Question

Find the equation of the tangent of x^2- xy + y^2 =7 at (-1, 2)

Answer 1

Explanation:

Here is an other way:

z=f(x,y)=x²-xy+y²-7=0

`(\partial f(x,y))/(\partial x) =2x-y\\\\(\partial f(x,y))/(\partial y) =-x+2y\\\\\\(dy)/(dx) =-((\partial f(x,y))/(\partial x) )/((\partial f(x,y))/(\partial y)) =-(2x-y)/(-x+2y) \\\\(-1;2)\\Slope=-(2*(-1)-2)/(1+2*2) =(4)/(5) \\\\y-2=(4)/(5) (x+1)\\\\\\\boxed{y=(4x)/(5) +(14)/(5) }`

Answer 2

It is 3y = 4x + 10

Explanation:

Let's first get the slope of the curve.

[ slope is the derivative of the equation ]

`{x}^(2) - xy + {y}^(2) = 7`

introduce dy/dx :

`(d)/(dx) ( {x}^(2) - xy + {y}^(2) ) = (d)/(dx) (7) \\ \\ 2x - (y + (dy)/(dx) ) + 2y (dy)/(dx) = 0`

make dy/dx the subject:

`2x - y - (dy)/(dx) + 2y (dy)/(dx) = 0 \\ \\ 2y (dy)/(dx) - (dy)/(dx) = y - 2x \\ \\ (dy)/(dx) (2y - 1) = y - 2x \\ \\ (dy)/(dx) = (y - 2x)/(2y - 1)`

At point (-1, 2):

`(dy)/(dx) = (2 - 2( - 1))/(2(2) - 1) \\ \\ slope = (4)/(3)`

but a tangent has the same slope as the curve:

`y = mx + c`

m is the slope

c is the y-intercept

At (-1, 2):

`2 = ( - 1 * (4)/(3) ) + c \\ \\ c = 2 + (4)/(3) \\ \\ c = (10)/(3)`

equation:

`y = (4)/(3) x + (10)/(3) \\ \\ { \boxed{3y = 4x + 10}}`