10-2. Parabola, Ellipse, Hyperbola
medium

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

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