Question

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

Answer 1

Hint – In this particular type of question use the general equation of the tangent to the hyperbola from any given general point which is given as ($\dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = 1$), so this equation and given equation of straight line is same so use these concepts to reach the solution of the question.

Complete step by step solution:
Proof –
The given curve is $\dfrac{{{x^2}}}{{{a^2}}} - \dfrac{{{y^2}}}{{{b^2}}} = 1$
As we all know that this curve represents the curve of hyperbola.
Given equation of straight line which touches the given curve, $x\cos \alpha + y\sin \alpha = p$
This is also written as,
 $x\dfrac{{\cos \alpha }}{P} + y\dfrac{{\sin \alpha }}{P} = 1$....... (1)
So the given straight line is a tangent to the given curve.
Now let us suppose that the straight line touches the curve at point (${x_1},{y_1}$).
Now as this point general equation of the tangent of the hyperbola is given as,
$ \Rightarrow \dfrac{{x{x_1}}}{{{a^2}}} - \dfrac{{y{y_1}}}{{{b^2}}} = 1$.................... (2)
Now equation (1) and (2) both are the tangent of the hyperbola so these two equations are the same.
As we know that the solution of the parallel lines (i.e. slope is same) or the same line (for example ax + by = c and lx + my = n) is given as $\dfrac{a}{l} = \dfrac{b}{m} = \dfrac{c}{n}$ so use this property in equation (1) and (2) the solution is given as,
\[ \Rightarrow \dfrac{{\dfrac{{\cos \alpha }}{P}}}{{\dfrac{{{x_1}}}{{{a^2}}}}} = \dfrac{{\dfrac{{\sin \alpha }}{P}}}{{ - \dfrac{{{y_1}}}{{{b^2}}}}} = 1\]
Now simplify this we have,
\[ \Rightarrow \dfrac{{\cos \alpha }}{P} = \dfrac{{{x_1}}}{{{a^2}}}\] and \[\dfrac{{\sin \alpha }}{P} = - \dfrac{{{y_1}}}{{{b^2}}}\]
$ \Rightarrow {x_1} = \dfrac{{{a^2}\cos \alpha }}{P}$,${y_1} = - \dfrac{{{b^2}\sin \alpha }}{P}$
Now these points satisfy the equation of straight line so substitute these values in the equation of straight line we have,
$ \Rightarrow {x_1}\cos \alpha + {y_1}\sin \alpha = p$
$ \Rightarrow \dfrac{{{a^2}\cos \alpha }}{P}\cos \alpha + \dfrac{{ - {b^2}\sin \alpha }}{P}\sin \alpha = p$
Now simplify this we have,
$ \Rightarrow \dfrac{{{a^2}{{\cos }^2}\alpha }}{P} - \dfrac{{{b^2}{{\sin }^2}\alpha }}{P} = p$
$ \Rightarrow {a^2}{\cos ^2}\alpha - {b^2}{\sin ^2}\alpha = {p^2}$
So this is the required expression we have to prove.
Hence proved.

Note – Whenever we face such types of questions the key concept we have to remember is that the solution of the parallel lines (i.e. slope is same) or the same line (for example ax + by = c and lx + my = n) is given as $\dfrac{a}{l} = \dfrac{b}{m} = \dfrac{c}{n}$, so first find out the equation of tangent to the hyperbola through any general point as above then using this property solve these equations as above we will get the required result.