At what point does the normal to
y=-5-1x+3x^2 at (1, -3 ) intersect the parabola a second time?
Answer:
Note: You should enter a cartesian coordinate.
The normal line is perpendicular to the tangent line. If two lines are perpendicular their slopes are negative reciprocals -- i.e. if the slope of the first line is m then the slope of the second line is - 1/m
y = 3x^2 - x - 5
y' = 6x-1
So, at (1,-3), the slope of the normal is -1/5, and the line there is
y+3 = -1/5 (x-1)
y = -1/5 x - 14/5
So, where does the line intersect the parabola?
-1/5 x - 14/5 = 3x^2-x-5
-x-14 = 15x^2-5x-25
15x^2-4x-11 = 0
(15x+11)(x-1) = 0
So, the line intersects at x=1 and x = -11/15
Now you just need y at x = -11/15
To find the equation of the normal line to the parabola at the point (1, -3), we first need to find the slope of the tangent line to the parabola at that point.
Given the equation of the parabola y = -5 - x + 3x^2, we can find the derivative of the equation with respect to x.
dy/dx = -1 + 6x
Now, substitute x = 1 into the derivative to find the slope of the tangent line at (1, -3):
dy/dx = -1 + 6(1) = 5
So, the slope of the tangent line at (1, -3) is 5.
Since the normal line is perpendicular to the tangent line, the slope of the normal line is the negative reciprocal of the slope of the tangent line. In this case, the slope of the normal line is -1/5.
Now, we have the slope of the normal line and the point (1, -3). We can use the point-slope form of a line to find the equation of the normal line.
y - y1 = m(x - x1)
where m is the slope of the line and (x1, y1) is a point on the line, which in this case is (1, -3). Substituting the values, we get:
y - (-3) = -1/5(x - 1)
Simplifying, we have:
y + 3 = -1/5x + 1/5
Now bring y to the other side:
y = -1/5x + 1/5 - 3
y = -1/5x - 14/5
To find the point of intersection between the normal line and the parabola a second time, we need to solve the system of equations formed by the parabola and the normal line.
Substitute the equation of the normal line into the equation of the parabola:
-1/5x - 14/5 = -5 - x + 3x^2
Now, solve for x. Rearrange the equation to get a quadratic equation in standard form:
3x^2 + x - 1/5x + 14/5 + 5 = 0
3x^2 + 4/5 x + 39/5 = 0
Using the quadratic formula:
x = (-b ± sqrt(b^2 - 4ac)) / (2a)
we can substitute in the values a = 3, b = 4/5, and c = 39/5 to solve for x.
x = (-4/5 ± sqrt((4/5)^2 - 4*3*(39/5))) / (2*3)
x = (-4/5 ± sqrt(16/25 - 12/5)) / 6
x = (-4/5 ± sqrt(-56/25)) / 6
Since the discriminant is negative, there are no real solutions for x. Therefore, the normal line to the parabola y = -5 - x + 3x^2 at (1, -3) does not intersect the parabola a second time.