Find an equation of the tangent line to the parabola at the given point, and find the x-intercept of the line.
17) x^2=2y , (-3,9/2)
18) y=-2x^2 , (-1,-2)
I'll do one of them. Use the same method for the other.
17) y = x^2/2 and the slope of the parabola is dy/dx = x
When x = -3, y = 9/2 and the slope is -3. A straight line going through (-3, 9/2) with slope -3 has the equation
y - 9/2 = -3[x -(-3)]
y = -3x + 9/2 -9 = -3x -9/2
The x intercept (where y = 0) is
x = -3/2
To find the equation of the tangent line to a parabola at a given point, we need to determine the slope of the tangent line and the coordinates of the given point. The slope of the tangent line is the derivative of the equation of the parabola evaluated at the x-coordinate of the given point.
Let's solve each question step by step:
17) The equation of the parabola is x^2 = 2y. To find the equation of the tangent line at the point (-3, 9/2), we first need to find the slope of the tangent line. This involves taking the derivative of the equation x^2 = 2y with respect to x.
Differentiating both sides with respect to x gives us:
2x = 2(dy/dx)
To find dy/dx, we can divide both sides by 2:
dy/dx = x
Now we can substitute the x-coordinate of the given point (-3) into the derivative to find the slope:
m = dy/dx = (-3)
The slope of the tangent line is -3.
Next, we substitute the given point (-3, 9/2) and the slope (-3) into the point-slope form of a linear equation:
y - y1 = m(x - x1)
Plugging in the values, we get:
y - (9/2) = -3(x - (-3))
Simplifying,
y - (9/2) = -3(x + 3)
y - (9/2) = -3x - 9
Finally, rearranging the equation to the standard form:
3x + y = (9/2) - 9
3x + y = -15/2
Thus, the equation of the tangent line to the parabola x^2 = 2y at the point (-3, 9/2) is 3x + y = -15/2. To find the x-intercept of the line, we set y = 0 in the equation:
3x + 0 = -15/2
3x = -15/2
x = -15/6
The x-intercept of the line is x = -15/6.
18) The equation of the parabola is y = -2x^2. To find the equation of the tangent line at the point (-1, -2), we first need to find the slope of the tangent line. Taking the derivative of the equation y = -2x^2 with respect to x gives us:
dy/dx = -4x
Now we substitute the x-coordinate of the given point (-1) into the derivative to find the slope:
m = dy/dx = -4(-1) = 4
The slope of the tangent line is 4.
Next, we substitute the given point (-1, -2) and the slope (4) into the point-slope form of a linear equation:
y - y1 = m(x - x1)
Plugging in the values, we get:
y - (-2) = 4(x - (-1))
Simplifying,
y + 2 = 4(x + 1)
y + 2 = 4x + 4
Rearranging the equation to the standard form:
4x - y = -2 - 4
4x - y = -6
Thus, the equation of the tangent line to the parabola y = -2x^2 at the point (-1, -2) is 4x - y = -6. To find the x-intercept of the line, we set y = 0 in the equation:
4x - 0 = -6
4x = -6
x = -6/4
The x-intercept of the line is x = -6/4.