Parabola Find a the value of a

[synkk]

The point P(2, 8) lies on the parabola C with equation y2=4ax. Find

a the value of a,

b an equation of the tangent to C at P.


the value of a is 8, so y^2 = 32x

when finding the tangent y = 4\sqrt{2} x^{\frac{1}{2}} so at P \frac{dy}{dx} = \frac{2\sqrt{2}}{\sqrt{2}} so the gradient of the tangent is 2, but why isn't it also -2? as when differentiating y^2 = 32x, you get also y = -4\sqrt{2} x^{\frac{1}{2}} Could anyone explain why they have only used the positive value?

 

[tiny-tim]

hi synkk!

(try using the X² button just above the Reply box )

i'm not really following you

from y² = 32x, for any positive x there are two values of y

P is (2,8) and let's call Q (2,-8), so both P and Q lie on the parabola

for P, the slope is 2, for Q it is -2 … where's the difficulty?

(btw, it's a lot easier to use implicit differentiation … 2ydy/dx = 32, which works easily for either value )​

 

[synkk]

hi synkk!

(try using the X² button just above the Reply box )

i'm not really following you

from y² = 32x, for any positive x there are two values of y

P is (2,8) and let's call Q (2,-8), so both P and Q lie on the parabola

for P, the slope is 2, for Q it is -2 … where's the difficulty?

(btw, it's a lot easier to use implicit differentiation … 2ydy/dx = 32, which works easily for either value )​

Just my stupidity really,

I was just wondering why in the solutions they didn't include -2, then you pointed out it was the other point, doh.

thanks.