- #1

chwala

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- Homework Statement
- See attached

- Relevant Equations
- Differentiation

My approach on this question is as follows;

##\dfrac{dy}{dx} = -\dfrac {8}{x^3}##

##\dfrac{dy}{dx} (x=a) = -\dfrac {8}{a^3}##

The tangent equation is given by;

##y= -\dfrac {8}{a^3}x+\dfrac{12+a^2}{a^2}##

when ##x=0##,

##y=\dfrac{12+a^2}{a^2}##

and when ##y=0##,

##\dfrac {8}{a^3} x = \dfrac{12+a^2}{a^2}##

##⇒\dfrac {8}{a^2}=y##

##⇒\dfrac {8}{a^2}=\dfrac{4}{a^2} +1##

##8=4+a^2##

##4=a^2##

##a=\sqrt{4}=2##

on using; ##y=\dfrac{12+a^2}{a^2}## and substituting ##a=2## yields

##y=\dfrac{12+2^2}{2^2}=\dfrac{16}{4}=4##

##b=4##.

I would appreciate any other method or insight. Cheers guys!