Question

if ( y^{3}+2x + 5-xy = 0 ) and it is known that ( \frac{dy}{dx}=\frac{-2 + y}{3y^{2}-x} ), find all coordinate points on the curve where ( y=-1 ) and the line tangent to the curve has slope ( -\frac{3}{5} ), or state that no such points exist.

Explanation:

Step1: Substitute \( y = -1 \) and \( \frac{dy}{dx}=-\frac{3}{5} \) into the derivative formula

We know that \( \frac{dy}{dx}=\frac{-2 + y}{3y^{2}-x} \). Substitute \( y=-1 \) and \( \frac{dy}{dx}=-\frac{3}{5} \) into this formula:

$$ -\frac{3}{5}=\frac{-2+(-1)}{3(-1)^{2}-x} $$

Step2: Simplify the right - hand side numerator and denominator

First, simplify the numerator of the right - hand side: \( -2+(-1)=-3 \).
The denominator of the right - hand side: \( 3(-1)^{2}-x = 3 - x \).
So the equation becomes:

$$ -\frac{3}{5}=\frac{-3}{3 - x} $$

Step3: Cross - multiply to solve for \( x \)

Cross - multiplying gives: \( -3(3 - x)=-3\times5 \)
Expand the left - hand side: \( -9 + 3x=-15 \)
Add 9 to both sides of the equation: \( 3x=-15 + 9=-6 \)
Divide both sides by 3: \( x=-2 \)

Step4: Check if \( (x,y)=(-2,-1) \) satisfies the original curve equation

The original curve equation is \( y^{3}+2x + 5-xy = 0 \).
Substitute \( x = -2 \) and \( y=-1 \) into the left - hand side of the equation:

$$ (-1)^{3}+2(-2)+5-(-2)(-1) $$

Calculate each term:
\( (-1)^{3}=-1 \), \( 2(-2)=-4 \), \( (-2)(-1)=2 \)
Then: \( -1-4 + 5-2=-2
eq0 \)

Answer:

No such points exist.