The equation of the normal at the point $(0, 3)$ of the ellipse $9x^2 + 5y^2 = 45$ is

If the line $x \cos \alpha + y \sin \alpha = p$ is normal to the ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$,then

An ellipse $\frac{(x-x_0)^2}{a^2} + \frac{(y-y_0)^2}{b^2} = 1$ with $a > b$ is tangent to both the $x$ and $y$ axes and is located in the first quadrant. Let $F_1$ and $F_2$ be the two foci of the ellipse and $O$ be the origin such that $OF_1 < OF_2$. Suppose the triangle $OF_1F_2$ is an isosceles triangle with $\angle OF_1F_2 = 120^{\circ}$. Then the eccentricity of the ellipse is:

The sum of the focal distances of any point on the conic $\frac{x^2}{25} + \frac{y^2}{16} = 1$ is

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 $(6, 2)$.

The equation of the tangent to the ellipse $\frac{x^2}{4} + \frac{y^2}{12} = 1$ at the point $(1, 3)$ is: