Question

#2- For the curve given by 4x2+y2=48+2xy show that

dy/dx=y-4x/y-x

#3- For the curve given by 4x2+y2=48+2xy, find the positive y coordinate given that the xcoordinate is 2.

#4- For the curve given by 4x2+y2=48+2xy, show that there is a point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal.

Answer 1

it is an implicit function, (for high school students)

1) d(4x2+y2)/dx=d(48+2xy)/dx, after the differentiation, we obtain
8x + 2y .dy/dx = 2y + 2x .dy/dx
8x - 2y = 2x .dy/dx - 2y .dy/dx, 4x - y = (x .- y ).dy/dx, and then
dy/dx = y-4x / y - x
2) when x cordinate is 2
we have 4(2^2) + y^2 = 48 + 2 . 2 y, and the y^2 - 4y - 32=0, after solving the quadratic equation we find y = - 8, or y = 4, this the positive y coordinate, the answer is y = 4
3) Let be P (xp, yp) the point, since P is the point at which the line tangent to the curve at P is horizontal, so we can write 4xp2+yp2=48+2xpyp, but xp=2, we have yp^2 - 4yp - 32=0, so P(2, 4)

Answer 2

Explanation:

A. The given curve is:

`4x^2+y^2=48+2xy`

`8x+2y(dy)/(dx)=2(x(dy)/(dx)+y)`

`8x+2y(dy)/(dx)=2x(dy)/(dx)+2y`

`8x-2y=2(x-y)(dy)/(dx)`

`(dy)/(dx)=(4x-y)/(x-y)`

`(dy)/(dx)=(y-4x)/(y-x)`

Hence proved.

B. The given curve is:

`4x^2+y^2=48+2xy`

Put x=2, we get

`4(2)^2+y^2=48+2(2)y`

`16+y^2=48+4y`

`y^2-4y-32=0`

`(y+4)(y-8)=0`

Thus, y=-4,8

C. The given curve is:

`4x^2+y^2=48+2xy`

To show point P with x-coordinate 2 at which the line tangent to the curve at P is horizontal, consider
`(dy)/(dx)=0`

⇒
`(y-4x)/(y-x)=0`

⇒
`y=4x`

At y=8,

⇒8=4x

⇒x=2

The equation of the tangent is:

y-8=0(x-2)

⇒y=8.

Thus,
`(dy)/(dx)=0` at (2,80, therefore with x-coordinate 2 the line tangent to the curve at P is horizontal.