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 $(6,5)$ ?

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Answer 1 · 2018-04-14T17:38:08

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

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 $y-5=3495/648(x-6)$

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