TS EAMCET · Maths · Parabola
Find the equation to the parabola, whose axis parallel to the \(y\)-axis and which passes through the points \((0,4),(1,9)\) and \((4,5)\) is
- A \(y=-x^2+x+4\)
- B \(y=-x^2+x+1\)
- C \(y=\frac{-19}{12} x^2+\frac{79}{12} x+4\)
- D \(y=\frac{-19}{12} x^2+\frac{89}{12}+1\)
Answer & Solution
Correct Answer
(C) \(y=\frac{-19}{12} x^2+\frac{79}{12} x+4\)
Step-by-step Solution
Detailed explanation
The equation of parabola parallel to \(y\)-axis is \(y=A x^2+B x+C\) ...(i) The point \((0,4),(1,9)\) and \((4,5)\) lies on Eq. (i). Then, \(\quad 4=0+0+C \Rightarrow C=4\) ...(ii) \(9=A+B+C\) or \(\quad 9=A+B+4 \quad(\because C=4)\) \(A+B=5\) ...(iii) and…
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