Find the equation of the normal to the curve y = 1/( x - 2 ) at the point P ( 1 , -1).
Find also the coordinates of the point Q where this normal meets the curve again
To find the equation of the normal to the curve y = 1/( x - 2 ) at the point P ( 1 , -1), we need to find the slope of the tangent line at P and then find the negative reciprocal of this slope to get the slope of the normal line.
1. Find the slope of the tangent line:
To find the slope of the tangent line at point P (1, -1), we need to find the derivative of the function y = 1/(x - 2).
Using the quotient rule, we have:
dy/dx = [(d/dx)(1)(x - 2) - 1(d/dx)(x - 2)] / (x - 2)^2
= [-1 - 1] / (x - 2)^2
= -2 / (x - 2)^2
Evaluate this at x = 1 to find the slope at point P:
m = -2 / (1 - 2)^2
m = -2
2. Find the slope of the normal line:
The slope of the normal line is the negative reciprocal of the slope of the tangent line. Therefore, the slope of the normal line is 1/2.
3. Find the equation of the normal line using the point-slope form:
Using the point-slope form y - y1 = m(x - x1), where (x1, y1) is the given point P (1, -1) and m is the slope of the normal line (1/2):
y - (-1) = (1/2)(x - 1)
y + 1 = (1/2)(x - 1)
2y + 2 = x - 1
x - 2y = 3
The equation of the normal to the curve at point P (1, -1) is x - 2y = 3.
To find the coordinates of the point Q where this normal meets the curve again, we substitute the equation of the normal into the equation of the curve and solve for x and y.
Substitute x - 2y = 3 into y = 1/(x - 2):
1/(x - 2) - 2y = 3
1/(x - 2) = 2y + 3
1 = (x - 2)(2y + 3)
1 = 2xy + 3x - 4y - 6
Simplify by combining like terms:
2xy - 4y + 3x = 7
Rearrange to isolate y:
2xy - 4y = -3x + 7
-4y = -3x + 7 - 2xy
y = (3x - 7) / (2x - 4)
To find the coordinates of the point Q, substitute this value of y back into the equation of the normal:
x - 2[(3x - 7) / (2x - 4)] = 3
x - (6x - 14) / (2x - 4) = 3
(x(2x - 4) - 6x + 14) / (2x - 4) = 3
(2x^2 - 4x - 6x + 14) / (2x - 4) = 3
(2x^2 - 10x + 14) / (2x - 4) = 3
Multiply both sides of the equation by (2x - 4) to remove the denominator:
2x^2 - 10x + 14 = 3(2x - 4)
2x^2 - 10x + 14 = 6x - 12
2x^2 - 16x + 26 = 0
This equation does not factor nicely, so we can solve it using the quadratic formula:
x = (-(-16) +/- sqrt((-16)^2 - 4(2)(26))) / 2(2)
x = (16 +/- sqrt(256 - 208)) / 4
x = (16 +/- sqrt(48)) / 4
x = (16 +/- 4sqrt(3)) / 4
x = 4 +/- sqrt(3)
Therefore, the x-coordinates of points Q are 4 + sqrt(3) and 4 - sqrt(3). To find the corresponding y-coordinates, substitute these x-values back into the equation of the normal:
For x = 4 + sqrt(3):
y = (3(4 + sqrt(3)) - 7) / (2(4 + sqrt(3)) - 4)
y = (12 + 3sqrt(3) - 7) / (8 + 2sqrt(3) - 4)
y = (5 + 3sqrt(3)) / (4 + 2sqrt(3))
Similarly, for x = 4 - sqrt(3):
y = (3(4 - sqrt(3)) - 7) / (2(4 - sqrt(3)) - 4)
y = (12 - 3sqrt(3) - 7) / (8 - 2sqrt(3) - 4)
y = (5 - 3sqrt(3)) / (4 - 2sqrt(3))
Therefore, the coordinates of the point Q where the normal meets the curve again are approximately:
Q₁ ≈ (4 + sqrt(3), (5 + 3sqrt(3)) / (4 + 2sqrt(3)))
Q₂ ≈ (4 - sqrt(3), (5 - 3sqrt(3)) / (4 - 2sqrt(3)))