# Max perpendicular distance between line and parabola

I am trying to find the greatest perpendicular between a parabola and line between its intersection between $x=0$ and $x=33$:

• Equation of the line : $y=-0.0175758x$
• Equation of the parabola: $y=0.0005326x^2-0.0351515x$

Of course it is easy to find the intersection of the line and the parabola, but find this value has been impossible for me to figure out. Digging through my calculus books for a problem like this, but to avail.

Any help would be appreciated.

Thanks, Retug

Find the intersection of the parabola to the tangent parallel to the line.

Consider the curve $y=ax^2$' where $a$ is a scale constant, and the line joining the origin to the fixed point $P(p,ap^2)$, which has equation $y-apx=0$.

Now consider a variable point $T(t, at^2)$ on the curve where $t<p$. The perpendicular distance from $T$ to the line is $$D=\left|\frac{at^2-apy}{\sqrt{1+a^2p^2}}\right|=\frac{a(pt-t^2)}{\sqrt{1+a^2p^2}}$$

Completing the square in the bracket gives the maximum $$D=\frac{ap^2}{4\sqrt{1+a^2p^2}}$$

You just have to apply your values of $a=0.0005326$ and $p=33$

• There are 3 constants, one more than $a,p$. – Narasimham Dec 29 '15 at 22:56

You first need to find the slope of the line, and then find the derivative of the parabola's formula. Find the point where the value of the derivative equals the slope, and you have your point.

HINTS:

Let the parabola and line be represented by:

$$y = a x^2 -b ; \, y = m \, x ;$$

Parabola has slope $$= 2 a x = m , \rightarrow x_p = m/2a, \, y_p= m^2/(4 a ) - b.$$

Let the slant line be parametrized

$$x = t , y = m \,t;$$

Find $t$ by differentiating the following with respect to $t$ and set it to zero.

$$(x_p-t)^2 + (y_p - m \,t)^2$$

Evaluate the above at this $t$ and find its square root. Plug in the constants.