Locus and normal (1 Viewer)

[Getteral09]

Hi ppl! I can't seem to get the answer to this maths problem:Find the equation of the normal to the curve x^2=12y at the point (6,3). This normal meets the parabola again at point P. Find the coordinates of P.Can anyone solve this problem?
Thanks, Getteral

 

[Riviet]

y=x²/12
dy/dx=x/6
at x=6,
m=6/6
=1
.'. m_normal=-1/1
=-1
Equation of normal:
y-y₁=m(x-x₁)
y-3=-1(x-6)
y-3=-x+6
x+y-9=0 (1)
substitute y=x²/12 into (1):
x+(x²/12)-9=0
x²+12x-108=0
(x+18)(x-6)=0
x=-18, 6
take x=-18 as x=6 gives the original point that the normal passes through.
When x=-18,
-18+y-9=0
y=27
.'. P(-18 , 27)

 

[alexvincent]

You answered the post in 6mins... you're like a sniper..

 

[Getteral09]

what about : Find the equatoins of the mormals to the curve x^2 =-8y at the points (-16, -32) and (-2, -1/2). Find their point of intersection ans show that this point lies on the parabola.

Can anyone help me?

Thanks heaps!

 

[pLuvia]

y=-x²/8
dy/dx=-x/4
Normal gradient=4/x
At those points Normal gradient is: -1/4 and -2
Find the equation of the normal then solve simultaneously

 

[jed.hamers]

I got P(18, -40.5), that right?

 

[Getteral09]

I got P(18, -40.5), that right?

Yeah, that is right.