Let the equations of two ellipses be {E_ | vedclass

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

Question

Given equationd of ellipses

${E_1}:\frac{{{x^2}}}{3} + \frac{{{y^2}}}{2} = 1$

$ \Rightarrow {e_1} = \sqrt {1 - \frac{2}{3}}  = \frac{1}{{\s

Vedclass · Accepted Answer

$4$

Vedclass · Answer

Given equationd of ellipses

${E_1}:\frac{{{x^2}}}{3} + \frac{{{y^2}}}{2} = 1$

$ \Rightarrow {e_1} = \sqrt {1 - \frac{2}{3}}  = \frac{1}{{\sqrt 3 }}$

and ${E_2}:\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{{b^2}}} = 1$

$ \Rightarrow {e_1} = \sqrt {\frac{{1 - {b^2}}}{{16}}}  = \sqrt {\frac{{16 - {b^2}}}{4}} $

Also, given ${e_1} 	imes {e_2} = \frac{1}{2}$

$ \Rightarrow \frac{1}{{\sqrt 3 }} 	imes \sqrt {\frac{{16 - {b^2}}}{4}}  = \frac{1}{2} \Rightarrow 16 - {b^2} = 12$

$ \Rightarrow {b^2} = 4$

$	herefore $ Length of minor axis of 

${E_2} = 2b = 2 	imes 2 = 4$

Let the equations of two ellipses be ${E_1}:\,\frac{{{x^2}}}{3} + \frac{{{y^2}}}{2} = 1$ and ${E_2}:\,\frac{{{x^2}}}{16} + \frac{{{y^2}}}{b^2} = 1,$ If the product of their eccentricities is $\frac {1}{2},$ then the length of the minor axis of ellipse $E_2$ is

Similar Questions

The foci of $16{x^2} + 25{y^2} = 400$ are

Let a tangent to the Curve $9 x^2+16 y^2=144$ intersect the coordinate axes at the points $A$ and $B$. Then, the minimum length of the line segment $A B$ is $.........$

The pole of the straight line $x + 4y = 4$ with respect to ellipse ${x^2} + 4{y^2} = 4$ is

Let $P$ be a variable point on the ellipse $x^2 + 3y^2 = 3$ , then the maximum perpendicular distance of $P$ from the line $x -y = 10$ is