The equation of the tangents to the hyperbola | vedclass

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

Question

(B) The given hyperbola is $3x^2 - 4y^2 = 12$,which can be written as $\frac{x^2}{4} - \frac{y^2}{3} = 1$.Here,$a^2 = 4$ and $b^2 = 3$.The equation of

Vedclass · Accepted Answer

$y - x = \pm 1$

Vedclass · Answer

(B) The given hyperbola is $3x^2 - 4y^2 = 12$,which can be written as $\frac{x^2}{4} - \frac{y^2}{3} = 1$.Here,$a^2 = 4$ and $b^2 = 3$.The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.Since the tangent cuts equal intercepts from the axes,its equation is of the form $\frac{x}{c} + \frac{y}{c} = 1$ or $\frac{x}{c} - \frac{y}{c} = 1$,which implies the slope $m = \pm 1$.For $m = 1$,the tangent is $y = x \pm \sqrt{4(1)^2 - 3} = x \pm \sqrt{1} = x \pm 1$,which gives $y - x = \pm 1$.For $m = -1$,the tangent is $y = -x \pm \sqrt{4(-1)^2 - 3} = -x \pm 1$,which gives $y + x = \pm 1$.However,checking the options provided,the correct equation is $y - x = \pm 1$.

The equation of the tangents to the hyperbola $3x^2 - 4y^2 = 12$ which cut equal intercepts from the axes are:

Similar Questions

The locus of the midpoints of the parallel chords with gradient $m$ of the rectangular hyperbola $xy = c^2$ is

Let the transverse axis of a hyperbola $H$ be parallel to the $X$-axis and $x^2+y^2-2x-4y+3=0$ be the equation of the auxiliary circle of $H$. If the asymptotes of $H$ are at right angles,then the equation of the hyperbola is

At which of the following points is the tangent to the hyperbola $x^2 - y^2 = 3$ parallel to the line $2x + y + 8 = 0$?

If $lx + my = 1$ is a normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,then $a^2 m^2 - b^2 l^2 =$

The reciprocal of the eccentricity of a rectangular hyperbola is

Vedclass Test Series

Exam Paper Generator

Online Exam Module