Posted by gt on Tuesday, March 1, 2011 at 6:18pm.
y=-2x-3, y=6x-11
We are given a fixed point through which the tangents pass.
On the curve (parabola), the slope of the tangent is dy/dx = 2x at the point (x,x²-2).
What we need here is a line that passes through a point (x,f(x)) and (1,-5) with a slope of 2x.
Thus we form the equation:
(y2-y1)/(x2-x1)=slope
(x²-2 - (-5))/(x-1) = 2x
We solve for x=3 or x=-1.
Substitute into equation
(y-y0)=m(x-x0)
to get the formulae alx posted.
y = x^2 - 2
f' = 2x
P(1,-5)
f' = slope m = 2(1) = 2
Equation of the line tangent at P(1,-5)
m = 2
y = mx + b
-5 = 2(1) + b
-5 = 2 + b
b = -7
Equation of the tangent line is,
y = 2x - 7
2x - y = 7
Equation of the normal line at P(1,-5) (The normal line is the line that is perpendicular to the tangent line at the point of tangency).
m = -1/2
y = mx + b
-5 = -1/2(1) + b
-5 = -1/2 + b
b = -10/2 + 1/2
b = -9/2
Equation of the normal line
y = -1/2 x - 9/2
2y = -x - 9
x + 2y = -9
I think I read the question wrong.
I thought the question wanted the slope of the tangent of f(x) at (1,-5) and the two equations at this point.
Ignore my response.
I completely misunderstood the question.(: