50 \end{aligned}\\ \end{aligned}}                             

Putting the value of a and b in\mathrm{ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1}, the equation of required ellipse is

\mathrm{ \frac{x^2}{100}+\frac{y^2}{50}=1 \text {, } }

\mathrm{ \begin{aligned} & x^2+2 y^2=100 \\ & \therefore \quad m=1, n=2 \end{aligned} }
Then, 

\mathrm{10^m+20^n=10+400=410}

Posted by

sudhir.kumar

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