Question
Download Solution PDFअतिपरवलय 16x2 – 9y2 = 1 की उत्केन्द्रता कितनी है?
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFI. x - अक्ष पर फोकस और केंद्र पर उत्पत्ति वाले अतिपरवलय का समीकरण निम्नानुसार है
\(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) और उत्केन्द्रता \(e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}}\)
II. y- अक्ष पर फोकस और केंद्र पर उत्पत्ति वाले अतिपरवलय का समीकरण निम्नानुसार है
\(\frac{{{y^2}}}{{{b^2}}} - \frac{{{x^2}}}{{{a^2}}} = 1\) और उत्केन्द्रता \(e = \sqrt {1 + \frac{{{a^2}}}{{{b^2}}}}\)
गणना:
दिया गया है: अतिपरवलय का समीकरण 16x2 – 9y2 = 1
दिए गए अतिपरवलय समीकरण को इस प्रकार लिखा जा सकता है: \(\frac{{{x^2}}}{{\frac{1}{{16}}}} - \frac{{{y^2}}}{{\frac{1}{9}}} = 1\)
जहाँ, a2 = 1/16 और b2 = 1/9.
जैसा की हम जानते है x - अक्ष पर फोकस और केंद्र पर उत्पत्ति वाले अतिपरवलय का समीकरण निम्नानुसार है
\(\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1\) और उत्केन्द्रता \(e = \sqrt {1 + \frac{{{b^2}}}{{{a^2}}}}\)
\(\Rightarrow e = \sqrt {1 + \frac{{\frac{1}{9}}}{{\frac{1}{{16}}}}} = \frac{5}{3}\)Last updated on Jul 11, 2025
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