Question

If \((1,2)\) is the focus, \(x+2y=0\) is the directrix and \(\sqrt{2}\) is the eccentricity of a hyperbola, then the equation of the hyperbola is

• A. \(x^2-y^2=a^2\)
• B. \(3x^2-8xy-3y^2-10x-20y+25=0\)
• C. \(xy=c^2\)
• D. \(3x^2-8xy-3y^2+10x-20y-25=0\)

Hint:

For a conic defined by a focus, directrix, and eccentricity, \[ \frac{\text{Distance from focus}} {\text{Distance from directrix}} =e. \] Squaring the resulting equation usually leads directly to the Cartesian equation of the conic.

Answer 1

The Correct Option is B

Step 1: Use the focus-directrix definition of a conic.
For any point \(P(x,y)\) on the conic, \[ \frac{\text{Distance from focus}}{\text{Distance from directrix}} =e. \] Given: \[ \text{Focus }F(1,2), \] \[ \text{Directrix }x+2y=0, \] and \[ e=\sqrt{2}. \] Since \(e\gt 1\), the conic is a hyperbola.

Step 2: Write the distance from the focus.
Distance of \(P(x,y)\) from \(F(1,2)\) is \[ PF=\sqrt{(x-1)^2+(y-2)^2}. \]

Step 3: Write the distance from the directrix.
Distance of \(P(x,y)\) from the line \[ x+2y=0 \] is \[ \frac{|x+2y|}{\sqrt{1^2+2^2}} = \frac{|x+2y|}{\sqrt5}. \]

Step 4: Apply the definition of eccentricity.
Using \[ \frac{PF}{PD}=\sqrt2, \] we get \[ \frac{\sqrt{(x-1)^2+(y-2)^2}} {\frac{|x+2y|}{\sqrt5}} =\sqrt2. \] Squaring both sides, \[ \frac{(x-1)^2+(y-2)^2} {\frac{(x+2y)^2}{5}} =2. \] Therefore, \[ (x-1)^2+(y-2)^2 = \frac{2}{5}(x+2y)^2. \] Multiplying by \(5\), \[ 5\left[(x-1)^2+(y-2)^2\right] = 2(x+2y)^2. \]

Step 5: Expand and simplify.
Expanding the left-hand side, \[ 5\left(x^2-2x+1+y^2-4y+4\right) \] \[ =5x^2+5y^2-10x-20y+25. \] Expanding the right-hand side, \[ 2(x^2+4xy+4y^2) \] \[ =2x^2+8xy+8y^2. \] Bringing all terms to one side, \[ 5x^2+5y^2-10x-20y+25 -2x^2-8xy-8y^2=0. \] \[ 3x^2-8xy-3y^2-10x-20y+25=0. \]

Step 6: Final conclusion.
Hence, the required equation of the hyperbola is \[ \boxed{3x^2-8xy-3y^2-10x-20y+25=0}. \]