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If a parabola having its axis parallel to $\mathrm{X}$-axis passes through the points $(0,-1),(6,1)$ and $(-2,-3)$, then the point at which this parabola cuts the $\mathrm{X}$-axis is
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The correct answer is:
$\left(\frac{5}{2}, 0\right)$
$\because$ Axis of parabola is $x$-axis.
$\therefore$ The equation of the parabola will of the form
$x=a y^2+b y+c$ ...(i)
$\because(0,-1),(6,1)$ and $(-2,-3)$ passes through parabola (i),
$0=a-b+c$ ...(ii)
$6=a+b+c$ ...(iii)
$-2=9 a-3 b+c$ ...(iv)
Solving eqns. (ii), (iii) and (iv), we get
$a=\frac{1}{2} ; b=3$ and $c=\frac{5}{2}$
Putting the above values in eqn. (i), we get
$x=\frac{1}{2} y^2+3 y+\frac{5}{2}$ ...(5)
For the point, at which parabola (v) cuts $x$-axis
$\begin{aligned} & y=0 \text { and } x=\frac{1}{2} \times 0+3 \times 0+\frac{5}{2} \Rightarrow x=\frac{5}{2} \\ & \therefore \text { Point is }\left(\frac{5}{2}, 0\right) .\end{aligned}$
$\therefore$ The equation of the parabola will of the form
$x=a y^2+b y+c$ ...(i)
$\because(0,-1),(6,1)$ and $(-2,-3)$ passes through parabola (i),
$0=a-b+c$ ...(ii)
$6=a+b+c$ ...(iii)
$-2=9 a-3 b+c$ ...(iv)
Solving eqns. (ii), (iii) and (iv), we get
$a=\frac{1}{2} ; b=3$ and $c=\frac{5}{2}$
Putting the above values in eqn. (i), we get
$x=\frac{1}{2} y^2+3 y+\frac{5}{2}$ ...(5)
For the point, at which parabola (v) cuts $x$-axis
$\begin{aligned} & y=0 \text { and } x=\frac{1}{2} \times 0+3 \times 0+\frac{5}{2} \Rightarrow x=\frac{5}{2} \\ & \therefore \text { Point is }\left(\frac{5}{2}, 0\right) .\end{aligned}$
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