Find the equation for the ellipse that satisfies the given conditions: Major axis on the $x-$ axis and passes through the points $(4,\,3)$ and $(6,\,2)$
since the major axis is on the $x-$ axis, the equation of the ellipse will be of the form
$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ .......... $(1)$
Where, a is the semi-major axis
The ellipse passes through points $(4,\,3)$ and $(6,\,2)$ . Hence,
$\frac{16}{a^{2}}+\frac{9}{b^{2}}=1$ .......... $(2)$
$\frac{36}{a^{2}}+\frac{4}{b^{2}}=1$ .......... $(3)$
On solving equations $(2)$ and $(3),$ we obtain $a^{2}=52$ and $b^{2}=13$
Thus, the equation of the ellipse is $\frac{x^{2}}{52}+\frac{y^{2}}{13}=1$ or $x^{2}+4 y^{2}=52$
Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(b < a)$, be a ellipse with major axis $A B$ and minor axis $C D$. Let $F_1$ and $F_2$ be its two foci, with $A, F_1, F_2, B$ in that order on the segment $A B$. Suppose $\angle F_1 C B=90^{\circ}$. The eccentricity of the ellipse is
Let the tangents at the points $P$ and $Q$ on the ellipse $\frac{x^{2}}{2}+\frac{y^{2}}{4}=1$ meet at the point $R(\sqrt{2}, 2 \sqrt{2}-2)$. If $S$ is the focus of the ellipse on its negative major axis, then $SP ^{2}+ SQ ^{2}$ is equal to.
If lines $3x + 2y = 10$ and $-3x + 2y = 10$ are tangents at the extremities of latus rectum of an ellipse whose centre is origin, then the length of latus rectum of ellipse is
The eccentricity of the ellipse ${\left( {\frac{{x - 3}}{y}} \right)^2} + {\left( {1 - \frac{4}{y}} \right)^2} = \frac{1}{9}$ is
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 $4 x ^{2}+9 y ^{2}=36$