Question

Prove that the line $x\cos \alpha + y\sin \alpha = p$ touches the parabola ${y^2} = 4a(x + a)$ if $p\cos \alpha + a = 0$.

Answer 1

Hint: In this question, we first let a point $({x_1},{y_1})$ at the point of contact. After this, the point satisfies both the parabola and line equation. So, using that we will find a quadratic equation in ${y_1}$ and use the property that the discriminant of the quadratic equation must be zero for the line to touch the parabola.

Complete step-by-step answer:
We have an equation of line as $x\cos \alpha + y\sin \alpha = p$ the equation of parabola as:${y^2} = 4a(x + a)$.
First let $P({x_1},{y_1})$ be the point where the line touches the parabola.
Since, P lies on the line. So, we can write:
. ${x_1}\cos \alpha + {y_1}\sin \alpha = p$
On further solving, we have:
$ \Rightarrow {x_1} = \dfrac{{ - (\sin \alpha {y_1} - p)}}{{\cos \alpha }}$ (1)
Also, the point lies on the parabola, so we have:
${y_1}^2 = 4a({x_1} + a)$
On putting the value of ${x_1}$ from equation 1, we get:
${y_1}^2 = 4a(\dfrac{{ - (\sin \alpha {y_1} - p)}}{{\cos \alpha }} + a)$
On further solving, we have:
${y_1}^2 = 4a(\dfrac{{ - \sin \alpha {y_1} + p + a\cos \alpha }}{{\cos \alpha }})$
On multiplying by $\cos \alpha $ on both sides, we have:
${y_1}^2\cos \alpha = - 4a\sin \alpha {y_1} + 4ap + 4{a^2}\cos \alpha $

On taking all the terms on LHS, we get:
${y_1}^2\cos \alpha + 4a\sin \alpha {y_1} - 4ap - 4{a^2}\cos \alpha = 0$. ..........(2)
This is a quadratic equation in${y_1}$. Now, to get the condition, we will use the property that:
For the line to touch the given parabola, the quadratic equation given by equation 2 must have equal roots i.e. discriminant D =0.
We know that the discriminant is given by formula:
D =${b^2} - 4ac$.
Since, we have D = 0
$\therefore {b^2} - 4ac = 0$ ..........(3)
From equation 2, a=$\cos \alpha $ , b = 4a$\sin \alpha $ and c = -$4ap - 4{a^2}\cos \alpha $.
Putting these values in equation 3, we have:
$16{a^2}{\sin ^2}\alpha - 4\cos \alpha ( - 4ap - 4{a^2}\cos \alpha ) = 0$
On further solving, we have:
\[16{a^2}{\sin ^2}\alpha + 16{a^2}{\cos ^2}\alpha = - 16ap\cos \alpha \]
Dividing both sides by 16a, we have:
\[a({\sin ^2}\alpha + {\cos ^2}\alpha ) = - p\cos \alpha \] ..........(4)
We know that \[{\sin ^2}\theta + {\cos ^2}\theta = 1\]
Therefore, equation 4 can be written as:
\[a = - p\cos \alpha \]
$ \Rightarrow a + p\cos \alpha = 0$
This proved that the line $x\cos \alpha + y\sin \alpha = p$ touches the parabola ${y^2} = 4a(x + a)$ if $p\cos \alpha + a = 0$.

Note: In this type of question, the usual step is to assume the touching point and then proceed by satisfying both the line and the given conic section equation by the assumed point. And then use the condition for tangency to get the result. The condition for tangency for the line y = mx + c to the parabola ${y^2} = 4ax$ is, $c = \dfrac{a}{m}$ .You should remember the condition for equal root of quadratic equation.