SOLUTION: The equation of a curve is y = x^2/(x+2). The tangent to the curve at the point where x = -3 meets the y- axis at M. The normal to the curve at the point where x = -3 meets the x

Algebra ->  Test -> SOLUTION: The equation of a curve is y = x^2/(x+2). The tangent to the curve at the point where x = -3 meets the y- axis at M. The normal to the curve at the point where x = -3 meets the x       Log On


   

Question 1191708: The equation of a curve is y = x^2/(x+2). The tangent to the curve at the point where x = -3 meets the
y- axis at M. The normal to the curve at the point where x = -3 meets the x - axis at N. Find the
area of the triangle MNO, where O is the origin.
Answer by MathLover1(20850) About Me  (Show Source):
You can put this solution on YOUR website!
We first locate the precise point on the curve where the tangent and normal lines are:
f%28+-+3%29=x%5E2+%2F+%28x+%2B+2%29+=+%28-3%29%5E2+%2F+%28-3+%2B+2%29+=+9%2F-1+=+-9
So the point we’re looking for is (-3, -9). Next, we find the derivative of the curve, and determine its slope at x+=+-3:
y’=%28x%5E2%2B4%29%2F%28x%2B2%29%5E2 ; ->m=f’%28-+3%29=+-+3

Equation of tangent at x=+-3 is : y=+-+3x+-18 ; M lie on y-axis, so x=0then the other coordinate of M is
y=+-+3%280%29+-18=-18
so M(0 ; - 18)
So the slope of the tangent line is -3, while the slope of the normal is -1%2F-3=1%2F3
Equation of normal at x=+-3 is ; y+=++%281%2F3%29x+-+8+; the x- coordinate of N is intersection of a line with x-axis
0=++%281%2F3%29x+-+8+
x=24+
N(24 ; 0)
In right triangle MON+: ; MO=18 ; NO=24
then the area S is :
S=%281%2F2%29%2AMO%2ANO
S=%281%2F2%29%2A18%2A24
S=216 square units

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