If l and m are variable real number such that `5l^(2)+6m^(2)-4lm+3l=0`, then the variable line lx+my=1 always touches a fixed parabola, whose axes is parallel to the x-axis.
Then find the vertex of the parabola is

Options

A. `(1//6,-7//6)`
B. `(5//3,4//3)`
C. `(3//2,-3//2)`
D. `(-3//4,3//4)`

Answer

B

Explanation

(2)
Any parabola whose axes whose axes is parallel to the x-axis will be of the from
`(y-a)^(2)=4b(x-c)` (1)
Now, lx+my=1 can be rewritten as
`y-a=-(l)/(m)(x-c)+(1-am-lc)/(m)` (2)
Equation (2) will touch (1) if
`(1-am-lc)/(m)=(b)/(-l//m)`
`or-(l)/(m)=(bm)/(1-am-lc)`
`orcl^(2)-bm^(2)+alm-l=0` (3)
But given that
`5l^(2)+6m^(2)-4lm+3l=0` (4)
Comparing (3) and (4), we get
`(c)/(5)=(-b)/(6)=(a)/(-4)=(-1)/(3)`
`orc=(-5)/(3),b=2,anda=(4)/(3)`
So, the parabola is
`(y-(4)/(3))^(2)=8(x+(5)/(3))`
whose focus is `(1//3,4//3)` and directrix is 3x+11=0.`