Question
Download Solution PDFअति-परवलय 9x2 − 16y2 = 144 के लिए सत्य कथन है :
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFप्रयुक्त 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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