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) The equation of the hyperbola is $x^2-y^2=8$. The asymptotes of the hyperbola $x^2-y^2=a^2$ are given by $x^2-y^2=0$,which implies $x-y=0$ and $x+

Vedclass · Accepted Answer

$4$

Vedclass · Answer

(C) The equation of the hyperbola is $x^2-y^2=8$. The asymptotes of the hyperbola $x^2-y^2=a^2$ are given by $x^2-y^2=0$,which implies $x-y=0$ and $x+y=0$. Let $P(x, y)$ be any point on the hyperbola. The length of the perpendicular from $P(x, y)$ to the line $x-y=0$ is $d_1 = \frac{|x-y|}{\sqrt{1^2+(-1)^2}} = \frac{|x-y|}{\sqrt{2}}$. The length of the perpendicular from $P(x, y)$ to the line $x+y=0$ is $d_2 = \frac{|x+y|}{\sqrt{1^2+1^2}} = \frac{|x+y|}{\sqrt{2}}$. The product of the lengths of the perpendiculars is $d_1 	imes d_2 = \frac{|x-y|}{\sqrt{2}} 	imes \frac{|x+y|}{\sqrt{2}} = \frac{|x^2-y^2|}{2}$. Since the point lies on the hyperbola $x^2-y^2=8$,we substitute the value: $\frac{8}{2} = 4$.

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

Find the angle between the pair of lines represented by the equation $x^2+4xy+y^2=0$. (in $^{\circ}$)

Which one is an incorrect statement?

Two wires are made of the same material and have the same volume. Wire $1$ has cross-sectional area $A$ and wire $2$ has cross-sectional area $3A$. If the length of wire $1$ increases by $\Delta x$ on applying force $F$,how much force is needed to stretch wire $2$ by the same amount $\Delta x$?

Isogamous condition with non-flagellated gametes is found in:

$A$ person is observing a bacteria through a compound microscope. For better observation and to improve its resolving power,he should:

Vedclass Test Series

Exam Paper Generator

Online Exam Module