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In the given figue vertices of DeltaABC ...

In the given figue vertices of `DeltaABC` lie on `y=f(x)=ax^(2)+bx+c`. The `DeltaAB` is right angled isosceles triangle whose hypotenuse `AC=4sqrt(2)` units.

Minimum valueof `y=f(x)` is

A

`-4sqrt(2)`

B

`-2sqrt(2)`

C

`0`

D

`2sqrt(2)`

Text Solution

Verified by Experts

The correct Answer is:
B
Given that `AC=4sqrt(2)` units
`:.AB=BC=(AC)/(sqrt(2))=4` units andd `OB=sqrt((BC)^(2)-(OC)^(2))`
`=sqrt((4)^(2)-(2sqrt(2))^(2))[ :' OC=(AC)/2]`
`=2sqrt(2)` units
`:.` Vertices are `A=(-2sqrt(2),0)`,
`B=-(0,-2sqrt(2))`
and `C=(2sqrt(2),0)`
Minimum value of `y=(x^(2))/(2sqrt(2))-2sqrt(2)` is at `x=0`
`:.(y)_("min")=-2sqrt(2)`
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