The area between x=y2 and x=4 is divided into two equal parts by the line x=a, find the value of a.
The line,x=a,divides the area bounded by the parabola and x=4 into two equal
parts.
∴Area OAD=Area ABCD
It can be observed that the given area is symmetrical about x-axis.
⇒Area OED=Area EFCD
Area OED=∫a0ydx
=∫a0√xdx
=[x3/2/3/2]a0
=2/3(a)3/2...(1)
Area of EFCD=∫40√xdx
=[x3/2/3/2]40
=2/3[8-a3/2]...(2)
From (1)and(2),we obtain
2/3(a)3/2=2/3[8-(a)3/2]
⇒2.(a)3/2=8
⇒(a)=3/2=4
⇒(a)=(4)2/3
Therefore,the value of a is (4)2/3.
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 \(P,O,Q\).