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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.

`y=f(x)` is given by

A

`y=-x^(2)-8`

B

`y=(x^(2))/(2sqrt(2))-2sqrt(2)`

C

`y=x^(2)-4`

D

`y=(x^(2))/2-sqrt(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)`
Since `y=f(x)=ax^(2)+bx+c` passes through A,B and C then
`0=8a-2sqrt(2)b+c-2sqrt(2)=c`
and `0=8a+2sqrt(2)b+c`
We get `b=0, a=1/(2sqrt(20)` and `c=-2sqrt(2)`
`:.y=f(x)=(x^(2))/(2sqrt(2))-2sqrt(2)`
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