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 normal to the hyperbola $\frac{x^2}{25} - \frac{y^2}{16} = 1$ that is perpendicular to the line $2x + y = 1$ is:

The length of the transverse axis of a hyperbola is $7$ and it passes through the point $(5, -2)$. The equation of the hyperbola is

Find the equation of the hyperbola satisfying the given conditions: Foci $(0, \pm 13)$,the conjugate axis is of length $24$.

Find the equation of the tangent to the hyperbola $\frac{x^2}{3} - \frac{y^2}{2} = 1$ which is equally inclined to the axes.

Let $P(6,3)$ be a point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If the normal at the point $P$ intersects the $x$-axis at $(9,0)$,then the eccentricity of the hyperbola is

Vedclass Test Series

Exam Paper Generator

Online Exam Module