Find the coordinates of the point where the tangent to the curve $\displaystyle y = x^2 + 1$ at the point (2, 5) meets the normal to the same curve at the point (1, 2).
How can I solve this problem? Please help me.
Given the curve $\displaystyle y = x^2 +1$ the slope of the tangent at $\displaystyle (2,5)$ is $\displaystyle 4$ and the slope of the normal at $\displaystyle (1,2)$ is $\displaystyle \frac{-1}{2}$.
Now write the equations of the two lines.
Find the point of their intersection.
You are probably wondering how Plato got those slopes.
$\displaystyle f(x) = x^2 +1$
Find the derivative of $\displaystyle f$
$\displaystyle f'(x) = 2x$
Slope at $\displaystyle (2, 5)$ is a simple plug and chug of the x-coordinate, so $\displaystyle f'(2) = 4$
Slope at $\displaystyle (1, 2)$ is just $\displaystyle f'(1) = 2$, so the line normal to this point is perpendicular. Perpendicular slopes are opposite signed reciprocals, so that's how Plato got $\displaystyle \frac{-1}{2}$
let f(x) = ax^2 + bx + c.
set up three simultaneous equations. since (1,2) and (2,1) are on the curve, we have:
f(1) = 2 and f(2) = 1. so form two equations from that.
but there are three unknowns, we need at least three equations. so use the information from the derivative. we are also told that f ' (2) = 0, that's your third equation