In the given figure vertices of ∆ ABC lie on y = f (x) = ax 2 + bx + c. The ∆ ABC is…

Mathematics · JEE Advanced · NTA Exams — Complex Numbers and Quadratic Equations

In the given figure vertices of ∆ ABC lie on y = f (x)
= ax² + bx + c. The ∆ ABC is right angled isosceles triangle whose hypotenuse AC = \(4 \sqrt{2}\) units, then
 
y = f (x) is given by

1. \(y=\frac{x^{2}}{2 \sqrt{2}}-2 \sqrt{2}\)
2. \(y=\frac{x^{2}}{2}-2\)
3. y = x² – 8
4. \(y=x^{2}-2 \sqrt{2}\)

Answer

(A) y= x^ 2 2 2 -2 2

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