For
y^2=8x at (3/2, 6) and x^2=24y at (6, 3)
the given points are NOT on the parabola!
for y^2=24x at (3/2, 6)
2y dy/dx = 24
dy/dx = 12/y
so for (3/2, 6) , dy/dx = 12/6 = 2
tangent equation
y - 6 = 2(x - 3/2)
y-6 = 2x - 3
y = 2x + 3
you do the last one
Find the equation to the tangent of the following parabola at the given point.
y^2=8x at (3/2, 6)
x^2=24y at (6, 3)
y^2=24x at (3/2, 6)
x^2= 5y at (5, 5)
1 answer
1 answer