Question

What is the slope of the tangent line of `(x^2/y-y^2)/(1+x)^2 =C`, where C is an arbitrary constant, at `(6,5)`?

Answer 1

Slope is `3495/648` and tangent is `y-5=3495/648(x-6)`

Explanation:

As we are seeking tangent at point `(6,5)`, the curve `(x^2/y-y^2)/(1+x)^2=C` passes through it and hence

`C=(x^2/y-y^2)/(1+x)^2=(6^2/5-5^2)/(1+5)^2=-89/(5*36)=-89/180`

i.e. curve is `(x^2/y-y^2)/(1+x)^2=-89/180`

or `x^2/y-y^2=-89/180(1+x)^2`

Slope of tangent is given by value of `(dy)/(dx)` at this point

and differentiating we get

`2x/y-x^2/y^2(dy)/(dx)=-89/180(2x+2)`

or at `(6,5)` we have `12/5-36/25(dy)/(dx)=-89/180*14`

or `(dy)/(dx)=25/36(12/5+483/90)=25/36*699/90=3495/648`

and tangent is `y-5=3495/648(x-6)`