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Question: The equation of the conic with focus at (1, – 1), directrix along \(x - y + 1 = 0\) and with eccentr...

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

A

x2y2=1x^{2} - y^{2} = 1

B

xy=1xy = 1

C

2xy4x+4y+1=02xy - 4x + 4y + 1 = 0

D

2xy+4x4y1=02xy + 4x - 4y - 1 = 0

Answer

2xy4x+4y+1=02xy - 4x + 4y + 1 = 0

Explanation

Solution

Here, focus (S) = (1, –1), eccentricity (5)=2\sqrt{2}

From definition , SP=ePMSP = ePM

(x1)2+(y+1)2=2.(xy+1)12+12\sqrt{(x - 1)^{2} + (y + 1)^{2}} = \frac{\sqrt{2}.(x - y + 1)}{\sqrt{1^{2} + 1^{2}}}(x1)2+(y+1)2(x - 1)^{2} + (y + 1)^{2} =

(xy+1)2(x - y + 1)^{2}2xy4x+4y+1=02xy - 4x + 4y + 1 = 0, which is the required equation of conic (Rectangular hyperbola)