Limit word problem, find area of triangle

[Aoiro]

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 independent 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

 

[courtrigrad]

$$y - \frac{4}{a} = -\frac{4}{a^{2}}x + \frac{4a}{a^{3}}$$ should be $$y - \frac{4}{a} = -\frac{4}{a^{2}}x + \frac{4a}{a^{2}}$$.

So you have $$y - \frac{4}{a} = -\frac{4}{a^{2}}(x-a)$$.

$$y - \frac{4}{a} = -\frac{4x}{a^{2}} + \frac{4a}{a^{2}}$$

$$y = -\frac{4x}{a^{2}} + \frac{4}{a} + \frac{4}{a}$$.

$$y = -\frac{4x}{a^{2}} + \frac{8}{a}$$.

 

[Aoiro]

Thanks courtrigrad