Let be the parabola . A chord of has midpoint . Find the slope of line .

Analytic Geometry Solution – Problem 38: Find the slope of line

Question

Let $C$ be the parabola $x^2=8y$ . A chord $AB$ of $C$ has midpoint $M(4,5)$ . Find the slope of line $AB$ .

Step-by-step solution

(1) Parameterize $C$ as $(x,y)=(4t,2t^2)$ . Let $A(4t_1,2t_1^2),\quad B(4t_2,2t_2^2).$ (2) The slope of chord $AB$ is $\frac{2t_2^2-2t_1^2}{4t_2-4t_1}=\frac{t_1+t_2}{2}.$ (3) The midpoint $M$ has x-coordinate $\frac{4t_1+4t_2}{2}=2(t_1+t_2)=4.$ So $t_1+t_2=2$ . Hence the slope is $\frac{t_1+t_2}{2}=1.$ (Only the $x$ -coordinate of the midpoint is needed; the given $y$ -coordinate is consistent but not required.) \]

Final answer

The slope of chord $AB$ is $1$ .

Marking scheme

Step 1 — Setup

Checkpoint: use the parameter form $A(4t_1,2t_1^2),B(4t_2,2t_2^2)$ (2 pts)

Step 2 — Key Calculation

Checkpoint: derive slope $m=\frac{t_1+t_2}{2}$ and midpoint x-coordinate $x_M=2(t_1+t_2)$ (3 pts)

Step 3 — Final Answer

Checkpoint: substitute $x_M=4\Rightarrow t_1+t_2=2\Rightarrow m=1$ (2 pts)

Zero credit if: assumes the midpoint lies on the parabola and uses tangent slope.

Deductions: -1 pt for algebra mistake in midpoint computation.

Analytic Geometry – Problem 38: Find the slope of line