If a tangent to the ellipse x^{2}+4 y^{2}=4 m | vedclass

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

Question

$\frac{x^{2}}{4}+\frac{y^{2}}{1}=1 \Rightarrow a=2 \& b=1, \quad$ Let $P(2 \cos 	heta, \sin 	heta)$

Equation of tangent at $P$ is $(\cos 	he

Vedclass · Accepted Answer

$(\sqrt{3}, 0)$

Vedclass · Answer

$\frac{x^{2}}{4}+\frac{y^{2}}{1}=1 \Rightarrow a=2 \& b=1, \quad$ Let $P(2 \cos 	heta, \sin 	heta)$

Equation of tangent at $P$ is $(\cos 	heta) x+2 \sin 	heta y=2$

$\mathrm{B}\left(-2, \frac{1+\cos 	heta}{\sin 	heta}ight)$

$C\left(2, \frac{1-\cos 	heta}{\sin 	heta}ight)$

$B\left(-2, \cot \frac{	heta}{2}ight)$

$C\left(2, 	an \frac{	heta}{2}ight)$

Equation of circle is

$(x+2)(x-2)+\left(y-\cot \frac{	heta}{2}ight)\left(y-	an \frac{	heta}{2}ight)=0$

$x^{2}-4+y^{2}-\left(	an \frac{	heta}{2}+\cot \frac{	heta}{2}ight) y+1=0$

so, $(\sqrt{3}, 0)$ satisfying equation $(1)$

If a tangent to the ellipse $x^{2}+4 y^{2}=4$ meets the tangents at the extremities of its major axis at $\mathrm{B}$ and $\mathrm{C}$, then the circle with $\mathrm{BC}$ as diameter passes through the point:

Similar Questions

Tangent is drawn to ellipse $\frac{{{x^2}}}{{27}} + {y^2} = 1\,at\,(3\sqrt 3 \cos \theta ,\sin \theta )$ where $\theta \in (0, \pi /2)$ . Then the value of $\theta$ such that sum of intercepts on axes made by this tangent is minimum, is

The equation of an ellipse whose eccentricity is $1/2$ and the vertices are $(4, 0)$ and $(10, 0)$ is

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

In an ellipse $9{x^2} + 5{y^2} = 45$, the distance between the foci is

The locus of point of intersection of two perpendicular tangent of the ellipse $\frac{{{x^2}}}{{{9}}} + \frac{{{y^2}}}{{{4}}} = 1$ is :-