The line lx + my + n = 0 will be a tangent to | vedclass

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

Question

(A) The condition for the line $y = Mx + C$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $C^2 = a^2M^2 - b^2$.Given the

Vedclass · Accepted Answer

$a^2l^2 - b^2m^2 = n^2$

Vedclass · Answer

(A) The condition for the line $y = Mx + C$ to be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $C^2 = a^2M^2 - b^2$.Given the line $lx + my + n = 0$,we can rewrite it as $y = -\frac{l}{m}x - \frac{n}{m}$.Comparing this with $y = Mx + C$,we get $M = -\frac{l}{m}$ and $C = -\frac{n}{m}$.Substituting these values into the condition $C^2 = a^2M^2 - b^2$:$(-\frac{n}{m})^2 = a^2(-\frac{l}{m})^2 - b^2$$\frac{n^2}{m^2} = \frac{a^2l^2}{m^2} - b^2$Multiplying both sides by $m^2$,we get $n^2 = a^2l^2 - b^2m^2$,which can be written as $a^2l^2 - b^2m^2 = n^2$.

The line $lx + my + n = 0$ will be a tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,if

Similar Questions

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ where $\theta + \phi = \frac{\pi}{2}$ be two points on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If $(h, k)$ is the point of intersection of the normals drawn at $P$ and $Q$,then $k=$

The area (in sq units) of the triangle formed by the tangent at $(\sqrt{3}, 0)$ to the hyperbola $x^2-3y^2=3$ with the pair of asymptotes of the hyperbola is

The transverse axis of a hyperbola is along the $x$-axis and its length is $2a$. The vertex of the hyperbola bisects the line segment joining the centre and the focus. The equation of the hyperbola is

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

The vertices of the hyperbola $7x^2 - 49y^2 = 343$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module