Limit word problem, find area of triangle

  • Thread starter Aoiro
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  • #1
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1. the tangent to the curve xy=4 at a point P in the first quadrent meets the x-axis at A and the y-axsis at B. Prove that the area of triangle AOB, where O is the origin, is independant of the position P.

y=4/x
Find thd slope of the Tangent
lim f(a+h)-f(a) /h
lim 4/a+h - 4/a /h
lim 4a - 4a -4h/ a(a+h) /h
lim -4h / h(a(a+h))
lim -4/a(a+h)
-4/h
Quardinents of point P

p(a,4/a)

the equation for the tangent

y-y1=m(x-x1)
y - 4/a = -4/a^2 (x - a)
y - 4/a = (-4/a^2)x + 4a/a^3
y = (-4/a^2)x + 4a/a^3 + 4/a
y = (-4/a^2)x + 8a/a^3

but the answer is y = (-4/a^2)x + 8/a can someone show me the steps to solving this?

I can do the rest I just need help with this. Thanks
 
Last edited:

Answers and Replies

  • #2
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[tex] y - \frac{4}{a} = -\frac{4}{a^{2}}x + \frac{4a}{a^{3}} [/tex] should be [tex] y - \frac{4}{a} = -\frac{4}{a^{2}}x + \frac{4a}{a^{2}} [/tex].

So you have [tex] y - \frac{4}{a} = -\frac{4}{a^{2}}(x-a) [/tex].

[tex] y - \frac{4}{a} = -\frac{4x}{a^{2}} + \frac{4a}{a^{2}} [/tex]

[tex] y = -\frac{4x}{a^{2}} + \frac{4}{a} + \frac{4}{a} [/tex].

[tex] y = -\frac{4x}{a^{2}} + \frac{8}{a} [/tex].
 
Last edited:
  • #3
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Thanks courtrigrad
 

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