To determine the equation (eqn.) of tangent (tgt.) line, say t,t, we
need, (1) slope of tt and (2) a point (pt.) on t.t.
We already have a pt. P(1,2) in t.P(1,2)∈t.
To find the slope, let us recall that slope tgt. to a curve gammaγ is
given by dy/dx" at "P, i.e., [dy/dx]_P.dydx at P,i.e.,[dydx]P.
So, let us start by differentiating the eqn. of the curve
gamma : x^2y^2+x^3+3=y^3.γ:x2y2+x3+3=y3.
:.d/dx[x^2y^2+x^3+3]=d/dx[y^3].
:. x^2d/dx(y^2)+y^2d/dx(x^2)+d/dx(x^3)+d/dx(3)=d/dx(y^3).
:. x^2*d/dy(y^2)*d/dx(y)+y^2*2x+3x^2+0=d/dy(y^3).dy/dx.
:. x^2*2y*dy/dx+y^2*2x+3x^2=3y^2*dy/dx.
:. (2x^2y-3y^2)dy/dx=-(2xy^2+3x^2).
:. dy/dx=-(x(2y^2+3x))/(y(2x^2-3y)).
:. [dy/dx]_{P(1,2)}=-{1(2*2^2+3*1)}/{2(2*1^2-3*2)}=(-11)/(-8).
To Sum up, the line t has slope 11/8 and, P(1,2) in t.
By the Slope-Point Form of line, we have, the eqn. of
t : y-2=11/8(x-1), or, 8y-16=11x-11, i.e.,
t : 11x-8y+5=0.
Enjoy Maths.!
