The coordinates of any point,in the parametric form,on the ellipse whose foci are $(-2,0)$ and $(8,0)$ and eccentricity is $\frac{1}{\sqrt{2}}$,is

  • A
    $(5 \sqrt{2} \cos \theta, 5 \sin \theta)$
  • B
    $(3+5 \sqrt{2} \cos \theta, 5 \sin \theta)$
  • C
    $(3+5 \cos \theta, 5 \sqrt{2} \sin \theta)$
  • D
    $(5 \cos \theta, 3+5 \sqrt{2} \sin \theta)$

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Similar Questions

If the length of the minor axis of an ellipse is equal to half of the distance between the foci,then the eccentricity of the ellipse is:

Consider the ellipse $\frac{x^2}{4}+\frac{y^2}{3}=1$. Let $H(\alpha, 0)$,$0 < \alpha < 2$,be a point. $A$ straight line drawn through $H$ parallel to the $y$-axis crosses the ellipse and its auxiliary circle at points $E$ and $F$ respectively,in the first quadrant. The tangent to the ellipse at the point $E$ intersects the positive $x$-axis at a point $G$. Suppose the straight line joining $F$ and the origin makes an angle $\phi$ with the positive $x$-axis.
$List-I$ $List-II$
$(I)$ If $\phi=\frac{\pi}{4}$,then the area of the triangle $FGH$ is $(P) \frac{(\sqrt{3}-1)^4}{8}$
$(II)$ If $\phi=\frac{\pi}{3}$,then the area of the triangle $FGH$ is $(Q) 1$
$(III)$ If $\phi=\frac{\pi}{6}$,then the area of the triangle $FGH$ is $(R) \frac{3}{4}$
$(IV)$ If $\phi=\frac{\pi}{12}$,then the area of the triangle $FGH$ is $(S) \frac{1}{2\sqrt{3}}$
  $(T) \frac{3\sqrt{3}}{2}$

The correct option is:

$A$ tangent to the ellipse $\frac{x^2}{25}+\frac{y^2}{16}=1$ intersects the coordinate axes at $A$ and $B$. Then the locus of the circumcentre of triangle $AOB$ (where $O$ is the origin) is:

Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $(\pm 3, 0)$,ends of minor axis $(0, \pm 2)$.

The length of the latus rectum of an ellipse is $\frac{18}{5}$ and eccentricity is $\frac{4}{5}$,then the equation of the ellipse is...

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