Question

Find the equation of the tangent to the following parabola
a) y^2=12x ; the tangent having slope -1/3

Answer 1

The first thing you need to know is - where does the parabola have a slope -1/3, so that the tangent line would have the slope -1/3 ?

Answer 2

\(y=\pm\sqrt{12x}\)

\(y'=\pm\dfrac{6}{\sqrt{12x}}\)

\(\dfrac{-1}{3}=\pm\dfrac{6}{\sqrt{12x}}\)

Answer 3

we would need to take the bottom half of the parabola.

Answer 4

where did u get that 6 frm \[\frac{ 6 }{ \sqrt{12x} }\]

Answer 5

-1 / 3 = -6 / √(12x)

1 / 3 = 6 / √(12x)

√(12x) = 18

12x = 324

x = 324/12 = 27

Answer 6

y' = 1/ ( 2 √(12x)) * 12 = 6/ √(12)

Answer 7

and ± comes from taking the square root on both sides

Answer 8

ok

Answer 9

so I get that it is going through

y = -(1/3) - 9

https://www.desmos.com/calculator/vpujilagiw

Answer 10

so thats the equation?

Answer 11

I forgot the x by the slope, oopsis

Answer 12

-(1/3)x - 9