At what points on the graph of y = x^2y=x2 does the tangent line pass through (3, −7)?

1 Answer
Mar 6, 2015

The points are (-1,1)(1,1) and (7,49)(7,49).

Method: Find the general form of the equation of a line tangent to the graph of y=x^2y=x2. Then find the particular points that satisfy the condition: the tangent line passes through (3,-7)(3,7).

I will continue to use xx and yy as variables.

Consider a particular value of xx, calls it aa.

The point an the graph at x=ax=a has coordinates (a,a^2)(a,a2) (The yy-coordinate must satisfy y=x^2y=x2 in order to be on the graph.)

The slope of the tangent to the graph is determined by differentiating: y'=2x, so at the point (a,a^2) the slope of the tangent is m=2a.

Use your favorite tehnique to find the equation of the line through (a,a^2) with slope 2a.

The tangent line has equation: y=2ax-a^2.

(One way to find the line: start with y-a^2=2a(x-a), so y-a^2=2ax-2a^2. Add a^2 to both sides to get y=2ax-a^2.)

We have been asked to make the point (3,-7) lie on the line. So we need, (-7)=2a(3)-a^2. Now, solve for a.

-7=6a-a^2 if and only if a^2-6x-7=0.
Solve by factoring: (a+1)(a-7)=0, which requires a=-1 or a=7.

The points we are looking for, then, are (-1,1) and (7,49).

You can now check the answers by verifying that the point (3,-7) lies on the lines:
y=-2x-1 (the tangent when a=-1),
and also on y=14x-49 (the tangent when a=7).