find the parametric equation of the line that is tangent to the parabola y=x^2 at the point(-2 , 4)
dy/dx = 2x
at (-2,4), slope = -4
equation of tangent:
y-4 = -4(x+2)
y = -4x -4
let x = t
y = -4t - 4
Why did the line go to the circus? Because it wanted to find its tangent to the parabola!
To find the parametric equation of the line tangent to the parabola y = x^2 at the point (-2, 4), we first need to find the derivative of the parabola.
The derivative of y = x^2 is dy/dx = 2x.
Now, let's plug in the x-coordinate of the point (-2, 4) into the derivative to find the slope of the tangent line:
m = dy/dx = 2(-2) = -4.
Using the point-slope form, the equation of the tangent line is y - y₁ = m(x - x₁), where (x₁, y₁) is the point on the line (-2, 4).
The equation becomes: y - 4 = -4(x - (-2)).
Simplifying it further, we get: y - 4 = -4(x + 2).
Let's rewrite it in parametric form:
x = t,
y = -4t + 4.
So, the parametric equation of the line tangent to the parabola y = x^2 at the point (-2, 4) is x = t and y = -4t + 4.
To find the parametric equation of the line that is tangent to the parabola y = x^2 at the point (-2, 4), we need to find the slope of the tangent line at that point.
The slope of the tangent line can be found by calculating the derivative of the parabola equation. The derivative of y = x^2 with respect to x is 2x.
At the point (-2, 4), the x coordinate is -2. Therefore, the slope of the tangent line is 2(-2) = -4.
Now, let's find the equation of the tangent line using the point-slope form.
The point-slope form of a line is given by: y - y1 = m(x - x1)
Plugging in the values for m, x1, and y1, we have:
y - 4 = -4(x - (-2))
Simplifying, we get:
y - 4 = -4(x + 2)
Expanding, we have:
y - 4 = -4x - 8
Rearranging the equation, we get:
y = -4x - 4
Therefore, the parametric equation of the line tangent to the parabola y = x^2 at the point (-2, 4) is:
x = t
y = -4t - 4
where t is a parameter.
To find the parametric equation of the line tangent to the parabola y = x^2 at the point (-2, 4), you can use the concept of the derivative.
The derivative of the parabola y = x^2 represents the slope of the tangent line at any given point on the curve. So, we need to find the derivative of y = x^2 and evaluate it at x = -2.
First, let's find the derivative of y = x^2 using the power rule. The power rule states that if y = x^n, then the derivative dy/dx is given by dy/dx = nx^(n-1).
For y = x^2, applying the power rule, we get dy/dx = 2x.
Now, let's evaluate the derivative at x = -2 to find the slope of the tangent line at that point.
Substituting x = -2 into the derivative, we get dy/dx = 2(-2) = -4.
So, the slope of the tangent line at the point (-2, 4) is -4.
To find the equation of the tangent line, we have a point (-2, 4) and a slope of -4. We can use the point-slope form of a linear equation, y - y1 = m(x - x1), where (x1, y1) is a known point on the line and m is the slope.
Substituting (-2, 4) and -4 into the formula, we get y - 4 = -4(x - (-2)).
Simplifying further, we have y - 4 = -4(x + 2).
Expanding, we get y - 4 = -4x - 8.
Finally, rearranging the equation, we get y = -4x - 4.
Therefore, the parametric equation of the line tangent to the parabola y = x^2 at the point (-2, 4) is:
x = t
y = -4t - 4, where t is a parameter that varies.