Tangent and Normal to an Ellipse

The equation of the normal at the point $(2,3)$ to the ellipse $9x^2+16y^2=180$ is 

Equations of the common tangent of the ellipse $\frac{(x+1{)}^2}{2^2}+\frac{(y-1{)}^2}{3^2}=1$ and the circle $(x+1{)}^2+(y-1{)}^2=4$ are

The product of the perpendiculars drawn from the foci of the ellipse $\frac{x^2}{9}+\frac{y^2}{25}=1$ upon the tangent to it at the point $\left(\frac{3}{2},\frac{5\sqrt{3}}{2}\right)$, is

Equation of the ellipse with axes coinciding the co-ordinate axes touches the straight lines $3\mathrm{x}\mathrm{ }-\mathrm{ }2\mathrm{y}\mathrm{ }-\mathrm{ }20=0$ and $\mathrm{x}\mathrm{ }+\mathrm{ }6\mathrm{y}\mathrm{ }-\mathrm{ }20=0$ , is:

If the normal at an end of a latus rectum of an ellipse passes through one extremity of the minor axis, then the eccentricity of the ellipse is given by

The equation of the tangents to the ellipse $9{\mathrm{x}}^2+\mathrm{ }16{\mathrm{y}}^2=144$ from the point $\left(2,\mathrm{ }3\right)$ are -

Equation of the ellipse with axes coinciding the co-ordinate axes touches the straight lines $3\mathrm{x}\mathrm{ }-\mathrm{ }2\mathrm{y}\mathrm{ }-\mathrm{ }20=0$ and $\mathrm{x}\mathrm{ }+\mathrm{ }6\mathrm{y}\mathrm{ }-\mathrm{ }20=0$ , is:

Equation of tangents to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$, which are perpendicular to the line $3x+4y=7$, are

The area (in sq. units) of the quadrilateral formed by the tangents at the end points of the latus ractum to the ellipse $\frac{x^2}{9}+\frac{y^2}{5}=1,$ is

The length of the minor axis (along $y$-axis) of an ellipse in the standard form is $\frac{4}{\sqrt{3}}$. If this ellipse touches the line, $x+6y=8$; then its eccentricity is:

Let the line $y=mx$ and the ellipse $2x^2+y^2=1$ intersect at a point $P$ in the first quadrant. If the normal to this ellipse at $P$ meets the co-ordinate axes at $\left(-\frac{1}{3\sqrt{2}},0\right)$ and $\left(0,\beta \right)$, then $\beta$ is equal to

The equation of the tangent to the ellipse $5x^2+9y^2=45$, and perpendicular to the line $3x+2y+1=0$ is

The product of the perpendicular distances drawn from the points $(3,0)$ and $(-3,0)$ to the tangent of an ellipse $\frac{x^2}{36}+\frac{y^2}{27}=1$ at $\left(3,\frac{9}{2}\right)$ is

For the ellipse $\frac{x^2}{18}+\frac{y^2}{32}=1,$ if a tangent with slope $\frac{-4}{3}$ intersects the major and minor axes at $P$ and $Q$ respectively, find $P$ and $Q$

Find the condition for the line $ax+by+c=0$ to be a normal to an ellipse $\frac{x^2}{4}+\frac{y^2}{36}=1$

Slope of normal to the ellipse at a point$\mathrm{P}$ is $\frac{3}{4}$ and eccentricity of ellipse is $\frac{1}{3}$. If this normal makes acute angle$\prime \mathrm{\beta }\prime$ with its focal chord through $\mathrm{P}$, then $\mathrm{sin\beta }$ is

Let the ellipse, $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b,$ pass through the point $(2,3)$ and have eccentricity equal to $\frac{1}{2}.$ Then equation of the normal to this ellipse at $(2,3)$ is

If a tangent of slope $2$ of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{1}=1$ passes through the point $\left(-\mathrm{2,0}\right)$, then the value of $a^2$ is equal to

Prove that the minimum length of the intercept made by the axes on the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ is equal to $a+b$.

Tangent at a point $P$ of ellipse $9x^2+16y^2-144=0$ having eccentric angle  $\theta =\frac{1}{2}{\sin }^{-1}\left(\frac{1}{7}\right)$ is drawn. Calculate $\angle PON$ if $N$ is the foot of perpendicular from centre $O$ of the ellipse to this tangent :