The given equation is:
\(16x^2 + y^2 = 16\)
or \(\frac {x^2}{1}+ \frac {y^2}{16} = 1\)
or \(\frac {x^2}{12} + \frac {y^2}{42} = 1\) .......(1)
Here, the denominator of \(\frac {y^2}{4^2}\) is greater than the denominator of \(\frac {x^2}{1^2}\).
Therefore, the major axis is along the y-axis, while the minor axis is along the x-axis.
On comparing the given equation (1) with \(\frac {x^2}{b^2} + \frac {y^2}{a^2} = 1\), we obtain b = 1 and a = 4.
∴ \(c = \sqrt {(a^2 – b^2)}\)
\(c= \sqrt {(16-1)}\)
\(c= \sqrt {15}\)
Therefore,
The coordinates of the foci are \((0, ±\sqrt {15})\).
The coordinates of the vertices are (0, ±4).
Length of major axis = 2a = 8
Length of the minor axis = 2b = 2
Eccentricity, \(e = \frac ca = \frac {\sqrt {15}}{4}\)
Length of latus rectum = \(\frac {2b^2}{a} = \frac {(2×1^2)}{4} = \frac 24 = \frac 12\)
Let each of the two ellipses $E_1:\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\;(a>b)$ and $E_2:\dfrac{x^2}{A^2}+\dfrac{y^2}{B^2}=1A$
A racing track is built around an elliptical ground whose equation is given by \[ 9x^2 + 16y^2 = 144 \] The width of the track is \(3\) m as shown. Based on the given information answer the following: 
(i) Express \(y\) as a function of \(x\) from the given equation of ellipse.
(ii) Integrate the function obtained in (i) with respect to \(x\).
(iii)(a) Find the area of the region enclosed within the elliptical ground excluding the track using integration.
OR
(iii)(b) Write the coordinates of the points \(P\) and \(Q\) where the outer edge of the track cuts \(x\)-axis and \(y\)-axis in first quadrant and find the area of triangle formed by points \(P,O,Q\).
An ellipse is a locus of a point that moves in such a way that its distance from a fixed point (focus) to its perpendicular distance from a fixed straight line (directrix) is constant. i.e. eccentricity(e) which is less than unity
Read More: Conic Section
The ratio of distances from the center of the ellipse from either focus to the semi-major axis of the ellipse is defined as the eccentricity of the ellipse.
The eccentricity of ellipse, e = c/a
Where c is the focal length and a is length of the semi-major axis.
Since c ≤ a the eccentricity is always greater than 1 in the case of an ellipse.
Also,
c2 = a2 – b2
Therefore, eccentricity becomes:
e = √(a2 – b2)/a
e = √[(a2 – b2)/a2] e = √[1-(b2/a2)]
The area of an ellipse = πab, where a is the semi major axis and b is the semi minor axis.
Let the point p(x1, y1) and ellipse
(x2 / a2) + (y2 / b2) = 1
If [(x12 / a2)+ (y12 / b2) − 1)]
= 0 {on the curve}
<0{inside the curve}
>0 {outside the curve}