Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{36}+\frac{y^2} {16}=1$
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse $\frac{x^{2}}{36}+\frac{y^2} {16}=1$
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 $=\frac{2 b^{2}}{a}=\frac{2 \times 16}{6}=\frac{16}{3}$
Similar Questions
If the distance between the foci of an ellipse be equal to its minor axis, then its eccentricity is
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Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $\mathrm{A}$ and $\mathrm{B}$.
$1.$ The coordinates of $\mathrm{A}$ and $\mathrm{B}$ are
$(A)$ $(3,0)$ and $(0,2)$
$(B)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$
$(C)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $(0,2)$
$(D)$ $(3,0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$
$2.$ The orthocentre of the triangle $\mathrm{PAB}$ is
$(A)$ $\left(5, \frac{8}{7}\right)$ $(B)$ $\left(\frac{7}{5}, \frac{25}{8}\right)$
$(C)$ $\left(\frac{11}{5}, \frac{8}{5}\right)$ $(D)$ $\left(\frac{8}{25}, \frac{7}{5}\right)$
$3.$ The equation of the locus of the point whose distances from the point $\mathrm{P}$ and the line $\mathrm{AB}$ are equal, is
$(A)$ $9 x^2+y^2-6 x y-54 x-62 y+241=0$
$(B)$ $x^2+9 y^2+6 x y-54 x+62 y-241=0$
$(C)$ $9 x^2+9 y^2-6 x y-54 x-62 y-241=0$
$(D)$ $x^2+y^2-2 x y+27 x+31 y-120=0$
Give the answer question $1,2$ and $3.$
Tangents are drawn from the point $P(3,4)$ to the ellipse $\frac{x^2}{9}+\frac{y^2}{4}=1$ touching the ellipse at points $\mathrm{A}$ and $\mathrm{B}$.
$1.$ The coordinates of $\mathrm{A}$ and $\mathrm{B}$ are
$(A)$ $(3,0)$ and $(0,2)$
$(B)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$
$(C)$ $\left(-\frac{8}{5}, \frac{2 \sqrt{161}}{15}\right)$ and $(0,2)$
$(D)$ $(3,0)$ and $\left(-\frac{9}{5}, \frac{8}{5}\right)$
$2.$ The orthocentre of the triangle $\mathrm{PAB}$ is
$(A)$ $\left(5, \frac{8}{7}\right)$ $(B)$ $\left(\frac{7}{5}, \frac{25}{8}\right)$
$(C)$ $\left(\frac{11}{5}, \frac{8}{5}\right)$ $(D)$ $\left(\frac{8}{25}, \frac{7}{5}\right)$
$3.$ The equation of the locus of the point whose distances from the point $\mathrm{P}$ and the line $\mathrm{AB}$ are equal, is
$(A)$ $9 x^2+y^2-6 x y-54 x-62 y+241=0$
$(B)$ $x^2+9 y^2+6 x y-54 x+62 y-241=0$
$(C)$ $9 x^2+9 y^2-6 x y-54 x-62 y-241=0$
$(D)$ $x^2+y^2-2 x y+27 x+31 y-120=0$
Give the answer question $1,2$ and $3.$
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