The acute angle between the pair of tangents | vedclass

Name: PDF Generator App | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

Equation of tangent to the ellipse $2 x ^{2}+3 y ^{2}=5$ is

$y = mx \pm \sqrt{\frac{5}{2} m ^{2}+\frac{5}{3}}$

It pass through $(1,3)$

$3=m \

Vedclass · Accepted Answer

$	an ^{-1}\left(\frac{24}{7 \sqrt{5}}ight)$

Vedclass · Answer

Equation of tangent to the ellipse $2 x ^{2}+3 y ^{2}=5$ is

$y = mx \pm \sqrt{\frac{5}{2} m ^{2}+\frac{5}{3}}$

It pass through $(1,3)$

$3=m \pm \sqrt{\frac{5}{2} m^{2}+\frac{5}{3}}$

$3\,m^{2}+12\,m-\frac{44}{3}=0$

Let $	heta$ be the angle between the tangents

$	an 	heta=\left|\frac{m_{1}-m_{2}}{1+m_{1} m_{2}}ight|$

$	an 	heta=\left|\frac{3 \sqrt{320}}{-35}ight|$

$	heta=	an ^{-1}\left(\frac{24}{7 \sqrt{5}}ight)$

The acute angle between the pair of tangents drawn to the ellipse $2 x^{2}+3 y^{2}=5$ from the point $(1,3)$ is.

Similar Questions

Let $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1, a>b$ be an ellipse, whose eccentricity is $\frac{1}{\sqrt{2}}$ and the length of the latus rectum is $\sqrt{14}$. Then the square of the eccentricity of $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is :

The eccentricity of the ellipse $4{x^2} + 9{y^2} + 8x + 36y + 4 = 0$ is

The equation of the ellipse whose vertices are $( \pm 5,\;0)$ and foci are $( \pm 4,\;0)$ is

The eccentricity of the ellipse ${\left( {\frac{{x - 3}}{y}} \right)^2} + {\left( {1 - \frac{4}{y}} \right)^2} = \frac{1}{9}$ is

The length of the latus rectum of an ellipse is $\frac{1}{3}$ of the major axis. Its eccentricity is