परवलय y2 = 4ax पर सामान्य जीवा के ध्रुवों का बिन्दुपथ _____ है। 

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  1. (x + 2a)y2 - 4a3 = 0
  2. (x + 2a)y2 + 4a3 = 0
  3. (x - 2a)y2 + 4a3 = 0
  4. (x - 2a)y2 - 4a3 = 0

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Option 2 : (x + 2a)y2 + 4a3 = 0
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माना 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

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