Write the equation of the secant to the parabola y² = 4x

Write the equation of the secant to the parabola $y^2 = 4x$ that passes through the origin and cuts off a region of area 9 from the given parabola.

We parametrize the secant by its slope, build the enclosed area as a function of that slope, and solve for the slope that gives area $9$.

Step 1: Set up the secant line

A line through the origin (not vertical) has the form $y=mx$.

Step 2: Find the intersection points

Substituting into $y^2=4x$: $m^2x^2=4x \implies x(m^2x-4)=0$, so $x=0$ or $x=\frac{4}{m^2}$. The second point is $\left(\frac{4}{m^2},\frac{4}{m}\right)$.

Step 3: Set up the area integral

Writing both boundaries as functions of $y$ — parabola: $x=\frac{y^2}{4}$; line: $x=\frac ym$ — and taking $m>0$ so the line lies to the right of the parabola for $y\in\left[0,\frac4m\right]$:

\[A(m) = \int_0^{4/m}\left[\frac ym-\frac{y^2}{4}\right]dy = \left[\frac{y^2}{2m}-\frac{y^3}{12}\right]_0^{4/m}\]

\[= \frac{1}{2m}\cdot\frac{16}{m^2}-\frac{1}{12}\cdot\frac{64}{m^3} = \frac{8}{m^3}-\frac{16}{3m^3} = \frac{8}{3m^3}\]

Step 4: Set the area equal to 9

\[\frac{8}{3m^3}=9 \implies m^3=\frac{8}{27} \implies m=\sqrt[3]{\frac{8}{27}}=\frac23\]

Step 5: Write the equation of the secant

\[\boxed{y = \frac{2x}{3} \quad\text{or equivalently}\quad 2x-3y=0}\]

(By the symmetry of the parabola about the $x$-axis, the mirror-image secant $y=-\frac{2x}{3}$, i.e. $2x+3y=0$, encloses the same area on the other side.)

Step 6: Verify

\[A = \frac{8}{3(2/3)^3} = \frac{8}{3\cdot\frac{8}{27}} = \frac{8\cdot27}{24}=9 \quad\checkmark\]

The intersection point is $\left(\frac{4}{(2/3)^2},\frac{4}{2/3}\right)=(9,6)$: indeed $6^2=36=4\cdot9$. ✓

Summary

\[\boxed{y=\frac{2x}{3} \ \ (\text{equivalently } 2x-3y=0), \text{ meeting the parabola again at } (9,6)}\]

Key insight: Because the enclosed area scales as $m^{-3}$, a modest change in slope produces a large change in area — here the cubic relationship pins down a unique positive slope $m=\frac23$ that cuts off exactly $9$ square units.