The equation of the common tangents to the tw | vedclass

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

Question

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

Vedclass · Accepted Answer

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

Vedclass · Answer

(B) The equation of a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.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$$a^2b^2m^2 + a^2c^2 - a^4 = 0$Dividing by $a^2$: $b^2m^2 + c^2 - a^2 = 0$,so $c^2 = a^2 - b^2m^2$.Equating the two expressions for $c^2$:$a^2m^2 - b^2 = a^2 - b^2m^2$$m^2(a^2 + b^2) = a^2 + b^2$$m^2 = 1 \implies m = \pm 1$.Substituting $m^2 = 1$ into $c^2 = a^2m^2 - b^2$:$c^2 = a^2(1) - b^2 = a^2 - b^2$.Thus,$c = \pm \sqrt{a^2 - b^2}$.The equation of the common tangents is $y = \pm x \pm \sqrt{a^2 - b^2}$.

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

Similar Questions

Let $A(\theta_1)$ and $B(\theta_2)$ be two points on the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ and $S$ be the focus of the hyperbola. If $A, S, B$ are collinear and $a \cos \left(\frac{\theta_1+\theta_2}{2}\right)=k \cos \left(\frac{\theta_1-\theta_2}{2}\right)$ then $k=$

$A$ hyperbola passes through the points $(3, 2)$ and $(-17, 12)$ and has its centre at the origin and its transverse axis along the $x$-axis. The length of its transverse axis is:

Let $P$ be a point on the hyperbola $H: \frac{x^2}{9}-\frac{y^2}{4}=1$,in the first quadrant such that the area of the triangle formed by $P$ and the two foci of $H$ is $2 \sqrt{13}$. Then,the square of the distance of $P$ from the origin is

Find the equation of the hyperbola satisfying the given conditions: Foci $(\pm 5, 0)$,the transverse axis is of length $8$.

If $2x - y + 1 = 0$ is a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{16} = 1$,then which of the following $CANNOT$ be sides of a right-angled triangle?

Vedclass Test Series

Exam Paper Generator

Online Exam Module