Question
Download Solution PDFअतिपरवलय \(\rm \frac{x^2}{cosec^2\alpha } - \frac{y^2}{\sec^2\alpha } = 1\) की उत्केंद्रता ज्ञात कीजिए।
Answer (Detailed Solution Below)
Detailed Solution
Download Solution PDFसंकल्पना:
अतिपरवलय का मानक रूप निम्न है,
\(\rm \frac{x^2}{a^2} - \frac{y^2}{b^2}\)= 1, जिसकी उत्केंद्रता (e) = \(\rm \sqrt{1 + \frac{b^2}{a^2}}\)
गणना:
दिया गया है: \(\rm \frac{x^2}{cosec^2\alpha } - \frac{y^2}{\sec^2\alpha } = 1\)
a2 = cosec2 α and b2 = sec2 α
हम जानते हैं कि,
b2 = a2 (e2 - 1)
sec2 α = cosec2 α (e2 - 1)
\(\rm \frac{sec^2α }{cosec^2α } = (e^2 - 1) \)
\(\rm \tan^2α + 1 = e^2 \)
\(\rm \sec^2α = e^2 \)
\(\rm \therefore e = \sec α \)
Last updated on Jul 4, 2025
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