Question

Sketch a graph of the parabola y=x^2+3. On the same graph, plot the point (0,−6). Note there are two tangent lines of y=x2+3 that pass through the point (0,−6).


The tangent line of the parabola y=x^2+3 at the point (a,a^2+3) passes through the point (0,−6) where a>0. The other tangent line that passes through the point (0,−6) occurs at the point (−a,a^2+3).


Find the number a

Answer 1

The slope at any point (x,y) is y'(x) = 2x

So, we want the line

y-(a^2+3) = 2a(x-a)

to pass through (0,-6)
-6-a^2-3 = -2a^2
a^2 - 9 = 0
a = 3

So, the lines tangent to x^2+3 at (3,12) and (-3,12) both pass through (0,-6)

Answer 2

The slopes of the tangent lines are equal to the derivative of the parabola at the points. y ' = 2x, so y'(a) = 2a and y'(-a) = -2a.

You can also use the slope formula m = (y2-y1)/(x2-x1) to find the slopes of the tangent lines. Therefore m = (a^2+3-(-6))/(a-0)=(a^2+9)/a. Setting this equal to 2a, and solving for a, you get a = 3 and a = -3