(x - a)2 + (y - b)2 + z2 = c2 से a और b को हटाने पर आंशिक अवकल समीकरण क्या है?

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  1. z2(p - q + 1) = c2
  2. z2(p2 + q2 + 1) = c2
  3. z2(p2 + q2) = c2
  4. z2(p - q) = c2

Answer (Detailed Solution Below)

Option 2 : z2(p2 + q2 + 1) = c2
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व्याख्या:

(x - a)2 + (y – b)2 + z2 = c2         ---(i)

'x' के सापेक्ष (i) का अवकलन करने पर

\(2\left( {x - a} \right) + 0 + 2z\frac{{\partial z}}{{\partial x}} = 0\)

\(2 \left( {x - a} \right) = - 2z\frac{{\partial z}}{{\partial x}}\)

\(\left( {x - a} \right) = - z\frac{{\partial z}}{{\partial x}} = \phi \)

जब,

'y' के सापेक्ष (i) का अवकलन करने पर

\(2 \left( {y - b} \right) +2z\frac{{\partial y}}{{\partial z}} = 0\)

\(2\left( {y - b} \right) = - 2z\frac{{\partial y}}{{\partial z}}\)

\(\left( {y - b} \right) = - z\frac{{\partial y}}{{\partial z}}\)

\({\left( { - z\frac{{\partial z}}{{\partial x}}} \right)^2} + {\left( { - z\frac{{\partial y}}{{\partial z}}} \right)^2} + {z^2} = {c^2}\)

(-zp)2 + (-zq)2 + z2 = c2

z2 (p2 + q2 + 1) = c2

 विकल्प (2) सही है। 

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