KCET · Maths · Parabola
If the parabola \(y=\alpha x^{2}-6 x+\beta\) passes through the point \((0,2)\) and has its tangent at \(x=\frac{3}{2}\) parallel to \(X\)-axis, then
- A \(\alpha=2, \beta=-2\)
- B \(\alpha=-2, \beta=2\)
- C \(\alpha=2, \beta=2\)
- D \(\alpha=-2, \beta=-2\)
Answer & Solution
Correct Answer
(C) \(\alpha=2, \beta=2\)
Step-by-step Solution
Detailed explanation
\(y=\alpha x^{2}-6 x+\beta...(i)\)
Equation of parabola passes through
\(\begin{aligned}
&(0,2) \Rightarrow 2=0-0+\beta \\
&\Rightarrow \beta=2
\end{aligned}\)
Differentiating Eq. (i) W.r.t. \(x\),
\(\frac{d y}{d x}=2 \alpha x-6\)
\(\begin{aligned}
\therefore \quad\left(\frac{d y}{d x}\right)_{x=\frac{3}{2}} &=2 \times \alpha \times \frac{3}{2}-6 \\
&=3 \alpha-6
\end{aligned}\)
As, the tangent is parallel to \(X\)-axis,
\(\begin{aligned}
& & \frac{d y}{d x} &=0 \\
\Rightarrow & & 3 \alpha-6 &=0 \\
\Rightarrow & \alpha &=2
\end{aligned}\)
Equation of parabola passes through
\(\begin{aligned}
&(0,2) \Rightarrow 2=0-0+\beta \\
&\Rightarrow \beta=2
\end{aligned}\)
Differentiating Eq. (i) W.r.t. \(x\),
\(\frac{d y}{d x}=2 \alpha x-6\)
\(\begin{aligned}
\therefore \quad\left(\frac{d y}{d x}\right)_{x=\frac{3}{2}} &=2 \times \alpha \times \frac{3}{2}-6 \\
&=3 \alpha-6
\end{aligned}\)
As, the tangent is parallel to \(X\)-axis,
\(\begin{aligned}
& & \frac{d y}{d x} &=0 \\
\Rightarrow & & 3 \alpha-6 &=0 \\
\Rightarrow & \alpha &=2
\end{aligned}\)
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