Question 977879
THe slope of the tangent line is equal to the value of the derivative at that point.
{{{y=4x^(-2)+1}}}
{{{dy/dx=-8x^(-3)=-8/x^3}}}
So at {{{x=a}}},
{{{m=dy/dx=-8/a^3}}}
Also when {{{x=a}}},
{{{y=4/a^2+1}}}
Using the point-slope form of a line to find the tangent line,
{{{y-(4/a^2+1)=(-8/a^3)(x-a)}}}
Multiply both sides by {{{a^3}}},
{{{a^3y-4a-a^3=-8x+8a}}}
{{{highlight_green(a^3y=-8x+12a+a^3)}}}
.
.
When {{{x=0}}}, {{{y=b}}},
{{{a^3b=12a+a^3}}}
When {{{x=b}}},{{{y=0}}},
{{{0=-8b+12a+a^3}}}
{{{8b=12a+a^3}}}
So then,
{{{a^3b=8b}}}
{{{a^3=8}}}
{{{highlight(a=2)}}}
Then,
{{{8b=12(2)+8}}}
{{{8b=32}}}
{{{highlight(b=4)}}}