Question

How do you find the equation of the chord of contact to the parabola `x^2=8y` from the point (3,-2)?

Answer 1

Start with the point-slope form of the equation of a line:

`y = m(x-x_0) + y_0`

where `(x_0,y_0) = (3,-2)`

`y = m(x-3) -2" [1]"`

Express the equation of the parabola as `y` in terms of `x`:

`y = x^2/8" [2]"`

We know that the equation for `m`, at the tangents, the first derivative with respect to x:

`dy/dx = x/4`

`m = x/4" [3]"`

Substitute equations [2] and [3] into equation [1]:

`x^2/8 = x/4(x-3) -2`

Solve the above equation for the values of x:

`x^2 = 2x(x-3) -16`

`x^2 = 2x^2-6x -16`

`0 = x^2-6x-16`

`0 = (x+2)(x-8)`

`x_1 = -2` and `x_2 = 8`

The above are the x-coordinates of the two points of tangency originating from point `(3,-2)`.

Use equation [2] to find the corresponding y values:

`y_1 = (-2)^2/8` and `y_2 = 8^2/8`

`y_1 = 1/2` and `y_2 = 8`

The equation of the chord of contact is the equation of the line that connects the points `(-2,1/2)` and `(8,8)`.

Compute the slope:

`m = (8-1/2)/(8--2)`

`m = 3/4`

Use the point-slope form of the equation of a line and the point `(8,8)`

`y = 3/4(x-8)+8`

`y = 3/4x+2, -2 <= x<= 8`

The above is the slope-intercept form of the equation of the chord of contact.

The following is a drawing of the parabola, the tangent lines, and the chord of contact: