Passage: The foci of the ellipse px^2+16y^2=16p and the foci of the hyperbola 25(81x^2-144y^2)=11664 coincide (assume p Question: What is the value of p ?

Step-by-step solution

The equation of the ellipse is x^2 16 + y^2 p = 1 . Since p The equation of the hyperbola is 25(81x^2 - 144y^2) = 11664 , which can be rewritten as x^2 144/25 - y^2 81/25 = 1 . For the hyperbola, a^2 = 144 25 and b^2 = 81 25 . The foci are given by ( a^2 + b^2 , 0) = ( 144 25 + 81 25 , 0 ) = ( 225 25 , 0 ) = ( 3, 0) . Since the foci of the ellipse and the hyperbola coincide, we have 16 - p = 3 . Squaring both sides, we get 16 - p = 9 p = 7 . Answer: 7