A hyperbola has its transverse axis along the | vedclass

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

Question

(C) The given conic is $\frac{x^2}{3} + \frac{y^2}{4} = 4$,which can be rewritten as $\frac{x^2}{12} + \frac{y^2}{16} = 1$.Here,$a^2 = 12$ and $b^2 =

Vedclass · Accepted Answer

$\left( 5, 2\sqrt{3} \right)$

Vedclass · Answer

(C) The given conic is $\frac{x^2}{3} + \frac{y^2}{4} = 4$,which can be rewritten as $\frac{x^2}{12} + \frac{y^2}{16} = 1$.Here,$a^2 = 12$ and $b^2 = 16$. Since $b^2 > a^2$,the major axis is along the $y$-axis.The eccentricity of the ellipse is $e = \sqrt{1 - \frac{12}{16}} = \sqrt{\frac{4}{16}} = \frac{1}{2}$.The foci of the ellipse are $(0, \pm be) = (0, \pm 4 	imes \frac{1}{2}) = (0, \pm 2)$.The hyperbola has its vertices at $(0, \pm 2)$,so its transverse axis is along the $y$-axis and $a = 2$.The equation of the hyperbola is $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,where $a = 2$.Given the eccentricity $e_h = \frac{3}{2}$,we use $b^2 = a^2(e_h^2 - 1) = 4(\frac{9}{4} - 1) = 4(\frac{5}{4}) = 5$.The equation of the hyperbola is $\frac{y^2}{4} - \frac{x^2}{5} = 1$.Checking the points:For $(0, 2)$: $\frac{4}{4} - 0 = 1$ (Lies on it).For $(\sqrt{5}, 2\sqrt{2})$: $\frac{8}{4} - \frac{5}{5} = 2 - 1 = 1$ (Lies on it).For $(\sqrt{10}, 2\sqrt{3})$: $\frac{12}{4} - \frac{10}{5} = 3 - 2 = 1$ (Lies on it).For $(5, 2\sqrt{3})$: $\frac{12}{4} - \frac{25}{5} = 3 - 5 = -2 
eq 1$ (Does not lie on it).

$A$ hyperbola has its transverse axis along the major axis of the conic $\frac{x^2}{3} + \frac{y^2}{4} = 4$ and its vertices at the foci of this conic. If the eccentricity of the hyperbola is $\frac{3}{2}$,then which of the following points does $NOT$ lie on it?

Similar Questions

The point of intersection of two tangents drawn to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{4} = 1$ lies on the circle $x^2 + y^2 = 5$. If these tangents are perpendicular to each other,then $a =$

If $(0, \pm 4)$ and $(0, \pm 2)$ are the foci and vertices of a hyperbola,respectively,then its equation is:

The equation of the hyperbola with asymptotes $3x - 4y + 7 = 0$ and $4x + 3y + 1 = 0$ and which passes through the origin is

The difference between the focal distances of any point on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is $6$. If $(\sqrt{13}, k)$ is an endpoint of a latus rectum of this hyperbola,then $k=$

If the product of the lengths of the perpendiculars from any point on the hyperbola $16x^2 - 25y^2 = 400$ to its asymptotes is $p$ and the angle between the two asymptotes is $\theta$,then $p \tan \frac{\theta}{2} =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module