Question

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

Explanation:

Step1: Recall horizontal tangent condition

A horizontal tangent line has a slope of 0. So, we set \(\frac{dy}{dx} = 0\). Given \(\frac{dy}{dx}=\frac{y + 2x}{-x + 3}\), we have \(\frac{y+2x}{-x + 3}=0\). A fraction is zero when its numerator is zero (and denominator is not zero). So, \(y+2x = 0\) (and \(-x + 3
eq0\)).

Step2: Substitute \(x = 1\) into the numerator equation

Substitute \(x = 1\) into \(y+2x = 0\). We get \(y+2(1)=0\), which simplifies to \(y+2 = 0\). Solving for \(y\), we subtract 2 from both sides: \(y=- 2\).

Step3: Check the denominator

Now, check the denominator \(-x + 3\) when \(x = 1\). Substitute \(x = 1\) into \(-x + 3\), we get \(-1 + 3=2
eq0\). So the denominator is not zero.

Step4: Verify the point on the original curve

Now, we need to check if the point \((1,-2)\) lies on the original curve \(5-xy - x^{2}=-3y\). Substitute \(x = 1\) and \(y=-2\) into the left - hand side (LHS) and right - hand side (RHS) of the equation.

• LHS: \(5-(1)(-2)-1^{2}=5 + 2-1=6\)
• RHS: \(-3(-2)=6\)

Since LHS = RHS, the point \((1,-2)\) lies on the curve.

Answer:

The coordinate point is \((1,-2)\)