Question
Download Solution PDFपरवलय y2 = 4ax पर सामान्य जीवा के ध्रुवों का बिन्दुपथ _____ है।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFमाना परवलय का ध्रुव y2 = 4ax, p(n, k) है
सामान्य जीवा का समीकरण निम्न है;
y = -tx + 2at + at3
y + tx – 2at – at3 = 0 …1)
P के सन्दर्भ में ध्रुवीय का समीकरण T = 0 है
ky = 2a (x + n)
2ax – ky + 2an = 0 …2)
अब समीकरण 1) और समीकरण 2) एक ही रेखा के समीकरण को निरूपित करते हैं
तो, दोनों समीकरणों की तुलना करने पर;
\(\frac{t}{{2a}} = \frac{1}{{ - k}} = \frac{{ - 2at - a{t^3}}}{{2an}}\)
\(\frac{t}{{2a}} = \frac{1}{{ - k}} = \frac{{ - 2t - {t^3}}}{{2n}}\)
\(\frac{t}{{2a}} = \frac{1}{{ - k}}\)
\(t = \frac{{ - 2a}}{k}\) …3)
साथ ही,
\(\frac{t}{{2a}} = \frac{{ - 2t - {t^3}}}{{2n}}\)
\(\frac{1}{a} = \frac{{ - 2 - {t^2}}}{n}\)
\(\frac{n}{a} + 2 = - {t^2}\)
t को 3 से प्रतिस्थापित करने पर
\(\frac{{n + 2a}}{a} = - {\left( { - \frac{{2a}}{k}} \right)^2}\)
\({k^2}\left( {n + 2a} \right) = - 4{a^2}\)
k2 (n + 2a) + 4a3 = 0
समीकरण का सामान्यीकरण करने पर, हम प्राप्त करते हैं;
y2 (x + 2a) + 4a3 = 0
Last updated on Jul 14, 2025
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