The product of the lengths of perpendiculars | vedclass

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

Question

(B) The given equation of the hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - \frac{y^2}{1} = 1$.Here,$a^2 = 2$ and $b^2 = 1$.T

Vedclass · Accepted Answer

$2/3$

Vedclass · Answer

(B) The given equation of the hyperbola is $x^2 - 2y^2 = 2$,which can be written as $\frac{x^2}{2} - \frac{y^2}{1} = 1$.Here,$a^2 = 2$ and $b^2 = 1$.The equations of the asymptotes are $\frac{x}{\sqrt{2}} - y = 0$ and $\frac{x}{\sqrt{2}} + y = 0$.Let $P(x_1, y_1)$ be any point on the hyperbola. The product of the lengths of the perpendiculars from $P$ to the asymptotes is given by the formula $\frac{a^2 b^2}{a^2 + b^2}$.Substituting the values,we get $\frac{2 	imes 1}{2 + 1} = \frac{2}{3}$.

The product of the lengths of perpendiculars drawn from any point on the hyperbola $x^2 - 2y^2 - 2 = 0$ to its asymptotes is:

Similar Questions

$A$ light and a heavy body have equal kinetic energy. Which one has a greater momentum?

Two rods of the same length and material transfer a given amount of heat in $12 \, s$,when they are joined in parallel. But when they are joined in series,then they will transfer the same amount of heat under the same conditions in ......... $s$.

The cubic unit cell of a metal (molar mass $= 63.55 \ g \ mol^{-1}$) has an edge length of $362 \ pm$. Its density is $8.92 \ g \ cm^{-3}$. The type of unit cell is

If $A = \begin{bmatrix} 1 & 2 & 2 \\ 2 & 1 & 2 \\ 2 & 2 & 1 \end{bmatrix}$,then $A^3 - 4A^2 - 6A$ is equal to:

The absolute value of the tangent of the difference of the angles made by the lines $4x^2 - 24xy + 11y^2 = 0$ with the $X$-axis is

Vedclass Test Series

Exam Paper Generator

Online Exam Module