Question

NEED HELP! Determining the equation of the tangent to the curve at the given point R(1,-2) : x^2y^2(1+xy)+4=0

Answer 1

Do implicit differentiation.
Gather up y' terms on the LHS and x & y terms on the RHS.
Put x = 1 and y = -2 to evaluate y' at (1, -2)
That will be the slope of the tangent 'm'
You know the slope and you know it passes through (1, -2).
You can find b in y = mx + b

Answer 2

Okayy i'll try it thanks !!

Answer 3

You are welcome.

Answer 4

Can you post what u did because I can't solve it

Answer 5

Haven't done it myself.

Answer 6

\[\Large x ^{2}y ^{2}(1 + xy) + 4 = 0\]\[\Large x ^{2}y ^{2} + x^3y^3 + 4 = 0\]Do implicit differentiation

Answer 7

\[\Large x^{2}(2yy') + (2x)y^{2} + x^{3}(3y^{2}y') + (3x^{2}y^{3}) = 0\]

Answer 8

\[\Large y'(2x^{2}y + 3x^{3}y^{2}) + 2xy^{2} + 3x^{2}y^{3} = 0\]

Answer 9

\[\Large y' = -\frac{ (2xy^{2} + 3x^{2}y^{3}) }{ (2x^{2}y + 3x^{3}y^{2}) }\]

Answer 10

Put x = 1, y = -2 in the above equation and find y' which is the slope 'm' of the tangent to the curve at the point (1, -2)

The tangent line is: y = mx + b
Put m from earlier calculation.
Also the line passes through (1, -2). Put x = 1, y = -2 and solve for b.
Knowing m and b you will have the equation of the tangent y = mx + b.

Answer 11

okk thanks a lot!

Answer 12

you are welcome.