Answer :

The given equation is $\dfrac{x^{2}}{36}+\dfrac{y^{2}}{16}=1$

Here, the denominator of $\dfrac{x^{2}}{36}$ is greater than the denominator of $\dfrac{y^{2}}{16}$

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

On comparing the given equation with $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, 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, $e=\dfrac{c}{a}=\dfrac{2 \sqrt{5}}{6}=\dfrac{\sqrt{5}}{3}$

Length of latus rectum $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 16}{6}=\dfrac{16}{3}$

Answer :

The given equation is $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{25}=1$ or $\dfrac{x^{2}}{2^{2}}+\dfrac{y^{2}}{5^{2}}=1$

Here, the denominator of $\dfrac{y^{2}}{25}$ is greater than the denominator of $\dfrac{x^{2}}{4}$

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

On comparing the given equation with $\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1$, we obtain $b=2$ and $a=5$

$\therefore \ \ c=\sqrt{a^{2}-b^{2}}=\sqrt{25-4}=\sqrt{21}$

Therefore,

The coordinates of the foci are $(0, \sqrt{21})$ and $(0,-\sqrt{21})$

The coordinates of the vertices are $(0,5)$ and $(0, - 5)$

Length of major axis $=2 a=10$

Length of minor axis $=2 b=4$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{21}}{5}$

Length of latus rectum $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 4}{5}=\dfrac{8}{5}$

Answer :

The given equation is $\dfrac{x^{2}}{16}+\dfrac{y^{2}}{9}=1$ or $\dfrac{x^{2}}{4^{2}}+\dfrac{y^{2}}{3^{2}}=1$

Here, the denominator of $\dfrac{x^{2}}{16}$ is greater than the denominator of $\dfrac{y^{2}}{9}$

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

On comparing the given equation with $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, we obtain $a=4$ and $b=3$

$\therefore \ \ c=\sqrt{a^{2}-b^{2}}=\sqrt{16-9}=\sqrt{7}$

Therefore,

The coordinates of the foci are $( \pm \sqrt{7}, 0)$

The coordinates of the vertices are $( \pm 4,0)$

Length of major axis $=2 a=8$

Length of minor axis $=2 b=6$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{7}}{4}$

Length of latus rectum $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 9}{4}=\dfrac{9}{2}$

Answer :

The given equation is $\dfrac{x^{2}}{25}+\dfrac{y^{2}}{100}=1$ or $\dfrac{x^{2}}{5^{2}}+\dfrac{y^{2}}{10^{2}}=1$

Here, the denominator of $\dfrac{y^{2}}{100}$ is greater than the denominator of $\dfrac{x^{2}}{25}$

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

On comparing the given equation with $\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1$, we obtain $b=5$ and $a=10$

$\therefore \ \ c=\sqrt{a^{2}-b^{2}}=\sqrt{100-25}=\sqrt{75}=5 \sqrt{3}$

Therefore,

The coordinates of the foci are $(0, \pm 5 \sqrt{3})$

The coordinates of the vertices are $(0, \pm 10)$

Length of major axis $=2 a=20$

Length of minor axis $=2 b=10$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{5 \sqrt{3}}{10}=\dfrac{\sqrt{3}}{2}$

Length of latus rectum $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 25}{10}=5$

Answer :

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

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

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

On comparing the given equation with $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, 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 $( \pm 7,0)$

Length of major axis $=2 a=14$

Length of minor axis $=2 b=12$

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

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

Answer :

The given equation is $\dfrac{x^{2}}{100}+\dfrac{y^{2}}{400}=1$ or $\dfrac{x^{2}}{10^{2}}+\dfrac{y^{2}}{20^{2}}=1$

Here, the denominator of $\dfrac{y^{2}}{400}$ is greater than the denominator of $\dfrac{x^{2}}{100}$

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

On comparing the given equation with $\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1$, we obtain $b=10$ and $a=20$

$\therefore \ \ c=\sqrt{a^{2}-b^{2}}=\sqrt{400-100}=\sqrt{300}=10 \sqrt{3}$

Therefore,

The coordinates of the foci are $(0, \pm 10 \sqrt{3})$

The coordinates of the vertices are $(0, \pm 20)$

Length of major axis $=2 a=40$

Length of minor axis $=2 b=20$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{10 \sqrt{3}}{20}=\dfrac{\sqrt{3}}{2}$

Length of latus rectum $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 100}{20}=10$

Answer :

The given equation is $36 x^{2}+4 y^{2}=144$

It can be written as

$36 x^{2}+4 y^{2}=144$

or, $\dfrac{x^{2}}{4}+\dfrac{y^{2}}{36}=1$

or, $\dfrac{x^{2}}{2^{2}}+\dfrac{y^{2}}{6^{2}}=1\qquad \ldots(1)$

Here, the denominator of $\dfrac{y^{2}}{6^{2}}$ is greater than the denominator of $\dfrac{x^{2}}{2^{2}}$

