The equation of the pair of asymptotes of the | vedclass

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

Question

(D) The given equation of the hyperbola is $4x^2 - 9y^2 - 24x - 36y - 36 = 0$. Rearranging the terms,we get $4(x^2 - 6x) - 9(y^2 + 4y) = 36$. Completi

Vedclass · Accepted Answer

$4x^2 - 9y^2 - 24x - 36y = 0$

Vedclass · Answer

(D) The given equation of the hyperbola is $4x^2 - 9y^2 - 24x - 36y - 36 = 0$. Rearranging the terms,we get $4(x^2 - 6x) - 9(y^2 + 4y) = 36$. Completing the square,$4(x^2 - 6x + 9) - 9(y^2 + 4y + 4) = 36 + 36 - 36$. This simplifies to $4(x - 3)^2 - 9(y + 2)^2 = 36$. Dividing by $36$,we get $\frac{(x - 3)^2}{9} - \frac{(y + 2)^2}{4} = 1$. The equation of the pair of asymptotes is obtained by setting the constant term to zero: $\frac{(x - 3)^2}{9} - \frac{(y + 2)^2}{4} = 0$. Multiplying by $36$,we get $4(x - 3)^2 - 9(y + 2)^2 = 0$. Expanding this,$4(x^2 - 6x + 9) - 9(y^2 + 4y + 4) = 0$. $4x^2 - 24x + 36 - 9y^2 - 36y - 36 = 0$. Thus,the equation is $4x^2 - 9y^2 - 24x - 36y = 0$.

The equation of the pair of asymptotes of the hyperbola $4x^2 - 9y^2 - 24x - 36y - 36 = 0$ is

Similar Questions

Let the foci of the hyperbola coincide with the foci of the ellipse $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$. If the eccentricity of the hyperbola is $5$, then the length of its latus rectum is:

The distance between the tangent lines to the hyperbola $x^2-2y^2=18$ which are perpendicular to the line $y=x$ is

Find the equation of the hyperbola with foci $(0, \pm 3)$ and vertices $(0, \pm \frac{\sqrt{11}}{2})$.

The equation of the hyperbola with focus $(1, 2)$,eccentricity $e = \sqrt{3}$,and directrix $2x + y = 1$ is given by:

The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^2 - 2y^2 - 2 = 0$ to its asymptotes is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module