Question

Show that the line $x\cos\alpha + y\sin\alpha = p$ touches the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ if $a^2\cos^2\alpha - b^2\sin^2\alpha = p^2$.

Answer 1

The statement is proven. The line $x\cos\alpha + y\sin\alpha = p$ touches the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ if $a^2\cos^2\alpha - b^2\sin^2\alpha = p^2$.

Answer 2

1. Rewrite the line $x\cos\alpha + y\sin\alpha = p$ in slope-intercept form: $y = (-\cot\alpha)x + \frac{p}{\sin\alpha}$. Here, $m = -\cot\alpha$ and $c = \frac{p}{\sin\alpha}$.

2. The condition for tangency of $y=mx+c$ to $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is $c^2 = a^2m^2 - b^2$.

3. Substitute $m$ and $c$: $\left(\frac{p}{\sin\alpha}\right)^2 = a^2(-\cot\alpha)^2 - b^2$ $\frac{p^2}{\sin^2\alpha} = a^2\frac{\cos^2\alpha}{\sin^2\alpha} - b^2$

4. Multiply by $\sin^2\alpha$: $p^2 = a^2\cos^2\alpha - b^2\sin^2\alpha$ This confirms the given condition.

5. For $\sin\alpha = 0$, the line is $x = \pm p$. The condition becomes $a^2 = p^2$, so $x = \pm a$. Substituting $x = \pm a$ into the hyperbola equation gives $y=0$, confirming tangency at $(\pm a, 0)$.