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10-2. Parabola, Ellipse, Hyperbola
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प्रतिबंधों को संतुष्ट करते हुए दीर्घवृत्त का समीकरण ज्ञात कीजिए
दीर्घ अक्ष की लंबाई $16,$ नाभियाँ $(0,\pm 6) .$
Option A
Option B
Option C
Option D
Solution
Length of minor axis $=16 ;$ foci $=(0,\,\pm 6)$
since the foci are on the $y-$ axis, the major axis is along the $y-$ axis.
Therefore, the equation of the ellipse will be of the form $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,$ where a is the semimajor axis.
Accordingly, $2 b=16 \Rightarrow b=8$ and $c=6$
It is known that $a^{2}=b^{2}+c^{2}$
$\therefore a^{2}=8^{2}+6^{2}=64+36=100$
$\Rightarrow a=\sqrt{100}=10$
Thus, the equation of the ellipse is $\frac{x^{2}}{8^{2}}+\frac{y^{2}}{10^{2}}=1$ or $\frac{x^{2}}{64}+\frac{y^{2}}{100}=1$
Standard 11
Mathematics