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Question: If \(x + 2 = 0\) and \(y = 1\)are the equation of asymptotes of rectangular hyperbola passing throug...

If x+2=0x + 2 = 0 and y=1y = 1are the equation of asymptotes of rectangular hyperbola passing through (1,0).\left( {1,0} \right).Then which of the following is (are) are not the equation(s) of the hyperbola:
A. xy+2y1=0xy + 2y - 1 = 0
B. xy+2y+1=0xy + 2y + 1 = 0
C. xy2y1=0xy - 2y - 1 = 0
D. xyx+2y+1=0xy - x + 2y + 1 = 0

Explanation

Solution

Hint : Center of rectangular hyperbola is the point of intersection of asymptotes. Equation of a rectangular hyperbola whose center is at origin can be written as xy=cxy = c

Complete step-by-step answer :
x+2=0x + 2 = 0and y=1y = 1are the asymptotes.
Therefore, the point of intersection of asymptotes will be
(x,y)=(2,1)(x,y) = ( - 2,1). . . . . (1)
A rectangular hyperbola with center at (h,k)(h,k)is written as
(xh)(yk)=c(x - h)(y - k) = c. . . . (2)
Where, ccis constant.
Since, point of intersection of asymptotes is the center of rectangular hyperbola,
We can of write
(xh)(yk)=c(x - h)(y - k) = c
(x(2))(y1)=c\Rightarrow (x - ( - 2))(y - 1) = c (From equation (1) and (2))
(x+2)(y1)=c\Rightarrow (x + 2)(y - 1) = c
By simplifying it, we get
x(y1)+2(y1)=cx(y - 1) + 2(y - 1) = c
xyx+2y2=c\Rightarrow xy - x + 2y - 2 = c . . . . . (3)
The hyperbola passes through the points (1,0)(1,0)
Therefore, equation (3) can be written as
1×01+2×02=c1 \times 0 - 1 + 2 \times 0 - 2 = c
12=c\Rightarrow - 1 - 2 = c
c=3\Rightarrow c = - 3
Put this value in equation (3)
xyx+2y2=3\Rightarrow xy - x + 2y - 2 = - 3
xyx+2y2+3=0\Rightarrow xy - x + 2y - 2 + 3 = 0 (Adding 3 to both the sides).
xyx+2y+1=0\Rightarrow xy - x + 2y + 1 = 0
Which is the equation of required rectangular hyperbola.
Therefore, (A), (B), (c) are not the equations of hyperbola.

Note : Read the question carefully. We have to find equations which do not meet the conditions given in the question. That means, the answer we get should not be the answer that we have to mark.