Find the equation of a TANGENT line & NORMAL line to the curve of x^2+y^2=20such that the tangent line is parallel to the line 7.5x – 15y + 21 = 0
you want a tangent line with slope = 1/2
x^2+y^2 = 20
2x + 2yy' = 0
y' = -x/y
For the tangent line, you want a point on the circle where y = -2x
for the normal, you want 2y = x
To find the equation of the tangent and normal lines to the curve given by the equation x^2 + y^2 = 20, we first need to find the point of tangency.
1. Find the derivative of the curve equation:
Differentiating both sides of x^2 + y^2 = 20 with respect to x, we get:
2x + 2yy' = 0, where y' represents the derivative of y with respect to x.
2. Rearrange the equation to solve for y':
2yy' = -2x
y' = -x/y
3. We need to find the slope of the tangent line that is parallel to the line 7.5x - 15y + 21 = 0.
The slope-intercept form of the given line is y = (7.5/15)x - 21/15, which simplifies to y = (1/2)x - 7/5.
We know that the slope of parallel lines is the same, so the slope of the tangent line is 1/2.
4. Equate the derivative of y with respect to x (-x/y) to the slope of the tangent line:
-x/y = 1/2
5. Solve for y:
Multiply both sides of the equation by y:
-x = (1/2)y
Solve for y:
y = -2x
6. Substitute the value of y into the curve equation to find the x-coordinate of the point of tangency:
x^2 + (-2x)^2 = 20
Simplify:
5x^2 = 20
Divide both sides by 5:
x^2 = 4
Taking the square root of both sides:
x = ±2
Therefore, the x-coordinate of the point of tangency is x = 2 and x = -2.
7. Substitute the values of x into the equation y = -2x to find the y-coordinate(s) of the point(s) of tangency:
For x = 2:
y = -2(2) = -4
So, the point of tangency is (2, -4).
For x = -2:
y = -2(-2) = 4
So, the other point of tangency is (-2, 4).
Now that we have the point(s) of tangency, we can find the equation(s) of the tangent and normal line(s).
Equation of the tangent line:
1. Use the point-slope form of a line: y - y1 = m(x - x1), where (x1, y1) is the point of tangency and m is the slope of the tangent line.
For the point (2, -4):
y - (-4) = (1/2)(x - 2)
Simplify:
y + 4 = (1/2)x - 1
y = (1/2)x - 5
For the point (-2, 4):
y - 4 = (1/2)(x - (-2))
Simplify:
y - 4 = (1/2)x + 1
y = (1/2)x + 5
Therefore, the equations of the tangent lines are y = (1/2)x - 5 and y = (1/2)x + 5.
2. The normal line to the curve at a given point is perpendicular to the tangent line at that point.
The slope of the normal line is the negative reciprocal (inverse) of the slope of the tangent line.
For the point (2, -4):
The slope of the tangent line is 1/2.
The slope of the normal line is -2 (negative reciprocal of 1/2).
Using the point-slope form with the point (2, -4) and slope -2:
y - (-4) = -2(x - 2)
Simplify:
y + 4 = -2x + 4
y = -2x
For the point (-2, 4):
The slope of the tangent line is 1/2.
The slope of the normal line is -2 (negative reciprocal of 1/2).
Using the point-slope form with the point (-2, 4) and slope -2:
y - 4 = -2(x - (-2))
Simplify:
y - 4 = -2x - 4
y = -2x
Therefore, the equations of the normal lines are y = -2x and y = -2x.