- #1
chwala
Gold Member
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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!