An ellipse inscribed in a semi-circle touches | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(d)

Let equation of ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$	herefore$ Equation of circle is

$x^2+(y+b)^2=r^2$

Put $x^2=a^2-\f

Vedclass · Accepted Answer

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

Vedclass · Answer

(d)

Let equation of ellipse is

$\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$	herefore$ Equation of circle is

$x^2+(y+b)^2=r^2$

Put $x^2=a^2-\frac{a^2 y^2}{b^2}$ in circle

$a^2-\frac{a^2 y^2}{b^2}+(y+b)^2=r^2$

$\Rightarrow\left(1-\frac{a^2}{b^2}ight) y^2+2 b y+\left(a^2+b^2-r^2ight)=0$

$D=0 \Rightarrow r^2=\frac{a^4}{a^2-b^2} \Rightarrow b=a \sqrt{1-\frac{a^2}{r^2}}$

Area of ellipse $=\pi a b$

$A=\pi a^2 \sqrt{1-\frac{a^2}{r^2}}$

$\frac{d A}{d a}=0 \Rightarrow a^2=\frac{2 r^2}{3} \Rightarrow a=\sqrt{\frac{2}{3} r}$

$	herefore \quad b=a \sqrt{1-\frac{2}{3}}=\frac{a}{\sqrt{3}}$

$\Rightarrow \quad e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{1}{3}}=\sqrt{\frac{2}{3}}$

An ellipse inscribed in a semi-circle touches the circular arc at two distinct points and also touches the bounding diameter. Its major axis is parallel to the bounding diameter. When the ellipse has the maximum possible area, its eccentricity is

Similar Questions

Let $'E'$ be the ellipse $\frac{{{x^2}}}{9}$$+$$\frac{{{y^2}}}{4}$ $= 1$ $\& $ $'C' $ be the circle $x^2 + y^2 = 9.$ Let $P$ $\&$ $Q$ be the points $(1 , 2) $ and $(2, 1)$ respectively. Then :

Latus rectum of ellipse $4{x^2} + 9{y^2} - 8x - 36y + 4 = 0$ is

Find the equation for the ellipse that satisfies the given conditions: Foci $(\pm 3,\,0),\,\, a=4$

If $P \equiv (x,\;y)$, ${F_1} \equiv (3,\;0)$, ${F_2} \equiv ( - 3,\;0)$ and $16{x^2} + 25{y^2} = 400$, then $P{F_1} + P{F_2}$ equals

If $a$ and $c$ are positive real numbers and the ellipse $\frac{{{x^2}}}{{4{c^2}}} + \frac{{{y^2}}}{{{c^2}}} = 1$ has four distinct points in common with the circle $x^2 + y^2 = 9a^2$ , then