# Question 6c148

Jan 11, 2017

The midpoint of the chord will start at one of the two points where the slope of the tangent line is equal to 1 and end at the other point where the slope of the tangent line is equal to 1.

#### Explanation:

Therefore, we need to find the two points on the ellipse where the slope of the tangent line is equal to 1.

Implicitly differentiate the given equation:

$\frac{d \left({x}^{2}\right)}{\mathrm{dx}} + \frac{d \left(4 {y}^{2}\right)}{\mathrm{dx}} = \frac{d \left(1\right)}{\mathrm{dx}}$

$2 x + 8 y \frac{\mathrm{dy}}{\mathrm{dx}} = 0$

$8 y \frac{\mathrm{dy}}{\mathrm{dx}} = - 2 x$

$\frac{\mathrm{dy}}{\mathrm{dx}} = - \frac{x}{4 y}$

Set $\frac{\mathrm{dy}}{\mathrm{dx}}$ = 1:

$- \frac{x}{4 y} = 1$

$y = - \frac{x}{4}$

Substitute $- \frac{x}{4}$ for y into the equation for the ellipse:

${x}^{2} + 4 {x}^{2} / 16 = 1$

$\frac{5}{4} {x}^{2} = 1$

${x}^{2} = \frac{4}{5}$

$x = \pm 2 \frac{\sqrt{5}}{5}$

Substitute $\frac{4}{5}$ for ${x}^{2}$ into the equation for the ellipse:

$\frac{4}{5} + 4 {y}^{2} = 1$

$4 {y}^{2} = \frac{1}{5}$

${y}^{2} = \frac{1}{20}$

$y = \pm \frac{\sqrt{5}}{10}$

The locus of the midpoint is a line segment between the points:

$\left(- 2 \frac{\sqrt{5}}{5} , \frac{\sqrt{5}}{10}\right) \mathmr{and} \left(2 \frac{\sqrt{5}}{5} , - \frac{\sqrt{5}}{10}\right)$

m = $\frac{- \frac{\sqrt{5}}{10} - \frac{\sqrt{5}}{10}}{2 \frac{\sqrt{5}}{5} - - 2 \frac{\sqrt{5}}{5}} = - \frac{1}{4}$

The equation of the locus is the line segment:

y = -1/4(x - 2sqrt(5)/5) - sqrt5/10; -2sqrt(5)/5< x < 2sqrt(5)/5#