10-2. Parabola, Ellipse, Hyperbola
medium

दीर्घवृत्त में नाभियों और शीर्षों के निर्देशांक, दीर्घ और लघु अक्ष की लंबाइयाँ, उत्केंद्रता तथा नाभिलंब जीवा की लंबाई ज्ञात कीजिए

$\frac{x^{2}}{49}+\frac{y^{2}}{36}=1$

Option A
Option B
Option C
Option D

Solution

The given equation is $\frac{x^2}{49}+\frac{y^{2}}{36}=1$ or $\frac{x^2} {7^{2}}+\frac{y^{2}}{6^{2}}=1$

Here, the denominator of $\frac{x^{2}}{49}$ is greater than the denominator of $\frac{y^{2}}{36}$

Therefore, the major axis is along the $x-$ axis, while the minor axis is along the $y-$ axis.

On comparing the given equation with $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,$ we obtain $a=7$ and $b=6$

$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{49-36}=\sqrt{13}$

Therefore,

The coordinates of the foci are $(\pm \,\sqrt{13}, 0)$

The coordinates of the vertices are $(±7,\,0)$

Length of major axis $=2 a =14$

Length of minor axis $=2 b =12$

Eccentricity, $e=\frac{c}{a}=\frac{\sqrt{13}}{7}$

Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 36}{7}=\frac{72}{7}$

Standard 11
Mathematics

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