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.
Let the foci of a hyperbola $ H $ coincide with the foci of the ellipse $ E : \frac{(x - 1)^2}{100} + \frac{(y - 1)^2}{75} = 1 $ and the eccentricity of the hyperbola $ H $ be the reciprocal of the eccentricity of the ellipse $ E $. If the length of the transverse axis of $ H $ is $ \alpha $ and the length of its conjugate axis is $ \beta $, then $ 3\alpha^2 + 2\beta^2 $ is equal to:
The correct IUPAC name of \([ \text{Pt}(\text{NH}_3)_2\text{Cl}_2 ]^{2+} \) is: