The product of the lengths of the perpendicul | vedclass

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

Question

(C) Let $(h, k)$ be any point on the hyperbola $x^2 - y^2 = 8$. Thus,$h^2 - k^2 = 8$. The equations of the asymptotes of the hyperbola $x^2 - y^2 = 8$

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) Let $(h, k)$ be any point on the hyperbola $x^2 - y^2 = 8$. Thus,$h^2 - k^2 = 8$. The equations of the asymptotes of the hyperbola $x^2 - y^2 = 8$ are $x + y = 0$ and $x - y = 0$. The perpendicular distance $d_1$ from $(h, k)$ to the line $x + y = 0$ is $d_1 = \frac{|h + k|}{\sqrt{1^2 + 1^2}} = \frac{|h + k|}{\sqrt{2}}$. The perpendicular distance $d_2$ from $(h, k)$ to the line $x - y = 0$ is $d_2 = \frac{|h - k|}{\sqrt{1^2 + (-1)^2}} = \frac{|h - k|}{\sqrt{2}}$. The product of the lengths of the perpendiculars is $d_1 d_2 = \frac{|h + k|}{\sqrt{2}} 	imes \frac{|h - k|}{\sqrt{2}} = \frac{|h^2 - k^2|}{2}$. Substituting $h^2 - k^2 = 8$,we get $d_1 d_2 = \frac{8}{2} = 4$.

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

Similar Questions

The distance between the directrices of a rectangular hyperbola is $10$ units. What is the distance between its foci?

If the equation of the tangent to the hyperbola $5x^2 - 9y^2 - 20x - 18y - 34 = 0$ which makes an angle of $45^{\circ}$ with the positive $X$-axis is $x + by + c = 0$,then $b^2 + c^2 =$

The equation of the director circle of the hyperbola $\frac{x^2}{16} - \frac{y^2}{4} = 1$ is given by

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

Let $P(a \sec \theta, b \tan \theta)$ and $Q(a \sec \phi, b \tan \phi)$ be two points such that $\theta+\phi=\frac{\pi}{2}$ 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 at $P$ and $Q$,then $k=$

Vedclass Test Series

Exam Paper Generator

Online Exam Module