Question

Determine the position and nature of the double points on the curve: 2x^4 – 8y^3 – 12y^2 – 4x^2 – 2 = 0

Answer 1

Given that,

The function of x and y is

$f(x,y)=2x^4-8y^3-12y^2-4x^2-2$

We need to find the value of $\dfrac{df}{dx}, \dfrac{df}{dy}, \dfrac{d^2f}{dx^2}, \dfrac{d^2f}{dy^2}$

Using given equation

$f(x,y)=2x^4-8y^3-12y^2-4x^2-2$

On differentiate w.r.to x

$\dfrac{df}{dx} =8x^3-8x$....(I)

On differentiate w.r.to y

$\dfrac{df}{dy} =-24y^2-24y$....(II)

On differentiate w.r.to x again equation (I)

$\dfrac{d^2f}{dx^2} =24x^2-8$

On differentiate w.r.to y again equation (II)

$\dfrac{d^2f}{dy^2} =-48y-24$

We need to calculate the position of double point on the curve

Using first derivative of x

$\dfrac{df}{dx}=0$

Put the value into the formula

$8x^3-8x=0$

$8x^2-8=0$

$x=(1,0)$

Using first derivative of y

$\dfrac{df}{dy}=0$

Put the value into the formula

$-24y^2-24y=0$

$-24y-24=0$

$y=(-1,0)$

We know that,

If the value of second derivative of x and y is positive then the minimum point.

If the value of second derivative of x and y is negative then the maximum point.

We need to find the nature of the double point on the curve

Using second derivative of x and y

For derivative x,

$\dfrac{d^2f}{dx^2}=24x^2-8$

Put the value of x

If x = 0,

$\dfrac{d^2f}{dx^2}=-8$

The point is maximum at x = 0

If x = 1,

$\dfrac{d^2f}{dx^2}=16$

The point is minimum at x = 1.

For derivative y,

$\dfrac{d^2f}{dx^2}=-48y^2-24$

Put the value of y

If y = 0,

$\dfrac{d^2f}{dx^2}=-24$

The point is maximum at y = 0

If y = 1,

$\dfrac{d^2f}{dx^2}=24$

The point is minimum at x = 1.

Hence, The position of the double points on the curve are (1,0) and (-1,0).

The nature of the double points on the curve are

The maximum point at x = 0 and minimum point at x = 1.

The maximum point at y = 0 and minimum point at y = 1.