find all points P on the parabola y=x^2 such that the tangent line at P passes through the point (0,-4).

Question

anonymous · Answer

I've got it never mind.

anonymous · Answer

$$
\ \ \ y(x)=x^2\
\text{First, differentiate with respect to }x.\
\ \ \ y'(x)=2x\
\text{We know that the point-slope form of a tangent}\
\text{line equation is as follows.}\
\ \ \ y-y_0=y'(x_0)(x-x_0)\
\ \ \ y-y_0=2x_0(x-x_0)\
\ \ \ y-y_0=2x_0x-2x_0^2\
\ \ \ y-x_0^2=2x_0x-2x_0^2\
\ \ \ y=2x_0x-x_0^2\
\text{We know that our lines must pass through}\
\ \ \ (x,y)=(0,-4)\
\text{So we may reduce the tangent line equation as follows.}\
\ \ \ -4=2(0)x_0-x_0^2\
\ \ \ 4=x_0^2\
\ \ \ x_0=\pm2\
\text{Now that we have the solutions for }x_0\text{ we can find }y_0.\
\ \ \ y_0 = y(\pm2)=(\pm2)^2=4\
\text{Therefore our solution is...}\
\ \ \ (x_0,y_0)\in\{(-2,4),(2,4)\}
$$