find the parametric equation of the line that is tangent to the parabola y=x^2 at the point(-2 , 4)
done, look back
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 determine the slope of the tangent line at that point.
The slope of the tangent line can be found by taking the derivative of the equation y = x^2 with respect to x. Let's find the derivative:
dy/dx = 2x
Plugging in the x-coordinate of the point of tangency (-2), we can determine the slope of the tangent line at that point:
m = dy/dx = 2(-2) = -4
Now that we have the slope of the tangent line, we can use the point-slope form of a linear equation to write the equation of the line:
y - y1 = m(x - x1)
Substituting (-2, 4) for (x1, y1) and -4 for m, we get:
y - 4 = -4(x + 2)
Expanding the equation:
y - 4 = -4x - 8
Simplifying:
y = -4x - 4
Therefore, the parametric equation of the line that is tangent to the parabola y = x^2 at the point (-2, 4) is:
x = t
y = -4t - 4, where t represents any real number.
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 determine the slope of the tangent line at that point.
Firstly, we can find the derivative of the parabola equation y = x^2 with respect to x:
dy/dx = 2x
Next, we substitute the x-coordinate of the given point into the derivative to find the slope at that point:
dy/dx = 2(-2) = -4
So, the slope of the tangent line at the point (-2, 4) is -4.
Now, we can use the point-slope form of a linear equation to determine the equation of the tangent line:
y - y1 = m(x - x1)
where m is the slope and (x1, y1) are the coordinates of the point of tangency.
Plugging in the values:
y - 4 = -4(x - (-2))
Simplifying:
y - 4 = -4(x + 2)
y - 4 = -4x - 8
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 the parameter that allows us to generate points on the line.