Question

How do you find the equations for the tangent to the curve in the point [-1,-2]?

`x^2y^2=4`

Answer 1

`y= -2x-4`

Explanation:

Differentiate the equation implicitly:

`x^2y^2 = 4`

`d/dx (x^2y^2) = 0`

`2xy^2+2x^2ydy/dx = 0`

For `x != 0`:

`y^2=-xydy/dx`

`(1) " " dy/dx = -y/x`

The general equation of the tangent line in the point `(x_0,y_0)` is:

`y = y_0+ [dy/dx]_(x_0,y_0)(x-x_0)`

that is, based on `(1)`

`y = y_0-y_0/x_0(x-x_0)`

and for `x_0 = -1`, `y_0=-2`:

`y = -2-2(x+1)`

`y= -2x-4`

Answer 2

You can do it like this:

Explanation:

`sf(x^2y^2=4)`

Using The Product Rule:

`sf((d[y^2])/(dx).x^2+y^2.(d[x^2])/(dx)=0)`

Differentiating implicitly:

`sf(2y.(dy)/(dx).x^2+y^(2).2x=0)`

`sf((dy)/(dx)=(-y^(cancel(2)).cancel(2x))/(cancel(2y).x^cancel(2)))`

`sf((dy)/(dx)=-y/x)`

This gives the gradient m of the tangent line:

`:.` `sf(m=-(-2)/(-1)=-2)`

The tangent line is of the form:

`sf(y=mx+c)`

`:.` `sf(-2=(-2xx-1)+c)`

`sf(c=-4)`

So the equation of the tangent line is:

`sf(y=-2x-4)`

The functions look like this:

graph{(x^2y^2-4)(-2x-4-y)=0 [-10, 10, -5, 5]}

Answer 3

`y=-2x-4`

Explanation:

`"differentiate "color(blue)"implicitly with respect to x"`

`"differentiate "x^2y^2" using the "color(blue)"product rule"`

`•color(white)(x)m_(color(red)"tangent")=dy/dx" at "(-1,-2)`

`rArr(x^2 .2ydy/dx+2xy^2)=0`

`rArr2x^2ydy/dx=-2xy^2`

`rArrdy/dx=(-2xy^2)/(2x^2y)=-y/x`

`dy/dx" at "(-1,-2)=-(-2)/(-1)=-2`

`rArry+2=-2(x+1)`

`rArry=-2x-4larrcolor(red)"equation of tangent"`
graph{(x^2y^2-4)(y+2x+4)=0 [-20, 20, -10, 10]}