# Math Help - Tangent line to parabola

1. ## Tangent line to parabola

y=1/2x^2
given point= (4,8)

I wasn't in my class when they did this but this is what I did.

Line form of point = y=m(x-4)+8
1/2x^2= m(x-4)+8
0=m(x-4)-1/2x^2+8
0=m(x-4)- x^2 +16
0=m(x-4) - (x-4) (x+4)
Here's where I get confused.. My friend told me at this point I do grouping or something and yeah... I got this
0=(m + (x+4) (x-4)
0=(m+(4+4)) (x-4)
m=8
then I plug the m into the equation and get y=8(x-4)+8 for the equation. Can anyone check for me if it's right or wrong?

2. Originally Posted by C.C
y=1/2x^2
given point= (4,8)

I wasn't in my class when they did this but this is what I did.

Line form of point = y=m(x-4)+8
1/2x^2= m(x-4)+8
0=m(x-4)-1/2x^2+8
0=m(x-4)- x^2 +16 ... you forgot to multiply m(x-4) by 2
0=m(x-4) - (x-4) (x+4)
Here's where I get confused.. My friend told me at this point I do grouping or something and yeah... I got this
0=(m + (x+4) (x-4)
0=(m+(4+4)) (x-4)
m=8
then I plug the m into the equation and get y=8(x-4)+8 for the equation. Can anyone check for me if it's right or wrong?
$\frac{1}{2}x^2 = m(x-4)+8$

$x^2 = 2m(x-4) + 16$

$x^2 - 16 - 2m(x-4) = 0$

$(x+4)(x-4) - 2m(x - 4) = 0$

$(x-4)[(x+4) - 2m] = 0$

$x = 4$ , $m = 4$

$y = 4(x - 4) + 8$