Math
posted by Christine .
a) determine the equation of the line that is perpendicular to the tangent to y=5x^2 at (1,5).
b) determine the equation of the line that passes through (2,2) and is parallel to the line tangent to y=3x^32x at (1,5).

(a)
y=5x²
dy/dx=5*2x=10x, slope of tangent
slope of line perpendicular to tangent
= 1/(10x)
At (1,5), slope of perpendicular line
= 1/(10*1) = 1/10
Line passing through (1,5)
(y5)=(1/10)(x1)
y =(1/10)x + 5.10
Test for perpendicularity at (1,5)
(1/10) * (10*1)
= 1 Line and tangent are perpendicular.
(b) y=3x^32x at (1,5).
dy/dx = 9x² 2 = 92 = 11 at (1,5)
Slope of tangent = 11
Line passing through (2,2) with slope of 11
(y2) = 11(x2)
y2 = 11x + 22
y = 11x + 24 
slope of 8y3x+5=10x
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