SOLUTION: Find the equation of the tangent to the curve y= [4/(x^2)]+1 at the point where x=a. This tangent meets the axes at P(b,0) and Q(0,b). Find the value of a and of b.

Algebra ->  Test -> SOLUTION: Find the equation of the tangent to the curve y= [4/(x^2)]+1 at the point where x=a. This tangent meets the axes at P(b,0) and Q(0,b). Find the value of a and of b.      Log On


   

Question 977879: Find the equation of the tangent to the curve y= [4/(x^2)]+1 at the point where x=a. This tangent meets the axes at P(b,0) and Q(0,b). Find the value of a and of b.
Answer by Fombitz(32388) About Me  (Show Source):
You can put this solution on YOUR website!
THe slope of the tangent line is equal to the value of the derivative at that point.
y=4x%5E%28-2%29%2B1
dy%2Fdx=-8x%5E%28-3%29=-8%2Fx%5E3
So at x=a,
m=dy%2Fdx=-8%2Fa%5E3
Also when x=a,
y=4%2Fa%5E2%2B1
Using the point-slope form of a line to find the tangent line,
y-%284%2Fa%5E2%2B1%29=%28-8%2Fa%5E3%29%28x-a%29
Multiply both sides by a%5E3,
a%5E3y-4a-a%5E3=-8x%2B8a
highlight_green%28a%5E3y=-8x%2B12a%2Ba%5E3%29
.
.
When x=0, y=b,
a%5E3b=12a%2Ba%5E3
When x=b,y=0,
0=-8b%2B12a%2Ba%5E3
8b=12a%2Ba%5E3
So then,
a%5E3b=8b
a%5E3=8
highlight%28a=2%29
Then,
8b=12%282%29%2B8
8b=32
highlight%28b=4%29