Find the equation of the line tangent to the parabola y=x^2 at the point (a,a^2)

Question

please reply with answer

Answer 1

(a, a²) is tangent point and slope of tangent line m = 2a( y = x², y' = 2x and m = 2a).

Then equation of tangent line y = a² + 2a(x - a); y = 2xa - a²

Answer 2

Start by differentiating y=x², which by the power rule gives

dy/dx=2x

Now you want to plug in x=a to get the slope at the point (a,a²)

dy/dx(a)=2a

Now you have the slope of the tangent line, solve for the y-intercept.

y=mx+b

where y=a², x=a, and m=2a

a²=(2a)(a)+b

-a²=b

So, the tangent line of y=x² at the point (a,a²) is:

y=2ax-a²