If minus, 5, x, y, minus, y, squared, minus, 3, x, cubed, equals, 1−5*xy*−*y*2−3*x*3=1 and it is known that start fraction, d, y, divided by, d, x, end fraction, equals, start fraction, 5, y, plus, 9, x, squared, divided by, minus, 5, x, minus, 2, y, end fraction, comma*dxdy*​=−5*x*−2*y*5*y*+9*x*2​, find all coordinate points on the curve where x, equals, 2*x*=2 and the line tangent to the curve is vertical, or state that no such points exist.

To find all coordinate points on the curve where \( x = 2 \) and the line tangent to the curve is vertical, we need to analyze the derivative and check for conditions when the tangent line is vertical.

Given:
1. The implicit equation for the curve:
   \[
   -5xy - y^2 - 3x^3 = 1
   \]
   
2. The derivative \( \frac{dy}{dx} \):
   \[
   \frac{dy}{dx} = \frac{5y + 9x^2}{-5x - 2y}
   \]

To identify points where the tangent line is vertical, we look for values of \( x \) and \( y \) where \( \frac{dx}{dy} \) is undefined. This occurs when the denominator of \( \frac{dy}{dx} \) is zero.

### Step 1: Set up the denominator condition for a vertical tangent
For a vertical tangent:
\[
-5x - 2y = 0
\]

### Step 2: Substitute \( x = 2 \) into the equation
Substitute \( x = 2 \) into \( -5x - 2y = 0 \):
\[
-5(2) - 2y = 0
\]
\[
-10 - 2y = 0
\]
\[
-2y = 10
\]
\[
y = -5
\]

Thus, the point \( (x, y) = (2, -5) \) is a candidate for a vertical tangent.

### Step 3: Verify that \( (2, -5) \) lies on the original curve
Substitute \( x = 2 \) and \( y = -5 \) into the implicit equation to check if it satisfies:
\[
-5(2)(-5) - (-5)^2 - 3(2)^3 = 1
\]
Calculate each term:
1. \(-5(2)(-5) = 50\)
2. \((-5)^2 = 25\)
3. \(3(2)^3 = 3 \cdot 8 = 24\)

Substitute these values:
\[
50 - 25 - 24 = 1
\]
\[
1 = 1
\]

The point \( (2, -5) \) satisfies the equation of the curve, so it is valid.

### Conclusion
The only coordinate point on the curve where \( x = 2 \) and the tangent line is vertical is:
\[
(2, -5)
\]

Would you like further details on any specific step or have additional questions?

Here are some related questions to consider:
1. How can we confirm that the tangent is indeed vertical at \( (2, -5) \)?
2. What would be the process to find points where the tangent line is horizontal?
3. How does implicit differentiation work in deriving \( \frac{dy}{dx} \) for implicit functions?
4. How would the solution change if the equation or the derivative expression were modified?
5. Can we determine the nature (concave up or down) of the curve at \( (2, -5) \)?

**Tip:** In implicit differentiation, always check if specific conditions, such as zero in the denominator or numerator, yield meaningful points on the curve.

If -5xy - y^2 - 3x^3 = 1 and it is known that dy/dx = (5y + 9x^2) / (-5x - 2y), find all coordinate points on the curve where x = 2 and the line tangent to the curve is vertical, or state that no such points exist.

Discover how to find points on the curve -5xy - y^2 - 3x^3 = 1 where x = 2 and the tangent line is vertical. This solution involves implicit differentiation and the condition for vertical tangency, suitable for advanced high school students in Grades 11-12.

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