Shortest distance of point $P$ on parabola from fixed point $Q$

Question

If $P(a,b)$ be any point in the curve $y^2=6x$. Then find shortest distance of point $P(a,b)$ from point $\displaystyle Q(3,\frac{3}{2})$ and also find $2(a+b)$

Given point $P(a,b)$ in curve $y^2=6x.$

Then we have $b^2=6a\cdots \cdots (1)$

And shortest distance is perpendicular to tangent at $P(a,b)$ and also passes through $\displaystyle Q(3,\frac{3}{2})$

So here $\displaystyle y^2=6x\Longrightarrow \frac{dy}{dx}=\frac{6}{2y}$

So slope of tangent line at $P(a,b)$ is $\displaystyle \frac{3}{b}$

and its Normal line slope is $\displaystyle -\frac{b}{3}$

Equation of Normal line $\displaystyle y-b=-\frac{b}{3}(x-a)$

It also passes through $\displaystyle (3,\frac{3}{2})$

$\displaystyle \frac{3}{2}-b=-\frac{b}{3}(3-a)=\frac{ab}{3}$

$\displaystyle b+\frac{ab}{3}=\frac{3}{2}\Longrightarrow b=\frac{9}{2(3+a)}\cdots (2)$

From $(1)$ and $(2)$

$\displaystyle \frac{81}{4(3+a)^2}=6a\Longrightarrow 8a(3+a)^2-27=0$

$\displaystyle 8a^3+48a^2+72a-27=0$

Now I am struck here

Please tell me How can I solve that cubic equation

and also please tell me is there is any better way to solve it

Answer 1

It turns out that$$\require{cancel}-\frac b3(3-a)=-b+\frac{ab}3\left(\ne\frac{ab}3\right).$$

In fact,$$\frac32-b=-\frac b3(3-a)\iff\frac32-\cancel b=-\cancel b+\frac{ab}3\iff ab=\frac92.$$But you also have $6a=b^2$, and therefore $\displaystyle\frac92=ab=\frac{b^3}6$, from which it follows that $b^3=27$ and that therefore $b=3$. So, $a=\dfrac32$; see the picture below.

Answer 2

The point $P(a,b)$ lies on the parabola: $$b^2=6a \ \ \ \ \ \ \ (1)$$ The tangent line to the parabola at the point $P(a,b)$: $$y-b=\frac3b(x-a)$$ The line passing through $P(a,b)$ and $Q(3,\frac32)$: $$\frac{y-b}{\frac32-b}=\frac{x-a}{3-a}\Rightarrow y=\frac{3-2b}{6-2a}x-\frac{3a-2ab}{6-2a}+b $$ The two lines must be perpendicular: $$\frac3b\cdot \frac{3-2b}{6-2a}=-1\Rightarrow 9-6b=-6b+2ab\Rightarrow 2ab=9 \ \ \ \ \ (2)$$ From $(1)$ and $(2)$: $$2\cdot \frac{b^2}6\cdot b=9\Rightarrow b^3=27\Rightarrow b=3\\ a=\frac{b^2}6=\frac96=\frac32 $$