Derivative problem (line to curve)

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NYC

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hello all, I'm doing some exam review, and am stuck on one problem, which should be fairly easy, yet I'm not sure how to do it.

Find all lines through (-1, 3) tangent to x^3+4y^2-4x-8y+3=0

I took the slope equation and set it equal to the derivative, then multiplied everything out, hoping for a cancellation somewhere...though I guess in hindsight, that probably wouldn't have happened.. Anyway I got as far as:
8y^2-32^y-3x^3+x^2+4x+20=0

any help would be appreciated
 
\(\displaystyle x^{3} = 4*x + 8*y - 4y^{2} - 3\)

Substituting into your equation gives

\(\displaystyle 11 - 4y^{2} + 8x -8y = 0\)

Are we getting anywhere?
 
I am trying:
Find the derivitive dy/dx useing an implicit derivitive.
I get dy/dx = (-3x²+4)/8(y-1)
I am using (y-3)=m(x+1) with m = dy/dx from above.
I am having trouble with it but I think that is the method.
-----------------
Gene
 
NYC said:
hello all, I'm doing some exam review, and am stuck on one problem, which should be fairly easy, yet I'm not sure how to do it.

Find all lines through (-1, 3) tangent to x^3+4y^2-4x-8y+3=0

I took the slope equation and set it equal to the derivative, then multiplied everything out, hoping for a cancellation somewhere...though I guess in hindsight, that probably wouldn't have happened.. Anyway I got as far as:
8y^2-32^y-3x^3+x^2+4x+20=0

any help would be appreciated
Click to expand...

We have x^3+4y^2-4x-8y+3=0

Using implicit differentiation:

3x^2+8ydy/dx-4-8dy/dx=0

(8y-8)dy/dx=4-3x^2

dy/dx=(4-3x^2)/(8y-8)

At (-1,3),

dy/dx=(4-3)/(24-8)=1/16

So the equation of our tangent line is

y-3=(1/16)(x+1)
 
Daniel_Feldman, have you assumed that the point (−1,3) is on the curve?
The point is not on the curve.

I think that the problem has a misprint in it.
If the x<SUP>3</SUP> were x<SUP>2</SUP> then there is a nice solution.
At the point (3,2) the tangent to the curve passes through (−1,3).
 
NYC said:
hello all, I'm doing some exam review, and am stuck on one problem, which should be fairly easy, yet I'm not sure how to do it.

Find all lines through (-1, 3) tangent to x^3+4y^2-4x-8y+3=0

I took the slope equation and set it equal to the derivative, then multiplied everything out, hoping for a cancellation somewhere...though I guess in hindsight, that probably wouldn't have happened.. Anyway I got as far as:
8y^2-32^y-3x^3+x^2+4x+20=0

any help would be appreciated
Click to expand...
I agree. There isn't that cancellation.

Let (p, q) be a point on the curve where the tangent passes through (-1, 3).

Equating slopes:

\(\displaystyle \mbox{ \frac{q - 3}{p + 1} = \frac{4 - 3p^2}{8q - 8}}\)

\(\displaystyle \mbox{ \Rightarrow (q - 3)(8q - 8) = (4 - 3p^2)(p + 1)\) for \(\displaystyle \mbox{p + 1 \neq 0}\) and \(\displaystyle \mbox{8q - 8 \neq 0}\).

Expanding:

\(\displaystyle \mbox{ 8q^2 - 32q + 24 = 4 + 4p - 3p^2 - 3p^3}\)

That is

\(\displaystyle \mbox{ 3p^3 + 8q^2 + 3p^2 - 32q - 4p + 20 = 0}\) [1]

And (p, q) lies on the curve:

\(\displaystyle \mbox{ p^3 + 4q^2 - 4p - 8q + 3 = 0}\) [2]

Comparing [1] and [2]

\(\displaystyle \mbox{ 3p^3 + 8q^2 + 3p^2 - 32q - 4p + 20 = 0}\)
\(\displaystyle \mbox{ p^3 + 4q^2 - 8q - 4p + 3 = 0}\)

We can't seem to eliminate a variable . . .
 
For what its worth the three tangent points are very near
(1.54,1.944)
(-1.88,0.823)
(-0.223,0.827)

That might be useful in checking your answers (if any.) I'm still using up scratch paper.
 
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