A line of fixed length 2 units moves so that its ends are on the positive x-axis and that part of the line `x+y=0` which lies in the second quadrant. Then the locus of the midpoint of the line has equation. `x^2+5y^2+4x y-1=0` `x^2+5y^2+4x y+1=0` `x^2+5y^2-4x y-1=0` `4x^2+5y^2+4x y+1=0`

Options

A. `x^(2) + 5y^(2) +4xy-1=0`
B. `x^(2) + 5y^(2) +4xy+1=0`
C. `x^(2) + 5y^(2) -4xy-1=0`
D. `x^(2) + 5y^(2) -4xy-1=0`

Answer

A

Explanation

As shown in the figure, line segment AB, of fixed length 2 units, slides such that one of its ends A is on x-axis and other end B is on line x+y=0.
Let P(h,k) be midpoint of AB.
Also, let `angleBAO = theta`.
So, in triangle AMB, BM = `2 "sin" theta " and " AM = 2 "cos" theta.`
In triangle BMO, MO=MB = `2"sin" theta`
`therefore A-=(2"cos "theta - 2 "sin" theta, 0) " and "B-=(-2" sin" theta, 2 "sin "theta)`
Now, P(h,k) is the midpoint of AB.
`therefore 2h=2"cos "theta - 2 "sin" theta +
(-2" sin" theta) " and " 2k= 2"sin "theta`
`rArr h="cos "theta - 2 "sin" theta " and " k= "sin "theta`
`rArr h+2k="cos "theta " and " k= "sin "theta`
Squaring and adding, we get
`(h+2k)^(2) + k^(2) = 1`
`"or "x^(2) + 5y^(2)+4xy-1 = 0`
This is the required locus.