# Thread: equation of tangent question

1. ## equation of tangent question

The straight line y = -x + 4 cuts the parabola with equation y = 16 - x^2 at the points A and B.

a) Find the coordinates of A and B.

Soo... since the line meets the parabola, I can solve by equating the equation of the line and the equation of the parabola/the equation of the gradient of the parabola. I ended up using the equation of the gradient.

So:

-x + 4 = -2x
x= -4

Subbing that into the equation of the parabola, I get y=0, so my first point is (-4, 0).

What is the exact method to find a second point?

2. I gave it a try by setting -x+4=16 - x^2
and i get (4,0) and (-3, 7)

3. Originally Posted by shawli
The straight line y = -x + 4 cuts the parabola with equation y = 16 - x^2 at the points A and B.

a) Find the coordinates of A and B.

Soo... since the line meets the parabola, I can solve by equating the equation of the line and the equation of the parabola/the equation of the gradient of the parabola. I ended up using the equation of the gradient.
No. This has nothing to do with the gradient, nor is this a tangent line. The line will cut the parabola when y= -x+ 4= 16- x^2. Solve that equation for x (two values since it is quadratic) and find the corresponding values of y.

So:

-x + 4 = -2x
x= -4

Subbing that into the equation of the parabola, I get y=0, so my first point is (-4, 0).

What is the exact method to find a second point?