Find the equation of the tangent to the hyper | vedclass

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

Question

(B) The given hyperbola is $x^2 - 4y^2 = 36$,which can be written as $\frac{x^2}{36} - \frac{y^2}{9} = 1$.Here,$a^2 = 36$ and $b^2 = 9$.The slope of t

Vedclass · Accepted Answer

$x + y \pm 3\sqrt{3} = 0$

Vedclass · Answer

(B) The given hyperbola is $x^2 - 4y^2 = 36$,which can be written as $\frac{x^2}{36} - \frac{y^2}{9} = 1$.Here,$a^2 = 36$ and $b^2 = 9$.The slope of the line $x - y + 4 = 0$ is $m_1 = 1$.Since the tangent is perpendicular to this line,its slope $m$ must satisfy $m 	imes 1 = -1$,so $m = -1$.The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.Substituting the values,we get $y = (-1)x \pm \sqrt{36(-1)^2 - 9}$.$y = -x \pm \sqrt{36 - 9}$.$y = -x \pm \sqrt{27}$.$y = -x \pm 3\sqrt{3}$.Thus,the equation of the tangent is $x + y \pm 3\sqrt{3} = 0$.

Find the equation of the tangent to the hyperbola $x^2 - 4y^2 = 36$ which is perpendicular to the line $x - y + 4 = 0$.

Similar Questions

The locus of the point of intersection of any two perpendicular tangents to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is a circle,which is called the director circle of the hyperbola. What is the equation of this circle?

$A$ hyperbola having the transverse axis of length $\sqrt{2}$ has the same foci as that of the ellipse $3x^{2} + 4y^{2} = 12$. Then this hyperbola does not pass through which of the following points?

If $\theta$ is the acute angle between the asymptotes of a hyperbola $7x^2 - 9y^2 = 63$,then $\cos \theta =$

Find the equation of the hyperbola satisfying the given conditions: Vertices $(\pm 2, 0)$,foci $(\pm 3, 0)$.

The straight line $x + y = \sqrt{2}p$ will touch the hyperbola $4x^2 - 9y^2 = 36$,if

Vedclass Test Series

Exam Paper Generator

Online Exam Module