Find $c$ such that $6y=c+4x$ is normal to $y=8x^3 - 12x^2 +1$

Question

The equation of a curve is $y=8x^3 - 12x^2 +1$. Find the value of $c$ for which the line $6y=c+4x$ is a normal to the curve.

Help me please. I think there must be a point where the curve meets the line but I don't know how to find that point

Answer 1

Differentiate $y=8x^3 - 12x^2 +1 \ \ \ $

You get $\ \ \frac{dy}{dx}=24x^2-24x=24x(x-1) \ \ $ .

You can see that your line has slope $ \ \ 2 \over 3 \ \ $. So ask when $ \ \frac{dy}{dx} = - \frac 3 2 ?$ Solve for $x$ when $$\frac {-3} 2 = 24x^2-24x$$

There are two $x$ values that will satisfy. They will each have a corresponding $y$ coordinate on the cubic which you can get by plugging in those $x$ values you solved for. Then you will have a slope (your original slope of $2/3$) and two points, which you can make two lines with. Then you can find $c$.

Answer 2

Hint:

This line has slope $\frac23$. If it intersects the cubic at some point and is normal to thtangent to the cubic at that point, the tangent will have slope $-\frac 32$. Hence you first have to find the points on the cubic at which $\;y'(x)=-\frac32$, then determine the corresponding value of $c$ from the equation of the normal.

Answer 3

Let the point at which the normal is drawn be $(t, 8t^3 - 12t^2 + 1)$

Since $y = 8x^3 - 12x^2 + 1$,

$-\frac{dx}{dy} = \frac{1} {24x (1-x)}$

At $x = t$, $-\frac{dx}{dy} = \frac{1} {24t (1-t)}$

Equation of the normal:

$y - (8t^3 - 12t^2 + 1) = \frac{1} {24t (1-t)} (x-t)$

Now if the above equation is identical to $6y = c + 4x$, we must have

$\frac{6}{1} = \frac{4}{\frac{1}{24t (1-t)}} = \frac{c}{8t^3 - 12t^2 + 1 - \frac{1}{24(1-t)}}$

Solving, $t = \frac{1}{2} \pm \frac{\sqrt{3}}{4}$ and $c = \frac{1}{4}(-32 \pm 31 \sqrt{3})$

[Please check the calculations]