Question 1

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 ?

Accepted Answer

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

Question 2

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 difference between the eccentricities of the hyperbola and

Accepted Answer

The equation of the ellipse can be written as x^2 16 + y^2 p = 1. Since p The foci of the ellipse are ( 4e_1, 0) = ( 4 1 - p 16 , 0 ) = ( 16 - p , 0). The equation of the hyperbola is 25(81x^2 - 144y^2) = 11664. Dividing by 11664, we get 25x^2 144 - 25y^2 81 = 1 x^2 144/25 - y^2 81/25 = 1. The eccentricity e_2 of the hyperbola is e_2 = 1 + 81/25 144/25 = 1 + 81 144 = 225 144 = 15 12 = 5 4 = 1.25. The foci of the hyperbola are ( A e_2, 0) = ( 12 5 5 4 , 0 ) = ( 3, 0). Since the foci of the ellips

Question 3

If any point on an ellipse is (3 , 5 ), then what is the eccentricity of the ellipse?

Accepted Answer

Let x = 3 and y = 5 . x 3 = and y 5 = Squaring and adding both equations: x^2 9 + y^2 25 = ^2 + ^2 = 1 This represents an ellipse with a^2 = 9 and b^2 = 25. Since b > a, the eccentricity e is given by a^2 = b^2(1 - e^2). 9 = 25(1 - e^2) 1 - e^2 = 9 25 e^2 = 16 25 e = 4 5 Answer: 4 5

Ellipse — Previous Year Questions & Solutions