If the parabolas $$y^2 = 4b(x-c)$$ and $$y^2 = 8ax$$ have a common normal, then which one of the following is a valid choice for the ordered triad $$(a, b, c)$$:

We have two parabolas

$$y^{2}=4b\,(x-c)\qquad\text{and}\qquad y^{2}=8ax.$$

To decide whether a common normal exists, we write each curve in its standard parametric form, find the slope of the normal at a general point and then force the two normals to coincide.

For the first parabola, compare $$y^{2}=4bX$$ (whose parameter is generally written as $$t$$). A mere translation $$X\to x-c$$ gives

$$x=c+bt^{2},\qquad y=2bt.$$

Differentiating with respect to $$t$$ gives

$$\frac{dx}{dt}=2bt,\qquad\frac{dy}{dt}=2b.$$

Hence the slope of the tangent is

$$\frac{dy}{dx}=\frac{dy/dt}{dx/dt}=\frac{2b}{2bt}=\frac1t,$$

so the slope of the normal is the negative reciprocal, namely

$$m_{1,\text{normal}}=-t.$$

Therefore the normal at $$(c+bt^{2},\,2bt)$$ is

$$y-2bt=-t\bigl(x-(c+bt^{2})\bigr).$$

For the second parabola, write it as $$y^{2}=4(2a)x.$$ With parameter $$s$$ we have

$$x=2a s^{2},\qquad y=4a s,$$

and again

$$\frac{dx}{ds}=4as,\qquad\frac{dy}{ds}=4a.$$

Thus

$$\frac{dy}{dx}=\frac{4a}{4as}=\frac1s,\qquad m_{2,\text{normal}}=-s.$$

The normal at $$(2a s^{2},\,4a s)$$ reads

$$y-4a s=-s\bigl(x-2a s^{2}\bigr).$$

Because the two normals must be the same line, their slopes must be equal, i.e.

$$-t=-s\;\Longrightarrow\;t=s=k\;(\text{say}).$$

Writing both normals with this common parameter $$k$$ gives

$$y-2bk=-k\bigl(x-(c+bk^{2})\bigr)\quad\text{and}\quad y-4ak=-k\bigl(x-2ak^{2}\bigr).$$

Putting both in the form $$y=-kx+\text{constant},$$ and equating the constants, we obtain

$$k(c+b k^{2})+2bk=k(2a k^{2})+4ak.$$

For $$k\neq0$$ we may divide by $$k$$ to get a linear equation in $$k^{2}$$:

$$\bigl(b-2a\bigr)k^{2}+c+2b-4a=0.$$

Rewriting,

$$k^{2}=\frac{4a-c-2b}{\,b-2a\,}. \quad -(★)$$

For a real common normal with $$k\neq0$$ we need the right-hand side to be non-negative. Let us test each proposed ordered triple $$(a,b,c)$$.

Option A: $$(a,b,c)=(1,1,3)$$.

$$4a-c-2b=4-3-2=-1,\qquad b-2a=1-2=-1,$$

so $$k^{2}=\frac{-1}{-1}=1\ge0.$$ A real $$k=\pm1$$ exists, hence a non-horizontal common normal is present.

Option B: $$(a,b,c)=\bigl(\tfrac12,\,2,\,0\bigr)$$.

$$4a-c-2b=2-0-4=-2,\qquad b-2a=2-1=1,$$

so $$k^{2}=\frac{-2}{1}=-2\lt 0,$$ giving no non-zero $$k$$. However, we must still examine the special case $$k=0$$ which we excluded by division. For $$k=0$$ the normal has slope $$0$$, i.e. the $$x$$-axis $$y=0$$.

• On the first parabola, $$t=0$$ gives the point $$(c,0)$$ and a horizontal normal $$y=0$$.

• On the second parabola, $$s=0$$ gives $$(0,0)$$ and again the horizontal normal $$y=0$$.

Thus $$y=0$$ is a common normal, so Option B is valid.

Option C: $$(a,b,c)=\bigl(\tfrac12,\,2,\,3\bigr)$$.

Exactly the same calculation as in Option B yields $$k^{2}=-2\lt 0$$ and hence no non-zero $$k$$, but $$k=0$$ once more produces the common normal $$y=0$$ (now passing through $$(3,0)$$ on the first parabola and $$(0,0)$$ on the second). Therefore Option C is also valid.

Since every individual choice A, B and C supplies at least one real common normal, all of them are correct simultaneously.

Hence, the correct answer is Option D.