An ellipse frac{x^2}{a^2}+frac{y^2}{b^2}=1, a>b and parabola x^2=4(y+b) are such that two foci of

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10-2. Parabola, Ellipse, Hyperbola

normal

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

$\frac{1}{\sqrt{13}}$

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

$\frac{1}{\sqrt{11}}$

$-\frac{2}{\sqrt{11}}$

(KVPY-2017)

Solution

(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

$A B=C D \Rightarrow 2 a e=4 \Rightarrow a e=2$

$B C=C D \Rightarrow B C^2=C D^2$

$(2-a e)^2+(1-b)^2=4^2$

$0+(1-b)^2=4^2$

$1-b=\pm 4$

$b=5, b=-3 \Rightarrow a=\sqrt{29}$ or $\sqrt{13}$

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

$e=\sqrt{1-\frac{9}{13}}=\frac{2}{\sqrt{13}}$ or $2$

Standard 11

Mathematics

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(JEE MAIN-2023)

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The eccentricity of an ellipse whose length of latus rectum is equal to distance between its foci, is

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If $ \tan\ \theta _1. tan \theta _2 $ $= -\frac{{{a^2}}}{{{b^2}}}$ then the chord joining two points $\theta _1 \& \theta _2$ on the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}$ $= 1$ will subtend a right angle at :

normal

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

Solution

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