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  • The equation of the ellipse whose axes are along the coordinate axes, vertices are <img alt="\mathrm{( \pm 5,0)}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B%28%20%5Cpm%205%2C

The equation of the ellipse whose axes are along the coordinate axes, vertices are \mathrm{( \pm 5,0)} and foci at \mathrm{( \pm 4,0)}, is \mathrm{\alpha x^2+\beta y^2=225}, then value of \mathrm{(\alpha+\beta)} is

Option: 1

32


Option: 2

34


Option: 3

37


Option: 4

38


Answers (1)

best_answer

Let the equation of the required ellipse be

\frac{x^2}{a^2}+\frac{y^2}{b^2}=1                   ...(i)

The coordinates of its vertices and foci are ( \pm a, 0) and ( \pm a e, 0) respectively.

\therefore \quad a=5 \text { and } a e=4

\Rightarrow e=\frac{4}{5}

Now, b^2=a^2\left(1-e^2\right)b^2=a^2\left(1-e^2\right)

\Rightarrow b^2=25\left(1-\frac{16}{25}\right)=9

Substituting the values of a^2 and b^2 in (i), we get

\begin{aligned} & \frac{x^2}{25}+\frac{y^2}{9}=1,9 x^2+25 y^2=225 \\ & \Rightarrow \alpha+\beta=34 \end{aligned}

Posted by

Sanket Gandhi

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