The number of tangents to the curve y^2-2x^3 -4y+8=0 that pass through (1,2) is?

Question

The number of tangents to the curve y^2-2x^3 -4y+8=0 that  pass through (1,2) is?

anonymous · Answer

Assume the first derivative exists at the given point. Take the first derivative 

$$2y \frac{ dy }{ dx }-6x ^{2}-4\frac{ dy }{ dx }=0$$
and substitute (1,2)in the above differential equation to obtain
$$4\frac{ dy }{ dx }-6-4\frac{ dy }{ dx }=0$$
which simplifies to
$$-6=0$$
now this cannot be true. Therefore there exists no derivatives at the given point.

That implies the number of tangents are zero

hope this helps :)

dls · Answer

answer is wrong :|

dls · Answer

we can't substitute (1,2) that is the fallacy in your method since (1,2) doesn't lie on the curve

anonymous · Answer

the answer might be one then. Since there can be a line parallel to Y axis going through the given point and that line will not have a defined gradient. Let me check :)

Guess how I wrote the proof is wrong, but the differential equation gives the equation of the tangent line at any point on the curve, and we can substitute the point (1,2) to find out whether there's a line going through that point, even though the point itself is not on the curve

dls · Answer

no again

anonymous · Answer

only one real solution...(and two cmplx solutions)....

dls · Answer

no :|

anonymous · Answer

maybe i messed up while solving.... anyway, the procedure is like this....
(1)find the expression for slope by taking dervative of the curve equation
(2)write the expression for the line with that particular slope
(3)give the condition that particular line touches the curve at one point
(4)now substitute point 1,2 to that equation and solve for cordinates on the curve..no of cordinates you obtain will give no of possible tangents

dls · Answer

i know the procdure

anonymous · Answer

yup...is the answer two..? tangents drawn from curves where x=-1 and x=2

whpalmer4 · Answer

$$y' = \frac{3x^2}{y-2}$$Tangent line equation through (1,2) in point slope form will be
$$y-2 = y'(x-1)$$Find the number of solutions, and that's your answer.

dan815 · Answer

basically the line from point (1,2) to a point on the curve must equal the tangent line at that point on curve

dan815 · Answer

hi okay due to laziness im using wolfram to simplify.. but im pretty sure its easy enuff to just simply this expression into y = f(x) after simplfying by completing the square and stuff
|dw:1371844608767:dw|