If the product of the perpendicular distances | vedclass

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

Question

(B) The equations of the asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \f

Vedclass · Accepted Answer

$6$

Vedclass · Answer

(B) The equations of the asymptotes of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \frac{y}{b} = 0$.Let $P(x_1, y_1)$ be any point on the hyperbola. The product of the perpendicular distances from $P$ to the asymptotes is given by:$p_1 p_2 = \left| \frac{\frac{x_1}{a} - \frac{y_1}{b}}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} ight| 	imes \left| \frac{\frac{x_1}{a} + \frac{y_1}{b}}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} ight| = \frac{|\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2}|}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{1}{\frac{a^2 + b^2}{a^2 b^2}} = \frac{a^2 b^2}{a^2 + b^2}$.Given $e = \sqrt{3}$,we have $b^2 = a^2(e^2 - 1) = a^2(3 - 1) = 2a^2$.Substituting $b^2 = 2a^2$ into the product formula:$p_1 p_2 = \frac{a^2(2a^2)}{a^2 + 2a^2} = \frac{2a^4}{3a^2} = \frac{2a^2}{3}$.Given $p_1 p_2 = 6$,we have $\frac{2a^2}{3} = 6$ $\Rightarrow a^2 = 9$ $\Rightarrow a = 3$.The length of the transverse axis is $2a = 2(3) = 6$.

If the product of the perpendicular distances from any point on the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ of eccentricity $e = \sqrt{3}$ from its asymptotes is equal to $6$,then the length of the transverse axis of the hyperbola is

Similar Questions

The locus of the point of intersection of the tangents at the endpoints of normal chords of the hyperbola $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ is

The foci of the hyperbola $9x^2 - 16y^2 = 144$ are

The equation of the hyperbola in the standard form (with transverse axis along the $x$-axis) having the length of the latus rectum = $9$ units and eccentricity = $5/4$ 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

The locus of the point of intersection of the lines $\frac{x}{a} + \frac{y}{b} = \lambda$ and $\frac{x}{a} - \frac{y}{b} = \frac{1}{\lambda}$ (where $\lambda$ is a parameter) is:

Vedclass Test Series

Exam Paper Generator

Online Exam Module