Tangents are drawn to the hyperbola frac{x^2} | vedclass

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

Question

(D) The slope of the line $2x - y = 1$ is $m = 2$.The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$

Vedclass · Accepted Answer

$(A, B)$

Vedclass · Answer

(D) The slope of the line $2x - y = 1$ is $m = 2$.The equation of the tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$.Here $a^2 = 9$ and $b^2 = 4$,so the tangent equations are $y = 2x \pm \sqrt{9(2)^2 - 4} = 2x \pm \sqrt{36 - 4} = 2x \pm \sqrt{32} = 2x \pm 4\sqrt{2}$.The point of contact $(x_1, y_1)$ for a tangent $y = mx + c$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\left(\frac{a^2m}{c}, \frac{b^2}{c}\right)$.For $c = 4\sqrt{2}$,the point is $\left(\frac{9(2)}{4\sqrt{2}}, \frac{4}{4\sqrt{2}}\right) = \left(\frac{18}{4\sqrt{2}}, \frac{1}{\sqrt{2}}\right) = \left(\frac{9}{2\sqrt{2}}, \frac{1}{\sqrt{2}}\right)$,which is point $(A)$.For $c = -4\sqrt{2}$,the point is $\left(\frac{9(2)}{-4\sqrt{2}}, \frac{4}{-4\sqrt{2}}\right) = \left(-\frac{9}{2\sqrt{2}}, -\frac{1}{\sqrt{2}}\right)$,which is point $(B)$.Thus,the points of contact are $(A)$ and $(B)$.

Similar Questions

The product of the lengths of the perpendiculars from any point on the hyperbola $x^2 - y^2 = 8$ to its asymptotes is

The equation of the pair of asymptotes of the hyperbola $4x^2 - 9y^2 - 24x - 36y - 36 = 0$ is

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

The line $2x + y = 1$ is tangent to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. If this line passes through the point of intersection of the nearest directrix and the $x$-axis,then the eccentricity of the hyperbola is

Vedclass Test Series

Exam Paper Generator

Online Exam Module