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 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 equ

Vedclass · Accepted Answer

$2/3$

Vedclass · Answer

(B) The 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 to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ are given by $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 0$,which implies $\frac{x}{a} - \frac{y}{b} = 0$ and $\frac{x}{a} + \frac{y}{b} = 0$.Let $P(x_1, y_1)$ be any point on the hyperbola. The product of the lengths of the perpendiculars from $P$ to the asymptotes $L_1: \frac{x}{a} - \frac{y}{b} = 0$ and $L_2: \frac{x}{a} + \frac{y}{b} = 0$ is given by:$P = \left| \frac{\frac{x_1}{a} - \frac{y_1}{b}}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} ight| 	imes \left| \frac{\frac{x_1}{a} + \frac{y_1}{b}}{\sqrt{\frac{1}{a^2} + \frac{1}{b^2}}} ight| = \frac{|\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2}|}{\frac{1}{a^2} + \frac{1}{b^2}}$.Since $(x_1, y_1)$ lies on the hyperbola,$\frac{x_1^2}{a^2} - \frac{y_1^2}{b^2} = 1$.Thus,the product is $\frac{1}{\frac{1}{a^2} + \frac{1}{b^2}} = \frac{1}{\frac{1}{2} + \frac{1}{1}} = \frac{1}{3/2} = \frac{2}{3}$.

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

If the extremities of a diagonal of a square are $(1, -2, 3)$ and $(2, -3, 5)$, then the length of its side is

If the line $2x + 5y = 12$ intersects the ellipse $4x^2 + 5y^2 = 20$ in two distinct points $A$ and $B$,then the mid-point of $AB$ is

$A$ compound $(A)$ is used in the preparation of washing soda to recover ammonia in the Solvay process. When $CO_2$ is bubbled through an aqueous solution of $(A)$,the solution turns milky. It is also used in whitewashing due to its disinfectant nature. What is the chemical formula of $(A)$?

$A$ light wave has a frequency of $4 \times 10^{14} \text{ Hz}$ and a wavelength of $5 \times 10^{-7} \text{ m}$ in a medium. The refractive index of the medium is

The enthalpy of combustion of benzene from the following data will be
$(i) \ 6C_{(s)} + 3H_{2(g)} \to C_6H_{6(l)}; \Delta H = + 45.9 \ kJ$
$(ii) \ H_{2(g)} + \frac{1}{2}O_{2(g)} \to H_2O_{(l)}; \Delta H = - 285.9 \ kJ$
$(iii) \ C_{(s)} + O_{2(g)} \to CO_{2(g)}; \Delta H = - 393.5 \ kJ$
........$kJ$

Vedclass Test Series

Exam Paper Generator

Online Exam Module