Yahoo Answers is shutting down on 4 May 2021 (Eastern Time) and, as of 20 April 2021 (Eastern Time), the Yahoo Answers website will be in read-only mode. There will be no changes to other Yahoo properties or services, or your Yahoo account. You can find more information about the Yahoo Answers shutdown and how to download your data on this help page.
Find an equation of the normal line to the curve y = √x that is parallel to the line 6x + y = 1.?
- alexLv 72 months ago
gradient of the normal to the curve = -6
--->gradient of the tangent = 1/6
- ?Lv 62 months ago
y = sqrt(x)
y' = -0.5/sqrt(x)
6x + y = 1
Gradient of normal 1/6
1/6 = -0.5/sqrt(x)
x = (6*-0.5)^2
- ?Lv 72 months ago
The slope of the given line is -6, so you need a point on y = sqrt(x) where the derivative is +1/6.
The derivative is dy/dx = 1/[2*sqrt(x)], and if you set this equal to 1/6, the value of x must be 9.
So the line you're looking for goes through the point (9,3), and its equation is 6x + y = 57.