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10-2. Parabola, Ellipse, Hyperbola
medium
प्रतिबंधों को संतुष्ट करते हुए दीर्घवृत्त का समीकरण ज्ञात कीजिए
शीर्षों $(0,\pm 13),$ नाभियाँ $(0,±5)$
Option A
Option B
Option C
Option D
Solution
Vertices $(0,\,\pm 13),$ foci $(0,\,±5)$
Here, the vertices are on 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, $a=13$ and $c=5$
It is known that $a^{2}=b^{2}+c^{2}$
$\therefore 13^{2}=b^{2}+5^{2}$
$\Rightarrow 169=b^{2}+25$
$\Rightarrow b^{2}=169-25$
$\Rightarrow b=\sqrt{144}=12$
Thus, the equation of the ellipse is $\frac{x^{2}}{12^{2}}+\frac{y^{2}}{13^{2}}=1$ or $\frac{x^{2}}{144}+\frac{y^{2}}{169}=1$
Standard 11
Mathematics