Find the equation for the ellipse that satisfies the given conditions: Centre at $(0,\,0),$ major axis on the $y-$ axis and passes through the points $(3,\,2)$ and $(1,\,6)$
since the centre is at $(0,\,0)$ and the major axis is on the $y-$ 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 $(3,\,2)$ and $(1,\,6) .$ Hence,
$\frac{9}{b^{2}}+\frac{4}{a^{2}}=1$ ........... $(2)$
$\frac{1}{b^{2}}+\frac{36}{a^{2}}=1$ ........... $(3)$
On solving equations $(2)$ and $(3),$ we obtain $b^{2}=10$ and $a^{2}=40$.
Thus, the equation of the ellipse is $\frac{x^{2}}{10^{2}}+\frac{y^{2}}{40}=1$ or $4 x^{2}+y^{2}=40$
The equation of the tangent to the ellipse ${x^2} + 16{y^2} = 16$ making an angle of ${60^o}$ with $x$ - axis is
Let the length of the latus rectum of an ellipse with its major axis long $x -$ axis and center at the origin, be $8$. If the distance between the foci of this ellipse is equal to the length of the length of its minor axis, then which one of the following points lies on it?
If the maximum distance of normal to the ellipse $\frac{x^2}{4}+\frac{y^2}{b^2}=1, b < 2$, from the origin is $1$ , then the eccentricity of the ellipse is:
If the area of the auxiliary circle of the ellipse $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1\left( {a > b} \right)$ is twice the area of the ellipse, then the eccentricity of the ellipse is
In a triangle $A B C$ with fixed base $B C$, the vertex $A$ moves such that $\cos B+\cos C=4 \sin ^2 \frac{A}{2} .$ If $a, b$ and $c$ denote the lengths of the sides of the triangle opposite to the angles $A, B$ and $C$, respectively, then
$(A)$ $b+c=4 a$
$(B)$ $b+c=2 a$
$(C)$ locus of point $A$ is an ellipse
$(D)$ locus of point $A$ is a pair of straight lines