The given equation is \displaystyle 16x^{2}+y^{2}= 1616x

2

+y

2

=16

It can be written as

\displaystyle 16x^{2}+y^{2}= 1616x

2

+y

2

=16

\displaystyle \frac{x^{2}}{1}+\frac{y^{2}}{16}= 1

1

x

2

+

16

y

2

=1

\displaystyle \frac{x^{2}}{1^{2}}+\frac{y^{2}}{4^{2}}= 1 ...(1)

1

2

x

2

+

4

2

y

2

=1...(1)

Herethe denominator of \displaystyle \frac{y^{2}}{4^{2}}

4

2

y

2

is greater than the denominator of \displaystyle \frac{x^{2}}{1^{2}}

1

2

x

2

Therefore, the major axis is along the yy-axis while the minor axis is along the xx-axis.

On comparing equation (1) with \displaystyle \frac{x^{2}}{B^{2}}+\frac{y^{2}}{a^{2}}= 1

b

2

x

2

+

a

2

y

2

=1, we obtain b = 1b=1 and a = 4a=4

\displaystyle \therefore ae=c = \sqrt{a^{2}-b^{2}}=\sqrt{16 - 1}=\sqrt{15}∴ae=c=

a

2

−b

2

=

16−1

=

15

Therefore, the coordinates of the foci are \displaystyle \left ( 0, \pm \sqrt{15} \right )(0,±

15

)

The coordinates of the VERTICES are \displaystyle \left ( 0, \pm 4 \right )(0,± 4)

LENGTH of major axis = 2a = 8=2a=8

Length of minor axis = 2b = 2=2b=2

Eccentricity e =e= \displaystyle \frac{c}{a}=\frac{\sqrt{15}}{4}

a

c

=

4

15

Length of latus rectum == \displaystyle \frac{2b^{2}}{a}=\frac{2\times 1}{4}=\frac{1}{2}

a

2b

2

=

4

2×1

=

2

1