Draw a diagram to show that there are two tangent lines to the parabola y=x^2 that pass through the point (0,-4). Find the coordinates of the points where these tangent lines intersect the parabola.

Question

So far I have taken the derivative and got y'=2x..

Using point slope form I got y=2x-4....

I am not sure what to do next...

Answer 1

BUT, the given point is not on the curve, so what you did is bogus.

look at your sketch, the point (0,-4) is below the parabola.
Draw the two tangents, you will see that it touches the curve at two different points, one in the first quad, the other in the 2nd quad.
Let the point of contact be (a,b)

since (a,b) lies on the parabola, b = a^2
alsy s' = 2x = the slope at any point on the parabola
so at (a,b) the slope is 2a
but using the grade 9 way of finding slope
= (b+4)/(a-0) = (b+4)/a

so (b+4)/a = 2a
b+4 = 2a^2
b = 2a^2 - 4, but b = a^2
so 2a^2 - 4 = a^2
a^2 = 4
a = ± 2
then b = 4

so the two points of contact are (2,4) and (-2,4)

look at your sketch to see the validity of this answer
A further part of this question would be to find the equations of these two tangents.