What is the Equation of the Tangent to the Curve x^2 - y^2 - 6y 0 at x 4?
To find the equation of the tangent to the curve x^2 - y^2 - 6y 0 at the point where the x-coordinate is 4, we need to follow a series of steps. This process involves finding the corresponding y-coordinate, differentiating the curve's equation, and then applying the point-slope form to obtain the tangent line's equation.
Step 1: Find the y-coordinate for x 4
First, we substitute x 4 into the given curve equation:
Starting with the equation: x^2 - y^2 - 6y 0 Substituting x 4: 4^2 - y^2 - 6y 0 Simplifying: 16 - y^2 - 6y 0 Shifting terms: -y^2 - 6y 16 0 Multiplying by -1: y^2 6y - 16 0Next, we use the quadratic formula to solve for y where a 1, b 6, and c -16:
Quadratic formula: y (-b ± sqrt(b^2 - 4ac)) / (2a) Plugging in the values: y ((-6) ± sqrt(6^2 - 4*1*(-16))) / (2*1) Calculating the discriminant: y ((-6) ± sqrt(36 64)) / 2 Simplifying: y ((-6) ± sqrt(100)) / 2 Sol dy: y ((-6) ± 10) / 2 Final values: y 2 and y -8Therefore, the points on the curve are (4, 2) and (4, -8).
Step 2: Differentiate the Equation
We differentiate the curve implicitly: d/dx(x^2 - y^2 - 6y) 0
Implicit differentiation: 2x - 2y(dy/dx) - 6(dy/dx) 0 Combining terms: 2x - 2y 6(dy/dx) 0 Solving for dy/dx: dy/dx (2x) / (2y 6)Now, we evaluate dy/dx at both points:
For the point (4, 2): dy/dx (2*4) / (2*2 - 6) (8) / (4 - 6) 8 / -2 -4 For the point (4, -8): dy/dx (2*4) / (2*(-8) - 6) (8) / (-16 - 6) 8 / -22 -4/5Step 3: Write the Equation of the Tangent Line
We use the point-slope form: y - y_1 m(x - x_1) to write the equation of the tangent line for each point.
For the point (4, 2) with slope -4: y - 2 -4(x - 4) y - 2 -4x 16 y -4x 18 For the point (4, -8) with slope -4/5: y - (-8) -4/5(x - 4) y 8 -4/5x 16/5 y -4/5x 16/5 - 40/5 y -4/5x - 24/5Final Results
The equations of the tangents to the curve at the points are:
At (4, 2): y -4x 18 At (4, -8): y -4/5x - 24/5By following these steps, we can determine the equation of the tangent line to the given curve at any x-coordinate.