अति-परवलय 9x2 − 16y2 = 144 के लिए सत्य कथन है :

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RPSC 2nd Grade Mathematics (Held on 18th Feb 2019) Official Paper
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  1. इसकी नाभियाँ (0, ± 5) हैं।
  2. अर्ध-लघु अक्ष की लम्बाई 4 है।
  3. नाभि लम्ब की लम्बाई \(\frac{9}{4}\) है।
  4. इसकी उत्केन्द्रता \(\frac{5}{4}\) है।

Answer (Detailed Solution Below)

Option 4 : इसकी उत्केन्द्रता \(\frac{5}{4}\) है।
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RPSC Senior Grade II (Paper I): Full Test 1
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प्रयुक्त संकल्पना:

अतिपरवलय \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)के लिए , a > b,

a अर्ध-दीर्घ अक्ष है और b अर्ध लघु अक्ष है।

नाभि (± c, 0) है  जहां \( c=\sqrt{a^2+b^2}\)

नाभिलम्ब की लंबाई \( \frac{2b^2}{a}\)है। 

और उत्केंद्रता e = \(\sqrt{1+\frac{b^2}{a^2}} =\frac{c}{a} \)

गणना:

9x2 − 16y2 = 144

दोनों पक्षों को 144 से विभाजित करें

\(\frac{x^2}{16}-\frac{y^2}{9}=1\) 

\(\frac{x^2}{4^2}-\frac{y^2}{3^2}=1\) 

यहां, a = 4, b = 3

नाभिलम्ब की लंबाई = \(\frac{2b^2}{a}\)

⇒ नाभिलम्ब की लंबाई = \(\frac{2\cdot3^2}{4}=\frac{9}{2}\)

और उत्केंद्रता e = \(\sqrt{1+\frac{b^2}{a^2}} =\sqrt{1+\frac{9}{16}} \)

⇒ e  \(=\sqrt{\frac{16+9}{16}} =\sqrt{\frac{25}{16}} =\frac{5}{4}\)

इस प्रकार, उत्केन्द्रता 5/4 है।

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