If the vertices and foci of a hyperbola are r | vedclass

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

Question

(D) Given that the vertices are $(\pm a, 0) = (\pm 3, 0)$,so $a = 3$. Given that the foci are $(\pm ae, 0) = (\pm 4, 0)$,so $ae = 4$. Substituting $a=

Vedclass · Accepted Answer

$x=3 \sec 	heta, y=\sqrt{7} 	an 	heta$

Vedclass · Answer

(D) Given that the vertices are $(\pm a, 0) = (\pm 3, 0)$,so $a = 3$. Given that the foci are $(\pm ae, 0) = (\pm 4, 0)$,so $ae = 4$. Substituting $a=3$,we get $3e = 4$,which implies $e = \frac{4}{3}$. We know the relation $b^2 = a^2(e^2 - 1)$. Substituting the values,$b^2 = 3^2 \left( (\frac{4}{3})^2 - 1 ight) = 9 \left( \frac{16}{9} - 1 ight) = 16 - 9 = 7$. Thus,$b = \sqrt{7}$. The standard equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,which is $\frac{x^2}{9} - \frac{y^2}{7} = 1$. The parametric equations for a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $x = a \sec 	heta$ and $y = b 	an 	heta$. Substituting $a=3$ and $b=\sqrt{7}$,we get $x = 3 \sec 	heta$ and $y = \sqrt{7} 	an 	heta$.

If the vertices and foci of a hyperbola are respectively $(\pm 3,0)$ and $(\pm 4,0)$,then the parametric equations of that hyperbola are:

Similar Questions

Let the tangent drawn to the parabola $y^2 = 24x$ at the point $(\alpha, \beta)$ be perpendicular to the line $2x + 2y = 5$. Then the normal to the hyperbola $\frac{x^2}{\alpha^2} - \frac{y^2}{\beta^2} = 1$ at the point $(\alpha + 4, \beta + 4)$ does $NOT$ pass through which of the following points?

If the normal drawn to the hyperbola $xy=16$ at $(8,2)$ meets the hyperbola again at a point $(\alpha, \beta)$,then $|\beta|+\frac{1}{|\alpha|}=$

Let the eccentricity $e$ of a hyperbola satisfy the equation $6e^2 - 11e + 3 = 0$. If the foci of the hyperbola are $(3, 5)$ and $(3, -4)$, then the length of its latus rectum is:

If the normals drawn to the hyperbola $xy=4$ at $(\alpha_i, \beta_i)$ for $i=1, 2, 3, 4$ are concurrent at the point $(a, b)$,then $\frac{(\alpha_1+\alpha_2+\alpha_3+\alpha_4)}{(\beta_1+\beta_2+\beta_3+\beta_4)}(\alpha_1 \alpha_2 \alpha_3 \alpha_4) =$

The combined equation of the asymptotes of the hyperbola $2x^2 + 5xy + 2y^2 + 4x + 5y = 0$ is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module