Question

How do you find the slope of the line tangent to `x^2+xy+y^2=7` at (1,2) and (-1,3)?

Answer 1

The tangent equations are:

At `(1,2) \ \ \ \ \=> y = -4/5x+14/5`

At `(-1,3) => y = -1/5x+14/5`

Explanation:

The gradient of the tangent to a curve at any particular point is given by the derivative of the curve at that point. The normal is perpendicular to the tangent so the product of their gradients is `-1`

We have:

`x^2 +xy+y^2 = 7`

First let us check that `(1,2)` and #(-1,3) lies on the curve:

`(1,2) \ \ \ \ \=> 1+2+4 = 7`
`(-1,3) => 1-3+9 = 7`

Then differentiating (implicitly) wrt `x`, and applying the product rule gives us:

`2x + xdy/dx + y + 2ydy/dx = 0`
`:. (2x + y) + (2y+x)dy/dx = 0`
`:. dy/dx = -(2x + y)/(2y+x)`

Coordinate `(1,2)`

When `x = 1` and `y=2` we have:

`dy/dx = -(2 + 2)/(4+1) = -4/5`

So the tangent passes through `(1,2)` and has gradient `m_T=-4/5`, so using the point/slope form `y-y_1=m(x-x_1)` the equation we seek are;

`y - 2 = -4/5(x-1)`
`:. y - 2 = -4/5x+4/5`
`:. y = -4/5x+14/5`

Coordinate `(-1,3)`

When `x = -1` and `y=3` we have:

`dy/dx = -(-2+3)/(6-1) = -1/5`

So the tangent passes through `(-1,3)` and has gradient `m_T=1/5`, so using the point/slope form `y-y_1=m(x-x_1)` the equation we seek are;

`y - 3 = -1/5(x+1)`
`:. y - 3 = -1/5x-1/5`
`:. y = -1/5x+14/5`

We can confirm this solution is correct graphically: