# If the straight line

Question:

If the straight line $x \cos \alpha+y \sin \alpha=p$ touches the curve $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$, then prove that

$a^{2} \cos ^{2} \alpha-b^{2} \sin ^{2} \alpha=\rho^{2}$

Solution:

Given:

The straight line $x \cos \alpha+y \sin \alpha=p$ touches the curve $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$.

Suppose the straight line $x \cos \alpha+y \sin \alpha=p$ touches the curve at $\left(x_{1}, y_{1}\right)$.

But the equation of tangent to $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$ at $\left(x_{1}, y_{1}\right)$ is

$\Rightarrow \frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}=1$

Thus, equation $\frac{\mathrm{xx}_{1}}{\mathrm{a}^{2}}-\frac{\mathrm{yy}_{1}}{\mathrm{~b}^{2}}=1$ and $x \cos \alpha+y \sin \alpha=p$ represent the same line.

$\therefore \frac{\frac{x_{1}}{a^{2}}}{\cos \alpha}+\frac{\frac{y_{1}}{b^{2}}}{\sin \alpha}=\frac{1}{p}$

$\Rightarrow x_{1}=\frac{a^{2} \cos \alpha}{p}, y_{1}=\frac{b^{2} \sin \alpha}{p}$

Since the point $\left(x_{1}, y_{1}\right)$ lies on the curve $\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1$

$\Rightarrow \frac{\left(\frac{a^{2} \cos \alpha}{p}\right)^{2}}{a^{2}}-\frac{\left(\frac{b^{2} \sin \alpha}{p}\right)^{2}}{b^{2}}=1$

$\Rightarrow \frac{a^{4} \cos \alpha^{2}}{p^{2} a^{2}}-\frac{b^{4} \sin \alpha^{2}}{p^{2} b^{2}}=1$

$\Rightarrow \frac{a^{2} \cos \alpha^{2}}{p^{2}}-\frac{b^{2} \sin \alpha^{2}}{p^{2}}=1$

$\Rightarrow a^{2} \cos ^{2} \alpha-b^{2} \sin ^{2} \alpha=p^{2}$

Thus proved.