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Okay here's the question:

Consider the parabola P of equation $y=x^2$, and the line $L$ of equation $y=x+6$. Let $P(x_p,y_p)$ be a point on the arc of the parabola P below L. Let A and B be the points of intersection of $P$ and $L$.

a) Find an equation of the line $L_p$ passing through $P$ which is perpendicular to $L$.

I'm having a really hard time finding the equation for the line $L_p$ that passes through the parabola $y=x^2$.
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I know that the slope is -1 which leaves me with $L_p = -x + b$.

However, i'm having trouble solving for $b$. I need a coordinate point on the graph of $L_p$ or some type of relationship to find $b$, and i cant figure out a way to work around it.

Please help. I've spent too long on this and my head hurts.

Everything is right so far, but I wouldn't turn straight to slope-intercept form.

We know that the slope of $L_p$ is -1, and it passes through a point on the parabola. This point $P$ you say has coordinates $(x_p, y_p)$. Because $y=x^2$, this point can be rewritten as $(x_p, x_p^2)$.

Then, we'll turn to point-slope form, i.e. $y-y_0=m(x-x_0)$.

$$y-x_p^2=(-1)(x-x_p)$$
$$y=-x+x_p+x_p^2$$

You have a point $p = (x_p,y_p)$, and a slope $m$. Then, the equation of the line is given in the point-slope form

$$ y-y_p = m(x-x_p) $$

Since the line $L$ has a slope of $1$, the line perpendicular has a slope of $-1$, so we must have $m=-1$ (you knew this already). The point is not chosen, but we know it is below the line, which is strictly between the points $a = (x_a,y_a)$ and $b = (x_b,y_b)$. So, $x_a < x_p < x_b$, and since $p$ is on the parabola $y=x^2$, $y_p^2 = x_p^2$. Thus, we have the line

$$ y - x_p^2 = -(x - x_p) text{ for } x_a < x_p < x_b $$

Now, the points $x_a$ and $x_b$ are the $x$-values of the intersections of $y=x^2$ and $y = x+6$, i.e. solutions to the equation $x^2 = x+6$. The solutions to this equation are $x=-2,3$. Since $a$ is the further left point, we have $x_a = -2$ and $x_b = 3$. Thus, the equation for the perpendicular line is

$$ y - x_p^2 = -(x - x_p) text{ for } -2 < x_p < 3 $$

If you want to rearrange it, you can write the slope-intercept form

$$ y = -x + (x_p+x_p^2) text{ or } y = -x+x_p(1+x_p) text{ for } -2 < x_p < 3 $$

If you have the point $(x_P,y_P)$, then you have to satisfy that with the given equation of line(which I don't think is the case with you).

In the other case, you cannot get a fixed line which satisfies the given condition, you will get infinite number of lines,as the values of $b$ will lie in a given range.

You can find the range by satisfying the equation of this line by the points of intersection of $y = x^2$ with $y = x+6$ (which are (3,9) and (-2,4) in fact).

$therefore$ the value of b ranges from 2 to 12.

Hope it is helpful:)