Question

Find the equations of the normal to the ellipse x² + 4y² = 65 at the points (1, 4) and (7, 2). Find the equation of the straight line joining the intersection of the normals to the origin. 

Answer 1

$$x^2 + 4y^2 = 65$$

Differentiate both sides with respect to $x$

$$2x+8yy'=0$$

$$y'=-\dfrac{x}{4y}$$

Then the slope of the normal to the ellipse at the point $(x_0, y_0)$ is

$$slope=m=\dfrac{4y_0}{x_0}$$

The equation of the normal in point-slope form is

$$y-y_0=\dfrac{4y_0}{x_0}(x-x_0)$$

The equation of the normal in slope-intercept form is

$$y=\dfrac{4y_0}{x_0}x-3y_0$$

Point $(1,4)$

$$y=\dfrac{4(4)}{1}x-3(4)$$

$$y=16x-12$$

Point $(7,2)$

$$y=\dfrac{4(2)}{7}x-3(2)$$

$$y=\dfrac{8}{7}x-6$$

Point of intersection

$$16x-12=\dfrac{8}{7}x-6$$

$$52x=21$$

$$x=\dfrac{21}{52}$$

$$y=16(\dfrac{21}{52})-12=-\dfrac{72}{13}$$

Point $(\dfrac{21}{52}, -\dfrac{72}{13})$

The equation of the straight line joining the intersection of the normals to the origin. 

$$y=kx$$

Substitute

$$-\dfrac{72}{13}=k(\dfrac{21}{52})$$

$$k=-\dfrac{96}{7}$$

The equation of the straight line joining the intersection of the normals to the origin.

$$y=-\dfrac{96}{7}x$$