Help with parabola problem?
I'm not sure where to start with this problem. I'm guessing this is using precalc math considering im going into calc 1 but I havent started yet.
The parabola y=x^2 has slope 6 at the point (3,9). Thus, the tangent line to the curve at this point has slope 6. Find all point where the line perpendicular to this tangent line intersects the parabola.
Comments
y = 6x + c is tangent line
therefore perpendicular line
y = -(1/6)x + c
we know
9 = -1/2 + c ie c = 19/2
y = -(1/6)x + 19/2
we want
x^2 = -(1/6)x + 19/2
ie
6x^2 + x - 57 = 0
since this line passes through (3, 9) we know x = 3 is one solution
ie (x - 3)(6x + 19) = 0
x = -19/6
y = 361/36
other point is (-19/6, 361/36)
Slope of the perpendicular line to the tangent given = - 1/6
Equation of the perpendicular line that goes through (3,9)
y-9 = (-1/6)(x-3)
6y-54 = -x + 3
6y+x-57=0
Now solve this line & parabola for intersection points
Then, x^2 = (57-x)/6
6x^2 +x -57=0
(x-3)(6x+19)=0
x=3 or x= -19/6
then y=9 or y =361/36
Therefore next intersection point (-19/6, 361/36)
y = x²
The slope at any point (x, y) is the function's derivative evaluated for the point (x, y); that is,
y'(x) = 2x is the slope for any x value. You'll learn about the derivative as your course continues, if you haven't already.
As you said, at x = 3, the slope of the tangent line is 6. Using this slope, the given point, and the point-slope formula you presumably learned prior to this class, you can find the equation of the tangent line:
y - 9 = 6 (x - 3)
y = 6x - 9
Now you have to find the equation of the line perpendicular to the tangent line. You'll learn that this perpendicular line is called the "normal line."
Lines that are perpendicular to each other have opposite reciprocal slopes. Suppose one line's slope is a/b. The slope of its perpendicular counterpart is then -b/a.
In this case, the slope of the tangent line is 6, so the slope of the perpendicular line is -1/6. This line also contains the point (3, 9), so using the point-slope formula again,
y - 9 = -1/6 (x - 3)
y = -1/6 x + 19/2
To find the intersection points of the curve y = x² and the line y = -1/6 x + 19/2, set the two equations equal to each other.
x² = -1/6 x + 19/2
x² + 1/6 x - 19/2 = 0
1/6 (6x² + x - 57) = 0
6x² + x - 57 = 0
Using the quadratic formula,
x = 36/12 and x = -38/12, or
x = 3 and x = -19/6
Substituting these x-values into either equation gives you the points
(x, y) = (3, 9) and (x, y) = (-19/6, 361/36).
Equation of perpendicular line:
y - 9 = -1/6 * (x - 3)
y = -x/6 + 1/2 + 9 => y = -x/6 + 19/2
-x/6 + 19/2 = x^2
-x + 57 = 6x^2
6x^2 + x - 57 = 0
x = (-1 +/- â(1 - 4(6)(-57))/12 = (-1 +/- 37)/12
So the points are (3, 9)(Obvious) and the not so obvious point (-19/6, 361/36).