Find the equation for the ellipse that satisfies the given conditions: Ends of major axis $(±3,\,0)$ ends of minor axis $(0,\,±2)$
Ends of major axis $(±3,\,0)$ , ends of minor axis $(0,\,±2)$
Here, the major axis is along the $x-$ axis.
Therefore, the equation of the ellipse will be of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ where a is the semi major axis.
Accordingly, $a=3$ and $b=2$
Thus, the equation of the ellipse is $\frac{x^{2}}{3^{2}}+\frac{y^{2}}{2^{2}}=1$ or $\frac{x^{2}}{9}+\frac{y^{2}}{4}=1$
If $\alpha $ and $\beta $ are the eccentric angles of the extremities of a focal chord of an ellipse, then the eccentricity of the ellipse is
The product of the lengths of perpendiculars from the foci on any tangent to the ellipse $3x^2 + 5y^2 = 1$, is
If the distance between a focus and corresponding directrix of an ellipse be $8$ and the eccentricity be $1/2$, then length of the minor axis is
A tangent having slope of $-\frac{4}{3}$ to the ellipse $\frac{{{x^2}}}{{18}}$ + $\frac{{{y^2}}}{{32}}$ $= 1$ intersects the major and minor axes in points $A$ and $ B$ respectively. If $C$ is the centre of the ellipse then the area of the triangle $ ABC$ is : .............. $\mathrm{sq. \,units}$
The equation of an ellipse, whose vertices are $(2, -2), (2, 4)$ and eccentricity $\frac{1}{3}$, is