(x− a)² + (y−b)² + z² = 1
(x - a)2 + (y – b)2 + z2 = c2 ........(1)
Differentiating (1) w.r.t ‘x’
2(x−a)+0+2zdz/dx=0
2(x-a) =-2zdz/dx
(x-a)=- zdz/dx .......(2)
Differentiating (1) with respect to y
0+2(y-b)+2zdz/dy =0
2(y-b)=-2zdz/dy
(y-b) = -zdz/dy.....(3)
Inserting 2 and 3 in the equation 1
We have, (-zdz/dx)2 + (-zdz/dy)2 +z2=c2
Let dz/dx be p and dz/dy be q
We have,
(-zp)2+(-zq)2+z2 =c2
z2 (p2 + q2 + 1) = c2
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