Posted by George on Friday, November 7, 2008 at 12:43am.
I solved the question in A but please check..
first of all you have to find y while x=4 and as you find y^2=25-16=9 and y=3
then you have to find slope. It's derivative of y function so
(slope)m=1/2(1/25-x^2)(2x).
if we write 3 instead of x we will get m(slope) as -4/3.
from
the tangent line equation
y-3=(-4/3)(x-4)
y=(-4/3)x+25/3.(result)
solution for part B
for the normal line equation firstly you have to find again slope. product of tangentline's slope and normal line's slope must be -1.
there fore
line's slope.(-4/3)=-1
line's slope= 3/4
then
the line equation is
y-3= =(3/4)(x-4)
we get
y= 3x/4(result)
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