Find the equations of all lines tangent to y = 9 - x^2 that pass through the point (1,12).
2007-07-05 07:23:11 · 2 answers · asked by salmonella_jr 3 in Science & Mathematics ➔ Mathematics
thanks, you guys. i graphed both of your answers on my graphing calculator, and santmann's answers look right. i don't understand why "The slope of a tangent is m= -2x" though, so i'm still confused.
2007-07-05 07:52:51 · update #1
i understand why the slope is -2x now. haha, i'm still working through the rest though.
2007-07-05 07:55:15 · update #2
got it! thanks!
2007-07-05 08:03:43 · update #3
The slope of a tangent is m= -2x
so if (xo,yo) is a point of the curve
y-(9-xo^2) =-2xo(x-xo) Passing through (1,12)
12-(9-xo^2)=-2xo(1-xo)
so 3+xo^2 =-2xo+2xo^2
so xo^2-2xo-3=0 xo=(( 2+-sqrt(16))/2
xo=3 and xo=-1
so
y=-6x+18 and y-8=2x+2 ==> y=2x+10
2007-07-05 07:37:29 · answer #1 · answered by santmann2002 7 · 0⤊ 0⤋
well, actually i'm getting only 1 equation of the tangent!
Given the equation of the curve y = 9 - (x^2).
Y = 9 - (X^2)
Y = -(X^2) + 9
Using differentiation (dY/dX) to get the gradient of the tangent
(dy/dx) = -2X
At point (1,12), X=1 & Y=12 & (dy/dx)=-2
General form of equation of line: Y= mX + c; where m is the gradient & c is the y-intercept
Substituting those values in this equation,
12= [(-2)(1^2)] + c
12= -2 + c
c = 14
therefore equation of tangent to the curve, Y= 9 - (X^2), at the point (1, 12) is: Y = -2X + 14
2007-07-05 07:46:54 · answer #2 · answered by shresht c 1 · 0⤊ 0⤋