Use implicit differentiation to find an equation of the tangent line to the curve at the given point.

x2 + 2xy − y2 + x = 17, (3, 5)
(hyperbola)

2x + 2y + 2xy' - 2yy' + 1 = 0

(2x+2y+1) + y'(2x-2y) = 0
y' = (2x+2y+1)/(2y-2x)
y' at (3,5) = (6+10+1)/(10-6) = 17/4

so, you want the line through (3,5) with slope 17/4:

(y-5) = 17/4 (x-3)

Oh, a hyperbola. That's quite a "hip" curve!

To find the equation of the tangent line to the hyperbola at the point (3, 5), let's use implicit differentiation.

First, let's differentiate the equation of the hyperbola with respect to x. We have:

2x + 2y(dy/dx) - 2y(dy/dx) - 2y + 1 = 0.

Now, let's find the slope of the tangent line by substituting x = 3 and y = 5 into the above equation:

2(3) + 2(5)(dy/dx) - 2(5)(dy/dx) - 2(5) + 1 = 0.

Simplifying this equation, we get:

6 + 10(dy/dx) - 10(dy/dx) - 10 + 1 = 0,
6 - 9 = 0,

-3 = 0.

Oops! Looks like we made a mistake somewhere. Either I made a joke that divided by zero, or we took a wrong turn in our calculations. Let me try again!

So, I'm afraid my jokes didn't quite brighten up the calculation. It seems there was an error along the way. But fear not! I'm here to entertain, even if it means taking another shot at the math.

To find the equation of the tangent line to the hyperbola at the point (3, 5) using implicit differentiation, follow these steps:

Step 1: Differentiate both sides of the equation with respect to x. Treat y as a function of x.
d/dx (x^2 + 2xy - y^2 + x) = d/dx (17)

Step 2: Apply differentiation rules on each term separately.
2x + 2y(dy/dx) - 2y(dy/dx) - 2yy' + 1 = 0

Step 3: Simplify the equation by combining like terms.
2x - 2yy' + 1 = 0

Step 4: Solve for dy/dx to find the slope of the tangent line.
2yy' = 2x - 1
y' = (2x - 1) / (2y)

Step 5: Substitute the coordinates of the given point into the equation obtained in Step 4 to find the value of dy/dx at that point.
y' = (2(3) - 1) / (2(5))
y' = 5/10
y' = 1/2

Step 6: Now we have the slope of the tangent line, and we also have a point on the line (3, 5). Use the point-slope form to find the equation of the tangent line.
y - y1 = m(x - x1)
y - 5 = (1/2)(x - 3)

Step 7: Simplify the equation obtained in Step 6 to find the final equation of the tangent line.
y - 5 = (1/2)x - 3/2
y = (1/2)x + 7/2

Therefore, the equation of the tangent line to the hyperbola at the point (3, 5) is y = (1/2)x + 7/2.

To find the equation of the tangent line to the curve given by the equation x^2 + 2xy - y^2 + x = 17, we can use implicit differentiation.

Step 1: Differentiate both sides of the equation with respect to x.
d/dx (x^2 + 2xy - y^2 + x) = d/dx (17)

Step 2: For each term on the left-hand side, we apply the chain rule and the product rule.
2x + 2y * dy/dx - 2y * dy/dx - 2y + 1 = 0

Step 3: Simplify the equation by combining like terms.
2x - 2y + 2y * dy/dx = -1

Step 4: Solve for dy/dx (the derivative of y with respect to x).
2y * dy/dx = 1 - 2x + 2y

Step 5: Divide through by 2y to solve for dy/dx.
dy/dx = (1 - 2x + 2y) / 2y

Step 6: Now, we need to find the slope of the tangent line at the point (3, 5). Substitute x = 3 and y = 5 into the equation for dy/dx.
dy/dx = (1 - 2(3) + 2(5)) / 2(5) = (1 - 6 + 10) / 10 = 5/10 = 1/2

Step 7: The slope of the tangent line is 1/2. We can use the point-slope form of a line to find the equation of the tangent line.
Using the point (3, 5) and the slope 1/2, we have:
y - y1 = m(x - x1)
y - 5 = 1/2(x - 3)

Step 8: Simplify the equation to get the equation of the tangent line.
y - 5 = 1/2x - 3/2

Multiplying through by 2 to eliminate the fraction:
2(y - 5) = x - 3

Simplifying further, we get:
2y - 10 = x - 3

Finally, rearrange to get the equation of the tangent line in standard form:
x - 2y = -7

Therefore, the equation of the tangent line to the curve x^2 + 2xy - y^2 + x = 17 at the point (3, 5) is x - 2y = -7.