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If `l,m` are variable real numbers such that `5l^2+6m^2 -4lm+3l=0` and the variable line `lx + my = 1` always touches a fixed parabola `P(x,y) = 0` whose axis parallel to x-axis , then A) vertex of P(x,y) is (-5/3, 4/3) B) Latus Rectum is equal to 4 C) Directrix of P(x,y) is parallel to y-axis (D) AB is a focal chord of P(x,y) =0 then `1/(SA)+1/(SB)=1/2` where S is the focus

A

`(-5//3,4//3)`

B

`(-7//4,3//4)`

C

`(-7//4,3//4)`

D

`(1//2,-3//4)`

Text Solution

Verified by Experts

The correct Answer is:
A

(1)
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.`
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