An ellipse with its minor and major axis parallel to coordinate axes passes through (0,0),(1,0) and

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10-2. Parabola, Ellipse, Hyperbola

normal

An ellipse with its minor and major axis parallel to the coordinate axes passes through $(0,0),(1,0)$ and $(0,2)$. One of its foci lies on the $Y$-axis. The eccentricity of the ellipse is

$\sqrt{3}-1$

$\sqrt{5}-2$

$\sqrt{2}-1$

$\frac{\sqrt{3}-1}{2}$

(KVPY-2017)

Solution

(c)

An ellipse passes through $(0,0)$, $(1,0)$ and $(2,0)$. Its minor and major axis are parallel to coordinates axes. One of its foci lies on $Y$-axis.

Let one foci is $(0, k)$.

$\therefore O B$ is the latusrectum of ellipse $F_1(0, k)$ is mid-point of $O B$

$F_1(0,1) \quad\left[\because k=\frac{2+0}{2}=1\right]$

Now $F_2(h, k)=F_2(h, 1)$

Now by definition of ellipse

$B F_1+B F_2=2 a =O F_1+O F_2$

$\therefore \quad B F_1+B F_2 =O F_1+O F_24$

$\sqrt{1+1}+\sqrt{(h-1)^2+1} =\sqrt{0+1}+\sqrt{h^2+1}$

$\sqrt{2}+\sqrt{(h-1)^2+1} =1+\sqrt{h^2+1}$

$\Rightarrow \quad h =1$

$\therefore F_1(0,1) \text { and } F_2(1,1)$

$F_1 F_2=2 a e =1$

$\therefore \quad O F_1+O F_2=2 a =1+\sqrt{2}$

$\Rightarrow \quad e=\frac{1}{\sqrt{2}+1} \Rightarrow e =\frac{\sqrt{2}-1}{2-1}$

$e =\sqrt{2}-1$

Standard 11

Mathematics

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(b < a)$, be a ellipse with major axis $A B$ and minor axis $C D$. Let $F_1$ and $F_2$ be its two foci, with $A, F_1, F_2, B$ in that order on the segment $A B$. Suppose $\angle F_1 C B=90^{\circ}$. The eccentricity of the ellipse is

hard

(KVPY-2020)

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medium

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normal

Let $S$ and $S\,'$ be the foci of an ellipse and $B$ be any one of the extremities of its minor axis. If $\Delta S\,'BS$ is a right angled triangle with right angle at $B$ and area $(\Delta S\,'BS) = 8\,sq.$ units, then the length of a latus rectum of the ellipse is

hard

(JEE MAIN-2019)

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hard

(JEE MAIN-2024)

An ellipse with its minor and major axis parallel to the coordinate axes passes through $(0,0),(1,0)$ and $(0,2)$. One of its foci lies on the $Y$-axis. The eccentricity of the ellipse is

Solution

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