# Thread: help with proof

1. ## help with proof

for the curve C x3 + xy + 2y3 =k , the y co-ordinates of the tangent parallel to the y axis is given by

216y6 +4y3 +k = 0

Show that k<= 1/54

2. ## Re: help with proof

Let $m = y^3$, so the second equation becomes

$216m^2 + 4m + k = 0$

The discriminant of this quadratic is $16 - 4(216)(k) = 16 - 864k$

For the quadratic to even have real solutions for m, the discriminant must be non-negative, i.e. $16 - 864k \ge 0 \Rightarrow k \le \frac{1}{54}$. No calculus needed.

3. ## Re: help with proof

but its a tangent shouldnt you would get a repeated root because its a tangent so should you use the the discrm = 0 to find k than puttin >=0

4. ## Re: help with proof

Can you prove that? Note that the quadratic is now in terms of $m$, not $y$.

5. ## Re: help with proof

thanks so you cannot assum repeated for m, thanks

6. ## Re: help with proof

one more find the possible values of k for which the line x=-6 is a tangent to C

No problem

8. ## Re: help with proof

For the other question, differentiate both sides of the original equation:

$3x^2 + (y + x\frac{dy}{dx}) + 6y^2 \frac{dy}{dx} = 0$

$\frac{dy}{dx} (x + 6y^2) + 3x^2 + y = 0$

$\frac{dy}{dx} = -\frac{3x^2 + y}{x + 6y^2}$

If the line $x = -6$ is a tangent at the point (x,y), it means that x = -6 and dy/dx is undefined, i.e. the denominator of dy/dx is zero. This occurs only when $y = \pm 1$. So try substituting (1,6) and (-1,6) into your original equation and find k.