Question

Obtain the partial differential equation from the relation 2z = x^2/a^2+y^2/b^2
by eliminating the
arbitrary constants a, b.
​

Answer 1

(x-a)^2 - (y-b)^2 +z^2 = a^2 +b^2….(1). Differentiating the above relation partially with respect to x and y separately and denoting the partial derivatives of z with respect to x and y by p and q respectively, we get

2(x-a)+2z.p=0 and

-2(y-b)+2z.q=0.

The above relations yield x-a = -z.p and

a=(x+z.p) and also y-b = z.q and b=(y-z.q). Replacing these values in the given relation (1), we get the required partial differential equation :

(z.p)^2 -(z.q)^2 = (x+z.p)^2 +(y-z.q)^2 or

x^2 +y^2 +2z(p-q) +2(z.q)^2 = 0.

Answer 2

Answer:

2z = x^2/a^2+y^2/b^2

Take partial derivative of the given equation w. r. t x

Let d here shows partial deravative

2dz/dx= 1/a^2 dx^2/dx + 1/b^2 dy^2/dx (1)

2dz/dx= 2x/a^2+ 0

Because y for partial deravtive w.r.t x is constant

dz/dx=p

2p=2x/a^2

1/a^2=2p/2x => 1/a^2=p/x (1)

Now take partial derivative w.r.t y of eq(1)

So,

2dz/dy=0+1/b^2dy^2/dy.

Because x is constant for partial derivative w.r.t y

Here, dz/dy=q

2q=2y/b^2

1/b^2=2q/2y => 1/b^2=q/y (3)

Now put (2) and (3) in (1)

2z= (1/a^2)x^2+(1/b^2)y^2

2z= (p/x) x^2+(q/y)y^2

2z=px+ qy which is the required solution