Question -
Answer -
Given:
The equation of the directrix of a hyperbola => x тАУ y + 3= 0.
Focus = (-1, 1) and
Eccentricity = 3
Now, let us find the equation of the hyperbola
Let тАШMтАЩ be the point on directrix and P(x, y) be any pointof the hyperbola.
By using the formula,
e = PF/PM
PF = ePM [where, e is eccentricity, PM is perpendicular fromany point P on hyperbola to the directrix]
So,
[We know that┬а(a тАУ b)2┬а=a2┬а+ b2┬а+ 2ab &(a + b + c)2┬а=a2┬а+ b2┬а+ c2┬а+ 2ab + 2bc + 2ac]
So, 2{x2┬а+ 1 + 2x + y2┬а+ 1 тАУ2y} = 9{x2┬а+ y2+ 9 + 6x тАУ 6y тАУ 2xy}
2x2┬а+ 2 + 4x + 2y2┬а+ 2 тАУ 4y= 9x2┬а+ 9y2+ 81 + 54x тАУ 54y тАУ 18xy
2x2┬а+ 4 + 4x + 2y2тАУ 4y тАУ 9x2┬атАУ9y2┬атАУ 81 тАУ 54x + 54y + 18xy = 0
тАУ 7x2┬атАУ 7y2┬атАУ 50x + 50y +18xy тАУ 77 = 0
7(x2┬а+ y2) тАУ 18xy + 50x тАУ 50y +77 = 0
тИ┤The equation of hyperbola is 7(x2┬а+ y2)тАУ 18xy + 50x тАУ 50y + 77 = 0