Prove that the curve y= x2 - 3x + 1 and x(y + 3) = 4 intersect at the right angles at their points of intersection.
y = x² - 3x + 1
x(y + 3) = 4
x = 4 / (y + 3)
y = x² - 3x + 1
y = (4 / (y + 3))² - 3(4 / (y + 3)) + 1
y = 16 / (y + 3)² - 12 / (y + 3) + 1
Multiplying everything by (y + 3)² ...
y(y + 3)² = 16 - 12(y + 3) + (y + 3)²
y(y² + 6y + 9) = 16 - 12y - 36 + y² + 6y + 9
y³ + 6y² + 9y = y² - 6y - 11
y³ + 6y² + 16y + 11 = 0
Any rational roots would be a factor of the constant over the factor of the cubed term's coefficient. This leaves ±1 or ±11. since everything is adding, 1 and 11 can't be it, so I'll test -1:
(-1)³ + 6(-1)² + 16(-1) + 11 = 0
-1 + 6(1) - 16 + 11 = 0
-1 + 6 - 5 = 0
0 = 0
TRUE
So if -1 is a root, then (y + 1) is a factor, determine the remaining quadratic to determine if there are other real roots, or just imaginary ones:
. . . . _y²_+_5y_+_11_____
y + 1) y³ + 6y² + 16y + 11
. . . . . y³ + 1y²
. . . . . -----------
. . . . . . . . 5y² + 16y + 11
. . . . . . . . 5y² + 5y
. . . . . . . . -------------------
. . . . . . . . . . . .. 11y + 11
. . . . . . . . . . . .. 11y + 11
. . . . . . . . . . . . . --------------
. . . . . . . . . . . . . . . . . . 0
Now we have:
(y + 1)(y² + 5y + 11) = 0
Now to see if we have two reals or immaginary roots:
b² - 4ac
5² - 4(1)(11)
25 - 44
-19
Negative number, so we have imaginary roots. So the only real solution is y = -1
Now we have y, we can solve for the x:
x = 4 / (y + 3)
x = 4 / (-1 + 3)
x = 4 / 2
x = 2
So the point of intersection is (2, -1).
Now to get the first derivative of each function and solve for f'(2) for each.
y = x² - 3x + 1
dy/dx = 2x - 3
dy/dx = 2(2) - 3
dy/dx = 4 - 3
dy/dx = 1
If the slope at this point is 1, I would expect the slope of the other equation to be -1 if this is true.
x(y + 3) = 4
y + 3 = 4/x
y = 4/x - 3
dy/dx = -4/x²
dy/dx = -4/(2²)
dy/dx = -4/4
dy/dx = -1
And it is.
Both slopes have opposite reciprocal slopes, so they are perpendicular at that point.
x(y + 3) = 4
x = 4 / (y + 3)
y = x² - 3x + 1
y = (4 / (y + 3))² - 3(4 / (y + 3)) + 1
y = 16 / (y + 3)² - 12 / (y + 3) + 1
Multiplying everything by (y + 3)² ...
y(y + 3)² = 16 - 12(y + 3) + (y + 3)²
y(y² + 6y + 9) = 16 - 12y - 36 + y² + 6y + 9
y³ + 6y² + 9y = y² - 6y - 11
y³ + 6y² + 16y + 11 = 0
Any rational roots would be a factor of the constant over the factor of the cubed term's coefficient. This leaves ±1 or ±11. since everything is adding, 1 and 11 can't be it, so I'll test -1:
(-1)³ + 6(-1)² + 16(-1) + 11 = 0
-1 + 6(1) - 16 + 11 = 0
-1 + 6 - 5 = 0
0 = 0
TRUE
So if -1 is a root, then (y + 1) is a factor, determine the remaining quadratic to determine if there are other real roots, or just imaginary ones:
. . . . _y²_+_5y_+_11_____
y + 1) y³ + 6y² + 16y + 11
. . . . . y³ + 1y²
. . . . . -----------
. . . . . . . . 5y² + 16y + 11
. . . . . . . . 5y² + 5y
. . . . . . . . -------------------
. . . . . . . . . . . .. 11y + 11
. . . . . . . . . . . .. 11y + 11
. . . . . . . . . . . . . --------------
. . . . . . . . . . . . . . . . . . 0
Now we have:
(y + 1)(y² + 5y + 11) = 0
Now to see if we have two reals or immaginary roots:
b² - 4ac
5² - 4(1)(11)
25 - 44
-19
Negative number, so we have imaginary roots. So the only real solution is y = -1
Now we have y, we can solve for the x:
x = 4 / (y + 3)
x = 4 / (-1 + 3)
x = 4 / 2
x = 2
So the point of intersection is (2, -1).
Now to get the first derivative of each function and solve for f'(2) for each.
y = x² - 3x + 1
dy/dx = 2x - 3
dy/dx = 2(2) - 3
dy/dx = 4 - 3
dy/dx = 1
If the slope at this point is 1, I would expect the slope of the other equation to be -1 if this is true.
x(y + 3) = 4
y + 3 = 4/x
y = 4/x - 3
dy/dx = -4/x²
dy/dx = -4/(2²)
dy/dx = -4/4
dy/dx = -1
And it is.
Both slopes have opposite reciprocal slopes, so they are perpendicular at that point.