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

On comparing equation (1) with $\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1$, we obtain $b=2$ and $a=6$

$\therefore \ \ c=\sqrt{a^{2}-b^{2}}=\sqrt{36-4}=\sqrt{32}=4 \sqrt{2}$

Therefore,

The coordinates of the foci are $(0, \pm 4 \sqrt{2})$

The coordinates of the vertices are $(0, \pm 6)$

Length of major axis $=2 a=12$

Length of minor axis $=2 b=4$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{4 \sqrt{2}}{6}=\dfrac{2 \sqrt{2}}{3}$

Length of latus rectum $=\dfrac{2 b^{2}}{a}=\dfrac{2 \times 4}{6}=\dfrac{4}{3}$

Answer :

The given equation is $16 x^{2}+y^{2}=16$

It can be written as

$16 x^{2}+y^{2}=16$

or, $\dfrac{x^{2}}{1}+\dfrac{y^{2}}{16}=1$

or, $\dfrac{x^{2}}{1^{2}}+\dfrac{y^{2}}{4^{2}}=1\qquad \ldots(1)$

Here, the denominator of $\dfrac{y^{2}}{4^{2}}$ is greater than the denominator of $\dfrac{x^{2}}{1^{2}}$

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

On comparing equation (1) with $\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1$, we obtain $b=1$ and $a=4$

$\therefore \ \ c=\sqrt{a^{2}-b^{2}}=\sqrt{16-1}=\sqrt{15}$

Therefore,

The coordinates of the foci are $(0, \pm \sqrt{15})$

The coordinates of the vertices are $(0, \pm 4)$

Length of major axis $=2 a=8$

Length of minor axis $=2 b=2$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{15}}{4}$

Length of latus rectum $ =\dfrac{2 b^{2}}{a}=\dfrac{2 \times 1}{4}=\dfrac{1}{2} $

Answer :

The given equation is $4 x^{2}+9 y^{2}=36$

It can be written as

$4 x^{2}+9 y^{2}=36$

or, $\dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=1$

or, $\dfrac{x^{2}}{3^{2}}+\dfrac{y^{2}}{2^{2}}=1$

Here, the denominator of $\dfrac{x^{2}}{3^{2}}$ is greater than the denominator of $\dfrac{y^{2}}{2^{2}}$

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

On comparing the given equation with $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, we obtain $a=3$ and $b=2$

$\therefore \ \ c=\sqrt{a^{2}-b^{2}}=\sqrt{9-4}=\sqrt{5}$

Therefore,

The coordinates of the foci are $( \pm \sqrt{5}, 0)$

The coordinates of the vertices are $( \pm 3,0)$

Length of major axis $=2 a=6$

Length of minor axis $=2 b=4$

Eccentricity, $e=\dfrac{c}{a}=\dfrac{\sqrt{5}}{3}$

Length of latus rectum $ =\dfrac{2 b^{2}}{a}=\dfrac{2 \times 4}{3}=\dfrac{8}{3} $

Answer :

Vertices $( \pm 5,0)$, foci $( \pm 4,0)$

Here, the vertices are on the $x$-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, where $a$ is the semi-major axis.

Accordingly, $a=5$ and $c=4$

It is known that $a^{2}=b^{2}+c^{2}$

$\therefore \ \ 5^{2}=b^{2}+4^{2}$

$\Rightarrow 25=b^{2}+16$

$\Rightarrow b^{2}=25-16$

$\Rightarrow b=\sqrt{9}=3$

Thus, the equation of the ellipse is $\dfrac{x^{2}}{5^{2}}+\dfrac{y^{2}}{3^{2}}=1$ or $\dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}=1$

Answer :

Vertices $(0, \pm 13)$, foci $(0, \pm 5)$

Here, the vertices are on the $y$-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1$, where $a$ is the semi-major 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 $\dfrac{x^{2}}{12^{2}}+\dfrac{y^{2}}{13^{2}}=1$ or $\dfrac{x^{2}}{144}+\dfrac{y^{2}}{169}=1$

Answer :

Vertices $( \pm 6,0)$, foci $( \pm 4,0)$

Here, the vertices are on the $x$-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, where $a$ is the semi-major axis.

Accordingly, $a=6, c=4$

It is known that $a^{2}=b^{2}+c^{2}$

$\therefore \ \ 6^{2}=b^{2}+4^{2}$

$\Rightarrow 36=b^{2}+16$

$\Rightarrow b^{2}=36-16$

$\Rightarrow b=\sqrt{20}$

Thus, the equation of the ellipse is $ \dfrac{x^{2}}{6^{2}}+\dfrac{y^{2}}{(\sqrt{20})^{2}}=1 \text{ or } \dfrac{x^{2}}{36}+\dfrac{y^{2}}{20}=1 $

Answer :

Ends of major axis $( \pm 3,0)$ , ends of minor axis $(0, \pm 2 )$

Here, the major axis is along the $x$-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, where $a$ is the semi-major axis.

