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

What is the condition called when the heart rate decreases below the normal range?

Assign $A, B, C, D$ from the given type of reaction.
$NaCl + AgNO_3 \longrightarrow AgCl \downarrow + NaNO_3$

If $\alpha$ and $\beta$ are imaginary cube roots of unity,then the value of $\alpha^4 + \beta^{28} + \frac{1}{\alpha\beta}$ is:

Which among the following reactions will take place spontaneously?

An object is attached to the bottom of a light vertical spring and set vibrating. The maximum speed of the object is $15 \, cm/s$ and the period is $628 \, ms$. The amplitude of the motion in centimeters is

Vedclass Test Series

Exam Paper Generator

Online Exam Module