1. ## Parabola QUestion

The question goes:

Find the y-intercept of the normal to the parabola y=x^2 at the point
(a,a^2), where a cannot equal 0. Find the limit of this y-interept as a approaches 0. For each b>0 determine the number of normals that can be drawn from the point (0,b) to the parabola y=x^2; justify your answer.

So far i have gotton that the y intercepts= 1/2+a^2, however, i'm not sure what the limit of the y intercept as a approaches 0. Additionally, i can't figure out how to determine the amount of normals.

Thanks a Lot

Willie Li

2. Hello, aussiekid90!

Find the y-intercept of the normal to the parabola $\displaystyle y = x^2$
at the point $\displaystyle (a,a^2)$, where $\displaystyle a \neq 0$

(a) Find the limit of this y-intercept as $\displaystyle a \to 0$.

You did the hard work . . . You found that the y-intercept is: .$\displaystyle \frac{1}{2} + a^2$

As $\displaystyle a\to0$, we see that $\displaystyle \left(\frac{1}{2} + a^2\right) \to \frac{1}{2}$

(b) For each $\displaystyle b>0$, determine the number of normals that can be drawn
from the point $\displaystyle (0,b)$ to the parabola $\displaystyle y=x^2$.

We already know that the y-intercept is: .$\displaystyle b \:=\:a^2 + \frac{1}{2}$
. . Then:. $\displaystyle a \:=\:\pm\sqrt{b - \frac{1}{2}}$

In general, for every $\displaystyle b$, there are two values of $\displaystyle a$.

Therefore, there are two normals that can be drawn . . .
except: when $\displaystyle b = \frac{1}{2}$, then one normal can be drawn
. . and when $\displaystyle b < \frac{1}{2}$ when there are no normals.