Write the equation of the tangent line to the parabola y^2=-1/2x at the point (-1,-2)
To find the equation of the tangent line to the parabola y^2 = -1/2x at the point (-1, -2), we need to find the slope of the tangent line and use the point-slope form of the equation of a line.
Step 1: Take the derivative of the equation y^2 = -1/2x with respect to x to find the derivative of y with respect to x.
Let's rewrite the equation in terms of y:
y^2 = -1/2x
y = ±√(-1/2x)
Now we can take the derivative:
dy/dx = ±d/dx(√(-1/2x))
dy/dx = ±(1/2)(-1/2x)^(-1/2)(-1/2) [Applying the chain rule]
dy/dx = ±1/4√(1/2x^3)
dy/dx = ±1/(4√2x^(3/2))
Step 2: Evaluate the derivative at the given point (x₀, y₀) = (-1, -2).
dy/dx = ±1/(4√2(-1)^(3/2))
dy/dx = ±1/(4√2(-1))
dy/dx = ±1/(4√2)
Step 3: Use the point-slope form of the equation of a line to write the equation of the tangent line.
Using the point-slope form y - y₀ = m(x - x₀), where (x₀, y₀) = (-1, -2) and m = ±1/(4√2):
For the positive slope:
y + 2 = (1/(4√2))(x + 1)
For the negative slope:
y + 2 = (-1/(4√2))(x + 1)
Simplifying these equations will give you the final equations of the tangent lines to the parabola y^2 = -1/2x at the point (-1, -2).