Question

3 normals to parabola $y^2 = 4x$ at $A(0, 0)$, $B$, $C$ intersect at $P(4, 0)$, then match the entries:

• A. Area of $\triangle ABC$
• B. Circumradius of $\triangle ABC$
• C. Inradius of $\triangle ABC$
• D. Area of $\triangle PBC$

Answer 1

P -> 2, Q -> 4, R -> 1, S -> 2

Answer 2

We begin with the parabola

$$y^2 = 4x,$$

which is conveniently parametrized as

$$(x,y)=(t^2,2t).$$

For a point on the parabola corresponding to parameter $t$, the equation of the normal is derived as follows. The tangent at $(t^2,2t)$ has slope

$$\frac{dy}{dx}=\frac{1}{t}\quad (t\neq0),$$

so the normal (perpendicular) has slope $-t$. Its equation (through $(t^2,2t)$) is

$$y-2t=-t\,(x-t^2) \quad\Longrightarrow\quad y=-tx+t^3+2t.$$

We are told that three normals drawn at the points $A$, $B$, and $C$ of the parabola meet at a common point $P=(4,0)$. One of these is given as $A=(0,0)$ (for which $t=0$). For the normal to pass through $P(4,0)$ we substitute into the normal’s equation:

$$0=-t\,(4)+t^3+2t = t^3-4t+2t = t^3-2t.$$

Thus,

$$t(t^2-2)=0.$$

The three solutions are:

$$t=0,\quad t=\sqrt{2},\quad t=-\sqrt{2}.$$

This gives the points:

$$A=(0^2,2\cdot 0)=(0,0),$$ $$B=((\sqrt{2})^2,2\sqrt{2})=(2,2\sqrt{2}),$$ $$C= ((-\sqrt{2})^2,2(-\sqrt{2}))=(2,-2\sqrt{2}).$$

Now we determine the required quantities.

(P) Area of $\triangle ABC$:

Use the determinant formula:

$$\Delta_{ABC}=\frac{1}{2}\left|\det\begin{pmatrix} x_B-x_A & y_B-y_A \\ x_C-x_A & y_C-y_A \end{pmatrix}\right|=\frac{1}{2}\Big| (2)(-2\sqrt{2})-(2)(2\sqrt{2}) \Big|.$$

Compute:

$$(2)(-2\sqrt{2})-(2)(2\sqrt{2})=-4\sqrt{2}-4\sqrt{2}=-8\sqrt{2}.$$

Taking absolute value and halving:

$$\Delta_{ABC}=\frac{1}{2}\cdot 8\sqrt{2}=4\sqrt{2}.$$

Thus, (P) matches the entry $4\sqrt{2}$.

(Q) Circumradius of $\triangle ABC$:

First, find the lengths of the sides:

$$AB=\sqrt{(2-0)^2+(2\sqrt{2}-0)^2}=\sqrt{4+8}=2\sqrt{3},$$ $$AC=\sqrt{(2-0)^2+(-2\sqrt{2}-0)^2}=2\sqrt{3},$$ $$BC=\sqrt{(2-2)^2+(2\sqrt{2}-(-2\sqrt{2}))^2}=\sqrt{(4\sqrt{2})^2}=4\sqrt{2}.$$

The circumradius $R$ for any triangle is given by

$$R=\frac{abc}{4\Delta}.$$

Substitute:

$$R=\frac{(2\sqrt{3})(2\sqrt{3})(4\sqrt{2})}{4\cdot(4\sqrt{2})}=\frac{(4\cdot3)(4\sqrt{2})}{16\sqrt{2}}=\frac{48\sqrt{2}}{16\sqrt{2}}=3.$$

So, (Q) is $3$.

(R) Inradius of $\triangle ABC$:

The inradius is

$$r=\frac{\Delta}{s},$$

where the semiperimeter $s$ is

$$s=\frac{AB+AC+BC}{2}=\frac{2\sqrt{3}+2\sqrt{3}+4\sqrt{2}}{2}=2\sqrt{3}+2\sqrt{2}.$$

Thus,

$$r=\frac{4\sqrt{2}}{2(\sqrt{3}+\sqrt{2})}=\frac{2\sqrt{2}}{\sqrt{3}+\sqrt{2}}.$$

Multiply numerator and denominator by $\sqrt{3}-\sqrt{2}$:

$$r=\frac{2\sqrt{2}(\sqrt{3}-\sqrt{2})}{3-2}=2\sqrt{2}(\sqrt{3}-\sqrt{2})=2\sqrt{6}-4.$$

This is written as $2(\sqrt{6}-2)$.

(S) Area of $\triangle PBC$:

The vertices are $P=(4,0)$, $B=(2,2\sqrt{2})$, and $C=(2,-2\sqrt{2})$. Notice that $B$ and $C$ lie vertically above and below $(2,0)$. Thus, the base $BC$ is vertical with length:

$$BC=|2\sqrt{2}-(-2\sqrt{2})|=4\sqrt{2}.$$

The horizontal distance from $P$ (with $x=4$) to the line $x=2$ is $2$. Hence,

$$\Delta_{PBC}=\frac{1}{2}\times (4\sqrt{2})\times 2=4\sqrt{2}.$$

So, (S) again is $4\sqrt{2}$.

Final Matching:

• (P) Area of $\triangle ABC$ → $4\sqrt{2}$
• (Q) Circumradius of $\triangle ABC$ → $3$
• (R) Inradius of $\triangle ABC$  → $2(\sqrt{6}-2)$
• (S) Area of $\triangle PBC$     → $4\sqrt{2}$