Answer
$c=-2.5$
Work Step by Step
The product of the slopes of the curves at their intersection point should be equal to $-1$:
The slope of the first curve is
$$y'=8x$$
Using the implicit differentiating, the slope of the second curve is:
$$1=4yy' \to y'=\frac{1}{4y}$$
so:
$$8x\cdot \frac{1}{4y}=-1$$
$$8x\cdot \frac{1}{4(4x^{2})}=-1$$
$$8x\cdot \frac{1}{4(4x^{2})}=-1 \to x=-\frac{1}{2}=-0.5$$
so the intersection point will be:
$$(x=-0.5,y=4(-0.5)^{2}=1)$$
Both curves intersects if their $x$ and $y$ coordinates are equal:
$x=c+2(4x^{2})^{2}$
$c=-0.5-2(4\cdot 0.25)^2=-2.5$
