VITEEE 2014 Paper

The equation of any normal to $\frac{x^2}{a^2} - \frac{y^2}{b^2}$ = 1 is ax cos Φ + by cot Φ = $a^2+b^2$ ⇒ ax cos Φ + by cot Φ - $(a^2+b^2)$ = ... (i) The straight line lx +my-n = 0 will be normal to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2}$ = 1 , then eq. (i) and lx + my-n = 0 represent the same line, ∴ $\frac{a\cos\Phi}{l}$ = $\frac{b\cot\Phi}{m}$ = $\frac{a^2+b^2}{n}$ ⇒ sec Φ = $\frac{na}{l(a^2+b^2)}$ and tan Φ = $\frac{nb}{m(a^{+}b^2)}$ $\frac{n^2 a^2}{l^2 (a^2+b^2)} - \frac{n^2 b^2}{m^2 (a^2+b^2)^2}$ = 1 [Since $\sec^2$ Φ - $\tan^2$ Φ = 1] ⇒ $\frac{a^2}{l^2} - \frac{b^2}{m^2}$ = $\frac{(a^2+b^2)^2}{n^2}$ But given equation of normal is $\frac{a^2}{l^2} - \frac{b^2}{m^2}$ = $\frac{(a^2+b^2)^2}{k}$ ∴ k = $n^2$