Finding a parameter for which a line is orthogonal to a curve

[Lancelot1]

Hiya again,

I am trying to solve this problem, I thought I got somewhere, but kinda stuck.

The graph of y^2=x^3

is called a semicubical parabola. Determine the constant b so that the line y = -(1/3)x+b meets this graph orthogonally.

I found the derivative of the curve by using implicit derivation. I have multiplied it by -(1/3) and made it equal to -1. There was no b in this equation, only x and y. I though of finding the intersecting points of the line and curve, but if I sub y by the line equation, I get an equation with two parameters. Can you help please ? Thank you !

 

[MarkFL]

What I would do is observe that the orthogonal slope to the line is 3, and so if we equate the slope of the semicubical parabola to 3, we obtain:

\(\displaystyle x^2=2y\) where $0<y$.

Now, the point(s) on the semicubical parabola where this is true can be found by:

\(\displaystyle \left(\frac{x^2}{2}\right)^2=x^3\)

\(\displaystyle x^4-4x^3=x^3(x-4)=0\)

Since $x\ne0$, we are left with $(x,y)=(4,8)$. So now, we are left to find the value of the parameter $b$ such that the line passes through that point. Can you continue? :D