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If the parabola y=(a-b)x^2+(b-c)x+(c-a)...

If the parabola `y=(a-b)x^2+(b-c)x+(c-a)` touches x- axis then the line `ax+by+c=0`

A

A. always passes through a fixed point

B

B. represents the family of parallel lines

C

C. is always perpendicular to x-axis

D

D. always has negative slope

Text Solution

Verified by Experts

The correct Answer is:
A
Solving equation of parabola with x-axis (y=0), we get
`(a-b)x^(2) + (b-c)x +(c-a) =0`,
which should have two equal values of x, as x-axis touches the parabola.
`:. (b-c)^(2) -4(a-b) (c-a) =0`
`rArr (b+c -2a)^(2) =0`
`rArr (b+c -2a)^(2) =0`
`rArr -2a +b + c=0`
Thus, `ax +by +c =0` always passes through `(-2,1)`.
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Knowledge Check

  • Let the parabolas y=x(c-x)and y=x^(2)+ax+b touch each other at the point (1,0), then-

    A
    a+b+c=0
    B
    a+b=2
    C
    b-c=1
    D
    a+c=-2

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