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#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.

Verified Answer :

Nirina7

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)

boffeemadrid

Answer:

Step-by-step explanation:

A. The given curve is:

\(4x^2+y^2=48+2xy\)

\(8x+2y\frac{dy}{dx}=2(x\frac{dy}{dx}+y)\)

\(8x+2y\frac{dy}{dx}=2x\frac{dy}{dx}+2y\)

\(8x-2y=2(x-y)\frac{dy}{dx}\)

\(\frac{dy}{dx}=\frac{4x-y}{x-y}\)

\(\frac{dy}{dx}=\frac{y-4x}{y-x}\)

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 \(\frac{dy}{dx}=0\)

⇒\(\frac{y-4x}{y-x}=0\)

⇒\(y=4x\)

At y=8,

⇒8=4x

⇒x=2

The equation of the tangent is:

y-8=0(x-2)

⇒y=8.

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

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