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दीर्घवृत्त में नाभियों और शीर्षों के निर्देशांक, दीर्घ और लघु अक्ष की लंबाइयाँ, उत्केंद्रता तथा नाभिलंब जीवा की लंबाई ज्ञात कीजिए
$\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$
Solution
The given equation is $\frac{x^{2}}{36}+\frac{y^{2}}{16}=1$
Here, the denominator of $\frac{x^{2}}{36}$ is greater than the denominator of $\frac{y^{2}}{16}$
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 $a=6$ and $b=4$
$\therefore c=\sqrt{a^{2}-b^{2}}=\sqrt{36-16}=\sqrt{20}=2 \sqrt{5}$
Therefore,
The coordinates of the foci are $(2 \sqrt{5}, 0)$ and $(-2 \sqrt{5}, 0)$
The coordinates of the vertices are $(6,\,0)$ and $(-6,\,0)$
Length of major axis $=2 a=12$
Length of minor axis $=2 b=8$
Eccentricity, $e=\frac{c}{a}=\frac{2 \sqrt{5}}{6}=\frac{\sqrt{5}}{3}$
Length of latus rectum $=\frac{2 b^{2}}{a}=\frac{2 \times 16}{6}=\frac{16}{3}$
