Question

How to determine the equation of the tangent line to `y^3+xy=22` at (7,2)?

Answer 1

Equation of tangent at `(7,2)` is `2x+19y=137`

Explanation:

Observe that `(7,2)` lies on the curve `y^3+xy=22`, as putting these values in the equation `y^3+xy=22` the equality is established as `2^3+7xx2=8+14=22`.

Now, as slope of tangent at any point is given by the value of first derivative `(dy)/(dx)` at that point. Hence, we need to first find `(dy)/(dx)` using implicit differentiation.

As `y^3+xy=22`

`3y^2(dy)/(dx)+y+x(dy)/(dx)=0`

and putting `x=7` and `y=2`,

`3*2^2*(dy)/(dx)+2+7(dy)/(dx)=0`

or `(dy)/(dx)(12+7)=-2`

i.e. `(dy)/(dx)=-2/19`

As slope of tangent is `-2/19` and it passes through `(7,2)`, its equation using point-slope form is

`y-7=-2/19(x-2)` i.e. `2x+19y=137`