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.
Best Answer
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. $y-y_0=m(x-x_0)$.
$$y-x_p^2=(-1)(x-x_p)$$ $$y=-x+x_p+x_p^2$$