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समतलों के निम्न कुल z2 = kxy, k ∈ के लंबवत समतलों का सामान्य हल है

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CSIR-UGC (NET) Mathematical Science: Held on (26 Nov 2020)
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  1. ϕ(x2 - y2, xz) = 0, ϕ ∈ C1 (2)
  2. ϕ(x2 - y2, x2 + z2) = 0, ϕ ∈ C1 (2)
  3. ϕ(x2 - y2, 2x2 + z2) = 0, ϕ ∈ C1 (2)
  4. ϕ(x2 + y2, 3x2 - z2) = 0, ϕ ∈ C1 (2)

Answer (Detailed Solution Below)

Option 3 : ϕ(x2 - y2, 2x2 + z2) = 0, ϕ ∈ C1 (2)
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व्याख्या:

दिया गया है z2 = kxy, k ∈ ℝ ← निकाय पृष्ठ

⇒ ........(i)

अब, हम इसे लैग्रेंज AE का उपयोग करके हल करेंगे -

(स्मरण करने पर: Pp + Qq = R)

इसलिए, (p = zx, q = zy)

⇒

⇒

⇒

⇒ (zy)p + (xz)q = -2xy (LCM लेने पर) (ii)

(zy)p → P

(xz)q → Q

-2xy → R

इसलिए, लैग्रेंज समीकरण से -

अब,

⇒ x dx = y dy

समाकलन करने पर,

⇒ x2 - y2 = c1

⇒ 2x dx = -z dz

समाकलन करने पर,

⇒ 2x2 + z2 = c2

इसलिए, सामान्य हल ϕ ​​(c1, c2) = 0 होगा

⇒ ϕ (x2 - y2, 2x2 + z2) = 0 ⇒ विकल्प (3) सही है।

हल का अन्य संभावित रूप और जब c1 और c2 की गणना की जाती है।

ϕ(c1, c2), c1 = ϕ(c2), c2 = ϕ(c2), ...

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