The equation of the hyperbola whose directrix is x+2 y=1, focus (2,1) and eccentricity 2 will be

Question

The equation of the hyperbola whose directrix is x+2 y=1, focus (2,1) and eccentricity 2 will be (A) x2-16 x y-11 y2-12 x+6 y+21=0 (B) 3 x2+16 x y+15 y2-4 x-14 y-1=0 (C) 3 x2+16 x y+11 y2-12 x-6 y+21=0 (D) None of these

Tardigrade Admin · Accepted Answer

PS = ePM ⇒ PS 2= e 2 PM 2 ⇒(x-2)2+(y-1)2=4 ((x+2 y-1)2/5) ⇒ x2+y2-4 x-2 y+5=(4/5)(x2+4 y2+1+4 x y-4 y-2 x) ⇒ 5 x2+5 y2-20 x-10 y+25=4 x2+16 y2+4+16 x y-16 y-8 x x2-16 x y-11 y2-12 x+6 y+21=0

Q. The equation of the hyperbola whose directrix is $x + 2 y = 1$ , focus $(2, 1)$ and eccentricity 2 will be

$x^{2} - 16 x y - 11 y^{2} - 12 x + 6 y + 21 = 0$

$3 x^{2} + 16 x y + 15 y^{2} - 4 x - 14 y - 1 = 0$

$3 x^{2} + 16 x y + 11 y^{2} - 12 x - 6 y + 21 = 0$

None of these

Q. The equation of the hyperbola whose directrix is x+2y=1, focus (2,1) and eccentricity 2 will be

x2−16xy−11y2−12x+6y+21=0

3x2+16xy+15y2−4x−14y−1=0

3x2+16xy+11y2−12x−6y+21=0

None of these

PS=ePM ⇒PS2=e2PM2 ⇒(x−2)2+(y−1)2=45(x+2y−1)2​ ⇒x2+y2−4x−2y+5=54​(x2+4y2+1+4xy−4y−2x) ⇒5x2+5y2−20x−10y+25=4x2+16y2+4+16xy−16y−8x x2−16xy−11y2−12x+6y+21=0

Q. The equation of the hyperbola whose directrix is $x + 2 y = 1$ , focus $(2, 1)$ and eccentricity 2 will be

$x^{2} - 16 x y - 11 y^{2} - 12 x + 6 y + 21 = 0$

$3 x^{2} + 16 x y + 15 y^{2} - 4 x - 14 y - 1 = 0$

$3 x^{2} + 16 x y + 11 y^{2} - 12 x - 6 y + 21 = 0$