The equations of the common tangents to the t | vedclass

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

Question

(B) Any tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.For this line to be a tangent to

Vedclass · Accepted Answer

$y = \pm x \pm \sqrt{a^2 - b^2}$

Vedclass · Answer

(B) Any tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $y = mx \pm \sqrt{a^2m^2 - b^2}$.For this line to be a tangent to the hyperbola $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$,we substitute $y = mx + c$ into the second equation:$\frac{(mx + c)^2}{a^2} - \frac{x^2}{b^2} = 1$$b^2(m^2x^2 + 2mcx + c^2) - a^2x^2 = a^2b^2$$x^2(b^2m^2 - a^2) + 2b^2mcx + (b^2c^2 - a^2b^2) = 0$.Since the line is a tangent,the discriminant of this quadratic equation must be zero:$D = (2b^2mc)^2 - 4(b^2m^2 - a^2)(b^2c^2 - a^2b^2) = 0$$4b^4m^2c^2 - 4b^2(b^2m^2 - a^2)(c^2 - a^2) = 0$$b^2m^2c^2 - (b^2m^2c^2 - a^2b^2m^2 - a^2c^2 + a^4) = 0$$a^2b^2m^2 + a^2c^2 - a^4 = 0$$b^2m^2 + c^2 - a^2 = 0 \Rightarrow c^2 = a^2 - b^2m^2$.Equating the two expressions for $c^2$:$a^2m^2 - b^2 = a^2 - b^2m^2$$m^2(a^2 + b^2) = a^2 + b^2$ $\Rightarrow m^2 = 1$ $\Rightarrow m = \pm 1$.Substituting $m^2 = 1$ into $c^2 = a^2m^2 - b^2$:$c^2 = a^2 - b^2 \Rightarrow c = \pm \sqrt{a^2 - b^2}$.Thus,the common tangents are $y = \pm x \pm \sqrt{a^2 - b^2}$.

The equations of the common tangents to the two hyperbolas $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ and $\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1$ are-

Similar Questions

The equation of the tangent parallel to $y - x + 5 = 0$ drawn to the hyperbola $\frac{x^2}{3} - \frac{y^2}{2} = 1$ is

The equation $\frac{1}{r} = \frac{1}{8} + \frac{3}{8} \cos \theta$ represents:

If a hyperbola has a transverse axis of length $2 \sin \theta$ and is confocal with the ellipse $3x^2 + 4y^2 = 12$,then its equation is:

Let the latus rectum of the hyperbola $\frac{x^2}{9}-\frac{y^2}{b^2}=1$ subtend an angle of $\frac{\pi}{3}$ at the centre of the hyperbola. If $b^2$ is equal to $\frac{l}{m}(1+\sqrt{n})$,where $l$ and $m$ are co-prime numbers,then $l^2+m^2+n^2$ is equal to . . . . . . .

Vedclass Test Series

Exam Paper Generator

Online Exam Module