The equation of the conic with focus at 1-1- directrix along x - y -1-0 and with eccentricity -2 isA- 2 x y-4 x-4 y-1-0B- x2-y2-1C- x y-1D- 2 x y-4 x-4 y-1-0

The correct option is D If (x_1, y_1) is a point on the conic, then √((x_1 1)^2 + (y_1 1)^2) = √(2) | x_1 y_1 + 1/√(2) | ⇒ (x_1 1)^2 + (y_1 + 1)^2 = ( ...

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Standard XII

Mathematics

General Equation of Hyperbola

The equation ...

Question

The equation of the conic with focus at (1, –1), directrix along x – y + 1 = 0 and with eccentricity $\sqrt{2}$ is

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Solution

The correct option is C

If $(x_{1}, y_{1})$ is a point on the conic, then

$\sqrt{(x_{1} - 1)^{2} + (y_{1} - 1)^{2}} = \sqrt{2} ∣ ∣ ∣ \frac{x_{1} - y_{1} + 1}{\sqrt{2}} ∣ ∣ ∣ \Rightarrow (x_{1} - 1)^{2} + (y_{1} + 1)^{2} = (x_{1} - y_{1} + 1)^{2} \Rightarrow x_{1}^{2} + 1 - 2 x_{1} + y_{1}^{2} + 1 + 2 y_{1} = x_{1}^{2} + y_{1}^{2} - 2 x_{1} - 2 y_{1} + 1 \Rightarrow 2 x_{1} y_{1} - 4 x_{1} + 4 y_{1} + 1 = 0$

$∴$ The equation of the conic is $2 x y - 4 x + 4 y + 1 = 0$

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The equation of the conic with focus at (1, –1), directrix along x – y + 1 = 0 and with eccentricity √2 is

The correct option is C If (x1,y1) is a point on the conic, then √(x1−1)2+(y1−1)2=√2∣∣∣x1−y1+1√2∣∣∣⇒(x1−1)2+(y1+1)2=(x1−y1+1)2⇒x21+1−2x1+y21+1+2y1=x21+y21−2x1−2y1+1⇒2x1y1−4x1+4y1+1=0 ∴ The equation of the conic is 2xy−4x+4y+1=0

The equation of the conic with focus at (1, –1), directrix along x – y + 1 = 0 and with eccentricity $\sqrt{2}$ is