Parametrics (1 Viewer)

[GaganDeep]

find the equation of the common tangent to the parabola y=(x+1)^2 and y=x(x-4)
and

find the gradients of the common tangents to the parabola y=x(x-4) and circle (x-2)^2 + y^2 =4

Thanks

 

[Mountain.Dew]

okay...when it means common tangent, it means that the gradient is the same. note that dy/dx = m = gradient

so: y₁ =(x+1)^2, (dy/dx)₁ = 2(x+1)

and y₂=x(x-4)=x^2-4x, (dy/dx)₂=2x-4

NOW, to find the common tangent, let (dy/dx)₁ = (dy/dx)₂ ==> you will get a value for x. <== THAT IS THE GRADIENT, m.

so, u know that any line is y=mx + c...

now, to find c, plug y=mx + c into the first y, and the 2nd y, already knowing what your m. then, solve simulatenously to find c, eliminating x in the process.

if ur still not sure bout the 2nd part, please PM or reply to this thread!

 

[GaganDeep]

Yea i think now i get it. thanks.

 

[Mountain.Dew]

pleasure

 

[ckckckck]

ok but im confused now.. haha sorry fat chime..but when you make
2(x+1)=2x-4, the x's cancel out dont they? so ur left with 0=-6 or something. damn i feel stupid rite now. please help

 

[abcd9146]

im not sure if what Mountain.Dew did is correct, but this what i would have done.
you have 2 lines:
y=(x+1)²
y=x²-4x

suppose that on the first graph, at P the x value is a
y=(x+1)²
y'=2x+2
at x=a, y=a²+2a+1
y'=2a-2

eqn of tangent, m=2a-2, (a,a²+2a+1)
y-a²-2a-1=(2a-2)(x-a)
y-a²-2a-1=2ax-2a²-2x+a²
y=2ax-2x+2a+1

on the second graph, at Q, x value b
y=x²-4x
y'=2x-4
at x=b, y=b²-4b
y'=2b-4

eqn of tangent, m=2b-4, (b,b²-4b)
y-b²+4b=(2b-4)(x-b)
y-b²+4b=2bx-2b²-4x+b²
y=2bx+4x-4b

now we have two tangents
y=2ax-2x+2a+1
y=2bx+4x-4b

gradients are equal
2a-2=2b+4

y intercepts are equal
2a+1=-4b

solve for a & b
a=13/2
b=7/2

sub back into tangents
y=13x-2x+13+1 -> y=11x+14
y=7x+4x-14 -> y=11x+14

weee!!! we get the same tangent!!!!

BUT, when i graphed all that, it didnt make sense!??!!??! so what's wrong?

 

[rachypoo]

u guys r doing it all wrong....
check the text on page 92, itll solve everything

 

[Mountain.Dew]

im not sure if what Mountain.Dew did is correct, but this what i would have done.
you have 2 lines:
y=(x+1)²
y=x²-4x

suppose that on the first graph, at P the x value is a
y=(x+1)²
y'=2x+2
at x=a, y=a²+2a+1
y'=2a-2

eqn of tangent, m=2a-2, (a,a²+2a+1)
y-a²-2a-1=(2a-2)(x-a)
y-a²-2a-1=2ax-2a²-2x+a²
y=2ax-2x+2a+1

on the second graph, at Q, x value b
y=x²-4x
y'=2x-4
at x=b, y=b²-4b
y'=2b-4

eqn of tangent, m=2b-4, (b,b²-4b)
y-b²+4b=(2b-4)(x-b)
y-b²+4b=2bx-2b²-4x+b²
y=2bx+4x-4b

now we have two tangents
y=2ax-2x+2a+1
y=2bx+4x-4b

gradients are equal
2a-2=2b+4

y intercepts are equal
2a+1=-4b

solve for a & b
a=13/2
b=7/2

sub back into tangents
y=13x-2x+13+1 -> y=11x+14
y=7x+4x-14 -> y=11x+14

weee!!! we get the same tangent!!!!

BUT, when i graphed all that, it didnt make sense!??!!??! so what's wrong?

ah yes, that is far superior than mine. thank you abcd9146!

u guys r doing it all wrong....
check the text on page 92, itll solve everything

welcome rachypoo! with all respect, please tell us of this 'text' u speak of.