Question

Chord whose midpoint of parabola is given derive the formula

Answer 1

The equation of the chord of the parabola $y^2 = 4ax$ whose midpoint is $(x_1, y_1)$ is:

$yy_1 - 2ax = y_1^2 - 2ax_1$

Answer 2

To derive the formula for the chord of a parabola whose midpoint is given, let's consider the standard equation of a parabola and follow the steps below.

1. Define the Parabola and the Chord's Midpoint Let the equation of the parabola be $y^2 = 4ax$.
Let the midpoint of the chord be $M(x_1, y_1)$.
Let the two endpoints of the chord be $P(x_2, y_2)$ and $Q(x_3, y_3)$.

2. Apply Midpoint Formula Since $M(x_1, y_1)$ is the midpoint of $PQ$, we have:
$x_1 = \frac{x_2 + x_3}{2} \quad \implies \quad x_2 + x_3 = 2x_1 \quad \dots(1)$
$y_1 = \frac{y_2 + y_3}{2} \quad \implies \quad y_2 + y_3 = 2y_1 \quad \dots(2)$

3. Use the Property that Endpoints Lie on the Parabola Since $P(x_2, y_2)$ and $Q(x_3, y_3)$ lie on the parabola $y^2 = 4ax$, they must satisfy its equation:
$y_2^2 = 4ax_2 \quad \dots(3)$
$y_3^2 = 4ax_3 \quad \dots(4)$

4. Subtract the Equations and Factor Subtract equation (4) from equation (3):
$y_2^2 - y_3^2 = 4ax_2 - 4ax_3$
Factor the difference of squares on the left side and factor out $4a$ on the right side:
$(y_2 - y_3)(y_2 + y_3) = 4a(x_2 - x_3)$

5. Substitute from Midpoint Formula Substitute $y_2 + y_3 = 2y_1$ from equation (2) into the above equation:
$(y_2 - y_3)(2y_1) = 4a(x_2 - x_3)$

6. Determine the Slope of the Chord The slope of the chord $PQ$ is $m = \frac{y_2 - y_3}{x_2 - x_3}$.
From the previous step, rearrange the terms to find the slope:
$m = \frac{y_2 - y_3}{x_2 - x_3} = \frac{4a}{2y_1} = \frac{2a}{y_1}$

7. Write the Equation of the Chord (Point-Slope Form) Now we have the slope $m = \frac{2a}{y_1}$ and a point on the chord $M(x_1, y_1)$. Using the point-slope form of a linear equation ($y - y_1 = m(x - x_1)$):
$y - y_1 = \frac{2a}{y_1}(x - x_1)$

8. Simplify to the Standard Form Multiply both sides by $y_1$:
$y_1(y - y_1) = 2a(x - x_1)$
$yy_1 - y_1^2 = 2ax - 2ax_1$
Rearrange the terms to get the final formula:
$yy_1 - 2ax = y_1^2 - 2ax_1$

This formula is often represented as $T = S_1$, where $S = y^2 - 4ax$ is the equation of the parabola, $T = yy_1 - 2a(x+x_1)$ is the tangent form, and $S_1 = y_1^2 - 4ax_1$ is the value of $S$ at $(x_1, y_1)$.
Let's verify:
$yy_1 - 2a(x+x_1) = y_1^2 - 4ax_1$
$yy_1 - 2ax - 2ax_1 = y_1^2 - 4ax_1$
$yy_1 - 2ax = y_1^2 - 4ax_1 + 2ax_1$
$yy_1 - 2ax = y_1^2 - 2ax_1$
This confirms the derived formula.