The eccentricity of an ellipse having centre | vedclass

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

Question

Centre at $(0,0)$

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

at point $(4,-1)$

$\frac{{16}}{{{a^2}}} + \frac{1}{{{b^2}}} = 1$

Vedclass · Accepted Answer

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

Vedclass · Answer

Centre at $(0,0)$

$\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$

at point $(4,-1)$

$\frac{{16}}{{{a^2}}} + \frac{1}{{{b^2}}} = 1$

$16{b^2} + {a^2} = {a^2}{b^2}\,\,\,\,\,\,\,\,.......\left( i \right)$ at point $(-2,2)$

$\frac{4}{{{a^2}}} + \frac{4}{{{b^2}}} = 1$

$4{b^2} + {a^2} = {a^2}{b^2}\,\,\,\,\,\,\,\,.......\left( {ii} \right)$

$ \Rightarrow 16{b^2} + {a^2} = 4{b^2} + {a^2}$

From equation $(i)$ and $(ii)$

$ \Rightarrow 3{a^2} = 12{b^2}$

$ \Rightarrow \boxed{{a^2} = 4{b^2}}$

${b^2} = {a^2}\left( {1 - {e^2}} \right)$

${e^2} = \frac{3}{4}$

$e = \frac{{\sqrt 3 }}{2}$

The eccentricity of an ellipse having centre at the origin, axes along the co-ordinate axes and passing through the points $(4,-1)$ and $(-2, 2)$ is

Similar Questions

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{16}+\frac {y^2} {9}=1$.

Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse $9 x^{2}+4 y^{2}=36$.

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 :

The eccentricity of the ellipse $9{x^2} + 5{y^2} - 30y = 0$, is

The minimum area of a triangle formed by any tangent to the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{{81}} = 1$ and the coordinate axes is