1. Analyzing the Hyperbola
The given equation is 25(81x2−144y2)=11664.
Dividing both sides by 11664:
1166425⋅81x2−1166425⋅144y2=1
116642025x2−116643600y2=1
5.76x2−3.24y2=1
(2.4)2x2−(1.8)2y2=1
Here, a2=5.76 and b2=3.24.
For a hyperbola, c2=a2+b2=5.76+3.24=9. Thus, c=3.
The foci are at (±c,0), which is (±3,0).
The eccentricity eh is:
eh=ac=2.43=1.25
2. Analyzing the Ellipse
The given equation is px2+16y2=16p.
Dividing by 16p:
16ppx2+16p16y2=1
16x2+py2=1
Since the foci of the ellipse coincide with the foci of the hyperbola (±3,0), we have c=3.
For an ellipse, c2=a2−b2 (given p < 16, the major axis is along the x-axis, so a2=16 and b2=p).
32=16−p
9=16−p⟹p=7
The eccentricity ee is:
ee=ac=163=43=0.75
3. Difference Between Eccentricities
We need to find the difference between the eccentricities of the hyperbola and the ellipse:
Difference=eh−ee=1.25−0.75=0.5
The correct option is (a) 0.5.
