1. Analyze the Ellipse:
The equation is px2+16y2=16p. Dividing by 16p:
16x2+py2=1
Since p < 16, the ellipse is elongated along the x-axis. Here, a2=16 and b2=p.
The eccentricity e1 is given by e1=1−a2b2=1−16p.
The foci are at (±ae1,0)=(±16−p,0).
2. Analyze the Hyperbola:
The equation is 25(81x2−144y2)=11664. Dividing by 11664:
11664/2581x2−11664/25144y2=1
11664/(25⋅81)x2−11664/(25⋅144)y2=1
144/25x2−81/25y2=1
Here, a2=25144 and b2=2581.
The foci are at (±ae2,0), where ae2=a2+b2:
ae2=25144+2581=25225=9=3
So, the foci of the hyperbola are at (±3,0).
3. Equate the Foci:
Since the foci coincide, we equate the focal distance ae1 from the ellipse to the focal distance ae2 from the hyperbola:
16−p=3
Squaring both sides:
16−p=9
p=7
Final Answer:
The correct option is (c) 7.
