question - calculus (1 Viewer)

[Petinga]

the normal to the curve y=(x + 2) squared at the point A(-3,1) meets the curve again at B.

Find:
a)the equation of the normal
b)coordinates of B
c)angle at B and between the curve and the chord Ab

 

[nick1048]

a) y = (x+2)^2 --> x^2 + 4x + 4
dy/dx = 2x + 4
when x = -3
y' = 2(-3) + 4
y = -2
y - 1 = -2(x + 3)
equation of normal --> y = -2x - 5

b) -2x - 5 = x^2 + 4x + 4
0 = x^2 + 6x + 9
0 = (x + 3)^2
x = -3... errr.... it touches at x = -3, how can there be another point B... hmmm I'm confused. You got the answers there?

 

[FinalFantasy]

the normal to the curve y=(x + 2) squared at the point A(-3,1) meets the curve again at B.

Find:
a)the equation of the normal
b)coordinates of B
c)angle at B and between the curve and the chord Ab

y=(x+2)²
y'=2(x+2)
b)eq. of normal at A:
(y-1)\(x+3)=-1\2(-3+2)=1\2
2y-2=x+3
2y=x+5
y=x\2+5\2
b)x\2+5\2=(x+2)²
x²+4x+4-x\2-5\2=0
2x²+8x+8-x-5=0
2x²+7x+3=0
(2x+1)(x+3)=0
x=-1\2 or x=-3
.: it meets the curve at B where x=-1\2, y=9\4
C)for angle u need to use gradient at B and gradient of the chord AB
den use tan@=|m1-m2|\|1+m1m2|

 

[nick1048]

dude i honestly can't make sense from your working !?!


ok excellent I've used the tangent gradient instead of the normal gradient... wow, really must be fatigued -_-. Sorry about that FinalFantasy just got a little confused with the notation. You are right however.

 

[shafqat]

I think you may have forgotten to take the negative reciprocal for the gradient for the normal.

 

[FinalFantasy]

dude i honestly can't make sense from your working !?!


ok excellent I've used the tangent gradient instead of the normal gradient... wow, really must be fatigued -_-. Sorry about that FinalFantasy just got a little confused with the notation. You are right however.

hehe it's k, my workin here is a bit dodgy