Question

Equation of the hyperbola whose eccentricity is 5/4 focus (a,0) and directrix 4x-3y=a

Answer 1

Equation of the hyperbola whose eccentricity is 5/4 focus (a,0) and directrix 4x-3y=a is 7y²+24xy-24ax-16ay+15a² = 0

Given, the eccentricity is 5/4

Focus is (a,0)

Directrix is 4x - 3y - a = 0

So, the equation of the hyperbola would stand as:

(x-a)² + (y-0)² = (5/4)² * [(4x-3y-a)/√(4² + 3²)]            [x coordinate of focus is a and y coordinate is 0]

So,

16(x²+a²+y²-2ax) = 16x² + 9y²+ a² -24xy -8ax +6ay

⇒7y²+24xy-24ax-16ay+15a² = 0

Required equation of hyperbola.