Question

7. Find an equation for the conic that satisfies the given conditions. Hyperbola foci (0,£6), vertices (0,+3) A. 6x2 = y (x-6)2 y2 B. 36 27 = 1 c. 2; x2 = y x2 y2 D. 36 + = 1 27 x2 y2 E. 9 = 1 27

Answer 1

Answer] Given foci (0,+6) vertices (0,+3) from conic definaton s verlex co, tan foci - (0itae). a = 3 a e=6 le=2 It is clearly hyperbola. As es 1 1 +62 92 62= 3x9=27. . equation of hyperbola. 4². g. 22 27 cs Scanned with Camscanner

Answer 2

SOLUTION :

Conic section is a hyperbola. 

Since foci are (0, +/- 6) and vertices are (0, +/- 3), 

foci and vertices are on the y-axis and directrix is x-axis.

So, equation of the hyperbole will be of the form :

y^2 / a^2 - x^2 / b^2 = 1

Here, 

a = 3 and c = 6

=> a^2 = 9 and c^2 = 336

So, 

b^2 = c^2 - a^2 = 6^2 - 3^2 = 27

Therefore, equation of the hyperbola is :

y^2 / 9 - x^2 / 27 = 1   (Option E) (ANSWER).