The line through a point P perpendicular to the polar of P with respect to the parabola $ y^2 = 4ax $ passes through the fixed point $(\alpha,\beta)$ then prove that the polar of P is given by :
$$ (x-2a+\alpha)^2 + 4 \beta y = 0 $$.
My attempt :
Let P be (h,k) Polar of P w.r.t $y^2 = 4ax$ $\equiv $ T=0 $\implies$ $$yk = 2a(x+h)………..(*)$$ is equation of polar.
Line perpendicular to polar passing through P is $$(y-k)=\frac{-k}{(2a)}(x-h)………(1)$$
Given that this line passes through $(\alpha , \beta)$ its evident that equation of line in …...(1) can be obtained by writing line passing through (h,k) and $(\alpha, \beta)$, which is $$(y-\beta) = \frac{k-\beta}{h-\alpha}(x-\alpha)…………(2)$$ Since ………(1) and ……(2) Represent the same line we have by comparison : $$\frac{-k}{2a} = \frac{k-\beta}{h-\alpha}……………(3) $$ and $$\frac{hk}{2a} +k = \frac{-\alpha(k-\beta)}{h-\alpha} + \beta…..(4) $$
My intuition is to figure out h and k in terms of $ \alpha $ and $\beta$ from (3) and (4) and substitute it in (*) to prove the desired result. . Is this the right way to proceed ? . Is the said fixed point $(\alpha, \beta)$ some special point , the awareness of which can simplify the approach ? . Please help.
Visualisation of the problem.
I may not have understood the question, but I took “fixed point” to mean that all of the perpendiculars would coincide in some point. So I set out to determine what this point might be, and found that either it is trivial or it does not exist.
Let us give the name $Q$ to the parabola $y^2=4ax$. (For this to be a parabola, we require that $a \ne 0$.)
A tangent to $Q$ at a point $(\frac{y_0^2}{4a},y_0)$ has gradient $\frac{dx}{dy} = \frac{y_0}{2a}$, so its equation is:
$$ \begin{align} x - \frac{y_0^2}{4a} &= \frac{y_0}{2a} (y - y_0) \\ \therefore x &= \frac{y_0}{2a} (y - y_0) + \frac{y_0^2}{4a} \\ &= \frac{y_0}{2a} y - \frac{y_0^2}{2a} + \frac{y_0^2}{4a} \\ &= \frac{y_0}{2a} y - \frac{y_0^2}{4a} \end{align} $$
Take two distinct points on $Q$, called $A\,(\frac{y_A^2}{4a},y_A)$ and $B\,(\frac{y_B^2}{4a},y_B)$. The tangents to $Q$ at these points intersect at a third point $P\,(x_P,y_P)$. Hence:
$$ \begin{align} x_P = \frac{y_A}{2a} y_P - \frac{y_A^2}{4a} &= \frac{y_B}{2a} y_P - \frac{y_B^2}{4a} \\ \therefore \frac{y_A - y_B}{2a} y_P &= \frac{y_A^2 - y_B^2}{4a} \\ \therefore y_P &= \frac{y_A^2 - y_B^2}{2(y_A - y_B)} \\ &= \frac{y_A + y_B}{2} \end{align} $$
(Note that $\frac{u^2 - v^2}{u - v} = u + v$ for $u \ne v$, which is the case here—since $A$ and $B$ are distinct, $y_A \ne y_B$.)
$$ \begin{align} \therefore x_P &= \frac{y_A}{2a} \frac{y_A + y_B}{2} - \frac{y_A^2}{4a} \\ &= \frac{y_A^2}{4a} + \frac{y_A y_B}{4a} - \frac{y_A^2}{4a} \\ &= \frac{y_A y_B}{4a} \end{align} $$
The line through $A$ and $B$ is the polar of $P$, and its equation is:
$$ \begin{align} x - \frac{y_A^2}{4a} &= \frac{y_A^2 - y_B^2}{4a(y_A - y_B)} (y - y_A) \\ &= \frac{y_A + y_B}{4a} (y - y_A) \\ \therefore x &= \frac{y_A + y_B}{4a} (y - y_A) + \frac{y_A^2}{4a} \\ &= \frac{y_A + y_B}{4a} y - \frac{y_A + y_B}{4a} y_A + \frac{y_A^2}{4a} \\ &= \frac{y_A + y_B}{4a} y - \frac{y_A^2}{4a} - \frac{y_A y_B}{4a} + \frac{y_A^2}{4a} \\ &= \frac{y_A + y_B}{4a} y - \frac{y_A y_B}{4a} \end{align} $$
A line perpendicular to the polar of $P$ has gradient $\frac{dx}{dy} = -\frac{4a}{y_A + y_B}$. This is undefined if $y_A = -y_B$; in that case, $x_A = x_B$, the polar of $P$ is vertical, $y_P = 0$, and the perpendicular through $P$ is the horizontal line $y = 0$. (This is all a consequence of $Q$ being symmetrical about the $x$ axis.)
If $y_A \ne -y_B$, then the equation of the perpendicular that passes through $P$ is:
$$ \begin{align} x - x_P &= -\frac{4a}{y_A + y_B} (y - y_P) \\ \therefore x &= x_P - \frac{4a}{y_A + y_B} (y - y_P) \\ &= x_P - \frac{4a}{y_A + y_B} y + \frac{4a}{y_A + y_B} y_P \\ &= \frac{y_A y_B}{4a} - \frac{4a}{y_A + y_B} y + \frac{4a}{y_A + y_B} \frac{y_A + y_B}{2} \\ &= \frac{y_A y_B}{4a} - \frac{4a}{y_A + y_B} y + \frac{2a}{y_A + y_B} (y_A + y_B) \\ &= \frac{y_A y_B}{4a} - \frac{4a}{y_A + y_B} y + 2a \end{align} $$
Now, if all such perpendiculars coincide at a fixed point as some parameter varies… what is the varying parameter? My first assumption was that it’s $P$ that varies. But in this case, the fixed point does not exist.
Proof: Having shown that one possible equation of the perpendicular is $y = 0$, it follows that if there is a fixed point $(\alpha,\beta)$, then $\beta = 0$.
But if $y = \beta = 0$, then:
$$ \begin{align} x = \alpha &= \frac{y_A y_B}{4a} - \frac{4a}{y_A + y_B} 0 + 2a \\ &= \frac{y_A y_B}{4a} + 2a \\ &= x_P + 2a \end{align} $$
And this value is not fixed as $P$ varies. ∎
The other option is that $a$ varies while $P$ remains constant. In this case, the fixed point is trivial: either $P$ is the fixed point (since distinct straight lines may only intersect once), or the perpendiculars are all the same line and coincide everywhere. (This is the case when $y_P = 0$; since $y_P = \frac{y_A + y_B}{2}$, we can say that $y_B = 2 y_P - y_A = -y_A$, giving us the case discussed earlier.)
The third possibility is that I misunderstood the meaning of “fixed point” in this case—that it doesn’t mean “a common point shared by all possible lines as some parameter varies”, but simply “a chosen point” (that constrains the values of the parameter).