Question

X^2 y^2=(z-c)^2 tan^2 a form a pde by eliminating arbitrary constant

Answer 1

ax^2 + by^2 + z^2 = 1. ……….(1)

Differentiate w.r.t 'x', partially

2ax + 2z(δz/δx) = 0

ax + zp = 0 ……….{(δz/δx) = p}

a = (-zp)/x ……….(2)

Differentiate w.r.t 'y', partially

2by + 2z(δz/δy) = 0

by + zq = 0. ……….{(δz/δy) = q}

b = (-zq)/y ……….(3)

Now put the value of 'a' and 'b' from equation (2) and (3) in equation (1) we get

-zpx - zqy + z2 = 1

zpx + zqy = z^2 - 1

z(px + qy) = z^2 - 1

Answer 2

Answer:

Step-by-step explanation:

x^2+y^2 = (z-c)^2tan^2a    -----> eq(1)

diff wrt x

2x+0 = 2(z-c) ∂Z/∂x tan^2a

2x = 2(z-c) p tan^2a

x = (z-c) p tan^2a

x/p = (z-c)  tan^2a    ----->eq(2)

now diff wrt y

0+2y = 2(z-c) ∂Z/∂y tan^2a

2y = 2(z-c) q tan^2a

y = (z-c) q tan^2a

y/q = (z-c) tan^2a     ----->eq(3)

from eq(2) and eq(3)

x/p = y/q

qx = py

qx - py = 0   Answer