If the line lx + my = 1 is a normal to the hy | vedclass

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

Question

(C) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec 	heta, b 	an 	heta)$ is given by $\fr

Vedclass · Accepted Answer

$(a^2 + b^2)^2$

Vedclass · Answer

(C) The equation of the normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ at the point $(a \sec 	heta, b 	an 	heta)$ is given by $\frac{ax}{\sec 	heta} + \frac{by}{	an 	heta} = a^2 + b^2$. Comparing this with the given line $lx + my = 1$,we can rewrite the normal equation as $\frac{a \cos 	heta}{a^2 + b^2} x + \frac{b \cot 	heta}{a^2 + b^2} y = 1$. Thus,$l = \frac{a \cos 	heta}{a^2 + b^2}$ and $m = \frac{b \cot 	heta}{a^2 + b^2}$. Now,$\frac{a^2}{l^2} - \frac{b^2}{m^2} = \frac{a^2 (a^2 + b^2)^2}{a^2 \cos^2 	heta} - \frac{b^2 (a^2 + b^2)^2}{b^2 \cot^2 	heta}$. $= (a^2 + b^2)^2 (\sec^2 	heta - 	an^2 	heta)$. Since $\sec^2 	heta - 	an^2 	heta = 1$,we get $\frac{a^2}{l^2} - \frac{b^2}{m^2} = (a^2 + b^2)^2$.

If the line $lx + my = 1$ is a normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,then $\frac{a^2}{l^2} - \frac{b^2}{m^2}$ is equal to

Similar Questions

Which of the following carries glucose from the digestive tract to the liver?

Along the $x$-axis,three charges $\frac{q}{2}, -q$ and $\frac{q}{2}$ are placed at $x=0, x=a$ and $x=2a$ respectively. The resultant electric potential at a point $P$ located at a distance $r$ from the charge $-q$ (where $r > a$) is ($\varepsilon_0$ is the permittivity of free space):

The conjugate base of $HBr$ is:

What is common between chloroplasts,chromoplasts and leucoplasts?

Sodium is heated in air at $300^{\circ} C$ to form $X$. $X$ absorbs $CO_2$ and forms $Na_2CO_3$ and $Y$. Which of the following is $Y$?

Vedclass Test Series

Exam Paper Generator

Online Exam Module