The condition that the straight line lx + my | vedclass

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

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$\frac{a^2}{l^2} - \frac{b^2}{m^2} = \frac{(a^2 + b^2)^2}{n^2}$

Vedclass · Answer

(A) The equation of the hyperbola is $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$.Any normal to the hyperbola at point $(a \sec 	heta, b 	an 	heta)$ is given by $\frac{ax}{\sec 	heta} + \frac{by}{	an 	heta} = a^2 + b^2$.Given the line $lx + my = n$,we can rewrite it as $\frac{lx}{n} + \frac{my}{n} = 1$.Comparing the coefficients of the normal equation $\frac{ax}{(a^2+b^2)\sec 	heta} + \frac{by}{(a^2+b^2)	an 	heta} = 1$ with $\frac{lx}{n} + \frac{my}{n} = 1$,we get:$\frac{l}{n} = \frac{a}{(a^2+b^2)\sec 	heta} \implies \sec 	heta = \frac{an}{l(a^2+b^2)}$$\frac{m}{n} = \frac{b}{(a^2+b^2)	an 	heta} \implies 	an 	heta = \frac{bn}{m(a^2+b^2)}$Using the identity $\sec^2 	heta - 	an^2 	heta = 1$,we substitute the values:$\left(\frac{an}{l(a^2+b^2)}ight)^2 - \left(\frac{bn}{m(a^2+b^2)}ight)^2 = 1$$\frac{a^2n^2}{l^2(a^2+b^2)^2} - \frac{b^2n^2}{m^2(a^2+b^2)^2} = 1$$\frac{a^2}{l^2} - \frac{b^2}{m^2} = \frac{(a^2+b^2)^2}{n^2}$.

The condition that the straight line $lx + my = n$ may be a normal to the hyperbola $b^2x^2 - a^2y^2 = a^2b^2$ is given by

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