A hyperbola, having the transverse axis of le | vedclass

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

Question

The given ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$

$\Rightarrow \mathrm{a}=2, \mathrm{~b}=\sqrt{3} \Rightarrow 3=4\left(1-\mathrm{e}^2\right) \Ri

Vedclass · Accepted Answer

$x^2 \operatorname{cosec}^2 	heta-y^2 \sec ^2 	heta=1$

Vedclass · Answer

The given ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$

$\Rightarrow \mathrm{a}=2, \mathrm{~b}=\sqrt{3} \Rightarrow 3=4\left(1-\mathrm{e}^2ight) \Rightarrow \mathrm{e}=\frac{1}{2}$

so that $\mathrm{ae}=1$

Hence the eccentricity $e_1$, of the hyperbola is given by

$ 1=e_1 \sin 	heta \Rightarrow e_1=\operatorname{cosec} 	heta $

$ \Rightarrow b^2=\sin ^2 	heta\left(\operatorname{cosec}^2 	heta-1ight)=\cos ^2 	heta$

Hence the hyperbola is $\frac{x^2}{\sin ^2 	heta}-\frac{y^2}{\cos ^2 	heta}=1$ or $x^2 \operatorname{cosec}^2 	heta-y^2 \sec ^2 	heta=1$

A hyperbola, having the transverse axis of length $2 \sin \theta$, is confocal with the ellipse $3 x^2+4 y^2=12$. Then its equation is

Similar Questions

If the line $y = 2x + \lambda $ be a tangent to the hyperbola $36{x^2} - 25{y^2} = 3600$, then $\lambda = $

If $A$ and $B$ are the points of intersection of the circle $x^2+y^2-8 x=0$ and the hyperbola $\frac{x^2}{9}-\frac{y^2}{4}=1$ and $a$ point $P$ moves on the line $2 x-3 y+4=0$, then the centroid of $\triangle P A B$ lies on the line :

Eccentricity of rectangular hyperbola is

If $ P(x_1, y_1), Q(x_2, y_2), R(x_3, y_3) $ and $ S(x_4, y_4) $ are $4 $ concyclic points on the rectangular hyperbola $x y = c^2$ , the co-ordinates of the orthocentre of the triangle $ PQR$ are :