The question is: a line $x \cos\theta + y\sin\theta = p$ is given such that $a^2\cos^2\theta - b^2\sin^2\theta =p^2$.
I have to prove that it touches a hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$. I do not see how to proceed. I thought I could proceed by eliminating $p$ by squaring the line equation and equate both but I don't get an idea what to do next. What do I need to show to prove that it touches that hyperbola?
The tangent at $(x_1,y_1)$ to the hyperbola is \begin{align*} \frac{xx_1}{a^2} - \frac{yy_1}{b^2} = 1 \end{align*} If this is the given line, we have \begin{align*} \frac{x_1/a^2}{\cos \theta} = \frac{-y_1/b^2}{\sin \theta} = \frac{1}{p} \end{align*} Hence $x_1 = \frac{a^2\cos\theta}{p}, y_1 = -\frac{b^2\sin\theta}{p}$. Since this lies on the curve, we have \begin{align*} a^2\cos^2\theta - b^2\sin^2\theta = p^2 \end{align*}
HINT:
Replace $x$ with $\dfrac{p-y\sin\theta}{\cos\theta}$ in $$\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$$ to form a Quadratic Equation in $y$
Each value of $y$ represents the ordinate of the respective intersection.
For tangency, the two value of the ordinates must be same i.e., the the Quadratic Equation of $y$ must have equal roots.
Can you take it from here?