Accordingly, $a=3$ and $b=2$

Thus, the equation of the ellipse is $\dfrac{x^{2}}{3^{2}}+\dfrac{y^{2}}{2^{2}}=1$ i.e., $\dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=1$

Answer :

Ends of major axis $(0, \pm \sqrt{5})$, ends of minor axis $( \pm 1,0)$

Here, the major axis is along the $y$-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1$, where $a$ is the semi-major axis.

Accordingly, $a=\sqrt{5}$ and $b=1$

Thus, the equation of the ellipse is

$ \dfrac{x^{2}}{1^{2}}+\dfrac{y^{2}}{(\sqrt{5})^{2}}=1 \text{ or } \dfrac{x^{2}}{1}+\dfrac{y^{2}}{5}=1 $

Answer :

Length of major axis $=26$; foci $=( \pm 5,0)$

Since the foci are on the $x$-axis, the major axis is along the $x$-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, where $a$ is the semi-major axis.

Accordingly, $2 a=26 \Rightarrow a=13$ 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 $\dfrac{x^{2}}{13^{2}}+\dfrac{y^{2}}{12^{2}}=1$ or $\dfrac{x^{2}}{169}+\dfrac{y^{2}}{144}=1$

Answer :

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 $\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1$, where $a$ is the semi-major 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 $\dfrac{x^{2}}{8^{2}}+\dfrac{y^{2}}{10^{2}}=1$ or $\dfrac{x^{2}}{64}+\dfrac{y^{2}}{100}=1$

Answer :

Foci $ ( \pm 3,0), a=4$

Since the foci are on the $x$-axis, the major axis is along the $x$-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, where $a$ is the semi-major axis.

Accordingly, $c=3$ and $a=4$

It is known that $a^{2}=b^{2}+c^{2}$

$\therefore \ \ 4^{2}=b^{2}+3^{2}$

$\Rightarrow 16=b^{2}+9$

$\Rightarrow b^{2}=16-9=7$

Thus, the equation of the ellipse is $\dfrac{x^{2}}{16}+\dfrac{y^{2}}{7}=1$

Answer :

It is given that $b=3, c=4$, centre at the origin; foci on the $x$ axis.

Since the foci are on the $x$-axis, the major axis is along the $x$-axis.

Therefore, the equation of the ellipse will be of the form $\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1$, where $a$ is the semi-major axis.

Accordingly, $b=3, c=4$

It is known that $a^{2}=b^{2}+c^{2}$

$\therefore \ \ a^{2}=3^{2}+4^{2}=9+16=25$

$\Rightarrow a=5$

Thus, the equation of the ellipse is $\dfrac{x^{2}}{5^{2}}+\dfrac{y^{2}}{3^{2}}=1$ or $\dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}=1$

Answer :

Since the centre is at $(0,0)$ and the major axis is on the $y$-axis, the equation of the ellipse will be of the form

$\dfrac{x^{2}}{b^{2}}+\dfrac{y^{2}}{a^{2}}=1 \qquad \ldots{(1)}$

Where, $a$ is the semi-major axis

The ellipse passes through points $(3,2)$ and $(1,6)$ Hence,

$ \dfrac{9}{b^{2}}+\dfrac{4}{a^{2}}=1 \qquad \ldots{(2)}$

$\dfrac{1}{b^{2}}+\dfrac{36}{a^{2}}=1 \qquad \ldots{(3)} $

On solving equations $(2)$ and $(3),$ we obtain $b^{2}=10$ and $a^{2}=40$

Thus, the equation of the ellipse is $\dfrac{x^{2}}{10}+\dfrac{y^{2}}{40}=1$ or $4 x^{2}+y^{2}=40$

Answer :

Since the major axis is on the $x$-axis, the equation of the ellipse will be of the form

$\dfrac{x^{2}}{a^{2}}+\dfrac{y^{2}}{b^{2}}=1 \qquad \ldots{(1)}$

Where, $a$ is the semi-major axis

The ellipse passes through points $(4,3)$ and $(6,2)$ Hence,

$\dfrac{16}{a^{2}}+\dfrac{9}{b^{2}}=1 \qquad \ldots{(2)}$

$ \dfrac{36}{a^{2}}+\dfrac{4}{b^{2}}=1 \qquad \ldots{(3)}$

On solving equations $(2)$ and $(3),$ we obtain $a^{2}=52$ and $b^{2}=13$

Thus, the equation of the ellipse is

$ \dfrac{x^{2}}{52}+\dfrac{y^{2}}{13}=1 \text{ or } x^{2}+4 y^{2}=52 $