An ellipse frac{x^2}{a^2}+frac{y^2}{b^2}=1, a | vedclass

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Question

(b)

Equation of ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a > b$

Equation of parabola $x^2=4(y+b)$

Foci of ellipse $(\pm a e, 0)$.

Vedclass · Accepted Answer

$\frac{2}{\sqrt{13}}$

Vedclass · Answer

(b)

Equation of ellipse

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a > b$

Equation of parabola $x^2=4(y+b)$

Foci of ellipse $(\pm a e, 0)$.

End of latusrectum of parabola

$=(\pm 2,1-b)$

$A B C D$ is a square

$A B=C D \Rightarrow 2 a e=4 \Rightarrow a e=2$

$B C=C D \Rightarrow B C^2=C D^2$

$(2-a e)^2+(1-b)^2=4^2$

$(2-a e)^2+(1-b)^2=4^2$

$0+(1-b)^2=4^2$

$1-b=\pm 4$

$b=5, b=-3 \Rightarrow a=\sqrt{29}$ or $\sqrt{13}$

$e=\sqrt{1-\frac{b^2}{a^2}}$

$e=\sqrt{1-\frac{9}{13}}=\frac{2}{\sqrt{13}}$ or $2$

An ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ and the parabola $x^2=4(y+b)$ are such that the two foci of the ellipse and the end points of the latusrectum of parabola are the vertices of a square. The eccentricity of the ellipse is

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