Question

if ( y = - 5 - 3xy - 5x^{2} ) and it is known that ( \frac{dy}{dx}=\frac{-3y - 10x}{1 + 3x} ), find all coordinate points on the curve where ( x=-1 ) and the line tangent to the curve has slope ( \frac{5}{2} ), or state that no such points exist.

Explanation:

Step 1: Substitute \( \frac{dy}{dx} = \frac{5}{2} \) and \( x = -1 \) into the derivative formula

We know that \( \frac{dy}{dx} = \frac{-3y - 10x}{1 + 3x} \). Substitute \( \frac{dy}{dx}=\frac{5}{2} \) and \( x = -1 \) into this formula:

$$ \frac{5}{2}=\frac{-3y-10\times(-1)}{1 + 3\times(-1)} $$

Step 2: Simplify the right - hand side denominator and numerator

First, simplify the denominator: \( 1+3\times(-1)=1 - 3=-2 \)
The numerator: \( -3y-10\times(-1)=-3y + 10 \)
So the equation becomes:

$$ \frac{5}{2}=\frac{-3y + 10}{-2} $$

Step 3: Cross - multiply to solve for \( y \)

Cross - multiplying gives us \( 5\times(-2)=2\times(-3y + 10) \)
Simplify the left - hand side: \( -10 = 2\times(-3y + 10) \)
Divide both sides by 2: \( -5=-3y + 10 \)
Subtract 10 from both sides: \( -5-10=-3y \)
\( -15=-3y \)
Divide both sides by - 3: \( y = 5 \)

Step 4: Check if \( (x=-1,y = 5) \) satisfies the original equation

The original equation is \( y=-5-3xy-5x^{2} \)
Substitute \( x=-1 \) and \( y = 5 \) into the right - hand side:
\( -5-3\times(-1)\times5-5\times(-1)^{2}=-5 + 15-5=5 \)
The left - hand side is \( y = 5 \). Since the left - hand side equals the right - hand side, the point \( (-1,5) \) is on the curve.

Answer:

The coordinate point is \( (-1,5) \)