find an equation of the normal line to y=x^{2}-5x+4 that is parallel to the line x-3y=5.
please help me with this!!
One way to work this problem:
Implicitly differentiate the parabola with respect to y and solve for $\displaystyle -\frac{dx}{dy}$. Equate this to the slope of the given line, allowing you to find the point of intersection of the normal line and the parabola. Then use the point-slope formula to find the equation of the normal line.
What is the slope of the line x- 3y= 5? Since you want a line parallel to that, you want a line with that same slope.
What is the derivative of $\displaystyle x^2- 5x+ 4$ for general x? That will be the slope, m, of the tangent line to the curve. A normal line will have slope -1/m. For what x is the derivative equal to -1 over the slope from the first question?
You don't need the derivative of the second line in this problem. First put the second line in y = mx + b form. It has a slope of $\displaystyle 1/3 $, a line normal to that function would have a slope that's the negative reciprocal $\displaystyle -3 $. The negative reciprocal or -1/m of 1/3 is -3. To answer HallofIvy's question,we can equate the negative reciprocal of the second line $\displaystyle -3 $ with the derivative of the first function (which is what MarkFL2 was saying). $\displaystyle -3 = 2x -5 $, this gives $\displaystyle x = 1 $. So, if you plug 1 into the derivative of the parabola you'll get $\displaystyle -3 $ which is the negative reciprocal of the second line's slope. Plug 1 into the equation of the parabola and we'll get y = 0, so the point where the normal line intersects the parabola is (1, 0), and using the point slope form with the exact same slope of the second line $\displaystyle (y-0) = 1/3(x-1) $ gives $\displaystyle y = 1/3x -1/3 $.For what x is the derivative equal to -1 over the slope from the first question?
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