SITUATIONAL PROBLEMS INVOLVING CONIC SECTIONS

Question

The vertices and foci of 5x² -4y² -40x +72y -424 = 0 are, respectively, the foci and vertices of an ellipse.

Find the standard equation of this ellipse. Use the mid-point, and perpendicular bisector method

Answer 1

5(x² - 8x + 16) - 80 - 4(y² - 18y + 81) + 324 - 424 = 0; 5(x - 4)² - 4(y - 9)² = 180; (x - 4)²/36 - (y - 9)²/45 = 1.

We have hyperbola with center at point (4, 9); a² = 36; b² = 45; c² = 36 + 45 = 81; c = 9.

Foci: (- 9 + 4, 9) = (- 5, 9) and (9 + 4, 9) = (13, 9)

Vertices of hyperbola: (- 6 + 4, 9) = (- 2, 9) and (6 + 4, 9) = (10, 9)

For ellipse we have: center the same point (4, 9); a = 9; c = 6, b² = 81 - 36 =45.

So, equation of ellipse: (x - 4)²/81 + (y - 9)²/45 = 1