The equation of the hyperbola with focus $(1,2), e=\sqrt{3}$ and directrix $2 x+y=1$ is given by

The equation of the hyperbola with focus $(1,2), e=\sqrt{3}$ and directrix $2 x+y=1$ is given by
  1. $2 y^2-12 x y-7 x^2+2 x-14 y+22=0$
  2. $2 y^2+12 x y+7 x^2-2 x+14 y-22=0$
  3. $2 y^2-12 x y-7 x^2-2 x-14 y-22=0$
  4. $2 y^2+12 x y+7 x^2+2 x+14 y+22=0$

Solution

Given, Focus $(S)=(1,2)$ Eccentricity $(e)=\sqrt{3}$ Equation of Directrix is $2 x+y=1$ Required equation of hyperbola is $S P=e \mathrm{PM}$ $ \sqrt{(x-1)^2+(y-2)^2}=\sqrt{3} \frac{|2 x+y-1|}{\sqrt{2^2+1^2}} $ Squaring on both sides, $ \begin{aligned} & (x-1)^2+(y-2)^2=\frac{3}{5}(2 x+y-1)^2 \\ \Rightarrow \quad & x^2+1-2 x+y^2+4-4 y \\ = & \frac{3}{5}\left(4 x^2+y^2+1+4 x y-2 y-4 x\right) \\ \Rightarrow \quad & 5\left(x^2+y^2-2 x-4 y+5\right) \\ = & 3\left(4 x^2+y^2+4 x y-4 x-2 y+1\right) \\ & 5 x^2+5 y^2-10 x-20 y+25 \\ & =12 x^2+3 y^2+12 x y-12 x-6 y+3 \\ \Rightarrow \quad & 5 x^2+5 y^2-10 x-20 y+25-12 x^2 \\ \Rightarrow \quad & -3 y^2-12 x y+12 x+6 y-3=0 \\ \therefore \quad & 2 y^2-7 x^2-12 x y-7 x^2+2 x-14 y+22=0 \end{aligned} $ Hence, (1) Option is correct

Asked in: AP EAMCET 2020 (22 Sep Shift 2)

Practice more Hyperbola questions on Aicharya