Let be the parabola with focus . The line passes through and has slope . It intersects at two points . Find the chord length .

Analytic Geometry Solution – Problem 36: Find the chord length

Question

Let $C$ be the parabola $x^2=8y$ with focus $F(0,2)$ . The line $l$ passes through $F$ and has slope $1$ . It intersects $C$ at two points $A,B$ .

Find the chord length $|AB|$ .

Step-by-step solution

(1) The line is $l: y=x+2$ .

(2) Intersect with $x^2=8y$ : $x^2=8(x+2)\Rightarrow x^2-8x-16=0.$ So $x=4\pm 4\sqrt{2}.$ Then $y=x+2$ , giving points $A\bigl(4+4\sqrt2,\,6+4\sqrt2\bigr),\quad B\bigl(4-4\sqrt2,\,6-4\sqrt2\bigr).$ (3) Compute the distance: $|AB|=\sqrt{(8\sqrt2)^2+(8\sqrt2)^2}=\sqrt{128+128}=16.$ (Equivalently, for $x^2=4py$ with $p=2$ , a focal chord with slope $k$ has length $4p(1+k^2)$ ; here $k=1$ gives $16$ .) \]

Final answer

The focal chord length is $|AB|=16$ .

Marking scheme

Step 1 — Setup

Checkpoint: write $l:y=x+2$ and set up intersection with $x^2=8y$ (2 pts)

Step 2 — Key Calculation

Checkpoint: solve the quadratic and compute $|AB|$ correctly (3 pts)

Step 3 — Final Answer

Checkpoint: state $|AB|=16$ (2 pts)

Zero credit if: substitutes the focus incorrectly or uses a non-focal chord formula.

Deductions: -1 pt for arithmetic error in the distance calculation.

Analytic Geometry – Problem 36: Find the chord length