SOLUTION: Determine equations for two lines that pass through (1, -5) and are tangent to the graph y= x^2 -2.

Algebra ->  Vectors -> SOLUTION: Determine equations for two lines that pass through (1, -5) and are tangent to the graph y= x^2 -2.       Log On


   

Question 1067725: Determine equations for two lines that pass through (1, -5) and are tangent to the graph y= x^2 -2.
Answer by josgarithmetic(39617) About Me  (Show Source):
You can put this solution on YOUR website!
%28dy%2Fdx%29%28x%5E2-2%29
m=2x, slope for any line tangent to y=x^2-2.


y%2B5=m%28x-1%29, a line equation
y=m%28x-1%29-5
y=%282x%29%28x-1%29-5
y=2x%5E2-2x-5, a line expected to be tangent to y=x^2-2.

Expression for the parabola, MINUS the expression for the tangent line, should be 0, where the line and parabola intersect. This line is not above the parabola.

x%5E2-2-%282x%5E2-2x-5%29=0, to solve for finding either of the intersection points.

x%5E2-2-2x%5E2%2B2x%2B5=0
-x%5E2%2B2x%2B3=0
x%5E2-2x-3=0
%28x%2B1%29%28x-3%29=0


If x=-1, then y=(-1)^2-2=-1.
Point on parabola, (-1,-1).

If x=3, then y=(3)^2-2=9-2=7.
Point on parabola, (3,7).


The Two Possible Lines Tangent to y=x^2-2, and containing (1,-5):
-
Line containing (1,-5) and (-1,-1).
slope %28-1%2B5%29%2F%28-1-1%29=4%2F%28-2%29=-2;
Equation starting point-slope form,
highlight%28y%2B5=-2%28x-1%29%29

-
Line containing (1,-5) and (3,7).
slope %2812%2F2%29=6;
Equation,
highlight%28y%2B5=6%28x-1%29%29