Question 1156951
Find the equation of the line tangent to the curve at the point defined by the given value of t.
{{{x=16sint}}}, {{{y=4cost}}}, {{{t=pi/4}}}
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{{{x=16sint}}} ---> @ pi/4 = 8sqrt(2)
{{{y= 4cost}}} ---> @ pi/4 = 2sqrt(2)

dx/dt = 16cos(t) ---> @ pi/4 = 8sqrt(2)
dy/dt = -4sin(t) ---> @ pi/4 = -2sqrt(2)
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dy/dx = -sin(t)/(4cos(t)) = -tan(t)/4
@pi/4:  dy/dx = -1/4
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{{{y - 2sqrt(2) = (-1/4)*(x - 8sqrt(2))}}}
{{{y - 2sqrt(2) = (-x/4) + 2sqrt(2))}}}
{{{y = (-x/4) + 4sqrt(2))}}}