If y=mx+4 (n0 is a typo,likely m0 or irrele | vedclass

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

Question

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

Vedclass · Accepted Answer

$\left(\frac{25}{4}, \frac{9}{4}\right)$

Vedclass · Answer

(B) The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.Given the tangent $y = mx + 4$,we have $c = 4$.Comparing $c^2 = a^2m^2 - b^2$,we get $16 = 25m^2 - 9$.$25m^2 = 25 \Rightarrow m^2 = 1$. Since $m$ is usually positive for such forms,$m = 1$.The point of contact for a tangent $y = mx + c$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\left(\frac{a^2m}{c}, \frac{b^2}{c}ight)$.Substituting $a^2 = 25, b^2 = 9, m = 1, c = 4$:Point of contact = $\left(\frac{25 	imes 1}{4}, \frac{9}{4}ight) = \left(\frac{25}{4}, \frac{9}{4}ight)$.

If $y=mx+4$ ($n>0$ is a typo,likely $m>0$ or irrelevant) is a tangent to the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$,then the point of contact of this tangent is

Similar Questions

The equation of the line passing through the points $\left(ct_1, \frac{c}{t_1}\right)$ and $\left(ct_2, \frac{c}{t_2}\right)$ is

Let $A(2 \sec \theta, 3 \tan \theta)$ and $B(2 \sec \phi, 3 \tan \phi)$ where $\theta+\phi=\frac{\pi}{2}$,be two points on the hyperbola $\frac{x^2}{4}-\frac{y^2}{9}=1$. If $(\alpha, \beta)$ is the point of intersection of normals to the hyperbola at $A$ and $B$,then $\beta$ is equal to

The distance between the foci of a hyperbola is $16$ and its eccentricity is $\sqrt{2}$. Its equation is

If the angle between the asymptotes of a hyperbola is $30^{\circ}$,then its eccentricity is

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 $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module