Question

At what point does the normal to y=0−1x−1x2 at (1,−2) intersect the parabola a second time?

Answer:
Note: You should enter a cartesian coordinate.

The normal line is perpendicular to the tangent line. If two lines are perpendicular their slopes are negative reciprocals -- i.e. if the slope of the first line is m then the slope of the second line is −1/m

Answer 1

multiple steps here. Have you found the expression for the slope for the tangent line to the parabola?

Answer 2

-1

Answer 3

No, the expression for the slope of the tangent line to the parabola will depend on \(x\). I want to see your work for that.

Answer 4

y-1=m(-1)+b

Answer 5

\[y=-x-x^2\]\[\frac{dy}{dx} = -1x^{1-1}-2x^{2-1} = \]

Answer 6

x-2x

Answer 7

No,
\[-1x^{1-1} - 2x^{2-1} = -x^0 - 2x^1 = -1-2x\]

So the slope of the tangent to the parabola at any given point can be found from \[m = \frac{dy}{dx} = -2x-1\]At the point \((1,-2)\), \(m = -2(1)-1 = -3\), so the slope of the normal will be \[m_{normal} = \frac{-1}{-3} = \frac{1}{3}\]Now use the point-slope formula to run a line through \((1,-2)\) with slope \(m=\frac{1}{3}\). Solve that equation for \(y\), and equate it with the equation for the parabola. Solve for the values of \(x\). The value of \(x\) which is not \(x=1\) will be the \(x\) coordinate of the other intersection. Plug it into the formula for the parabola to find the \(y\) coordinate.

Answer 8

Here's a graph of the parabola (blue), tangent at (1,-2) (purple) and normal (olive). Sorry, I seem to have forgotten the axes :-)