The equation of the parabola whose axis is pa | vedclass

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(B) The general equation of a parabola with its axis parallel to the $X$-axis is given by $x = Ay^2 + By + C$. Since the parabola passes through the p

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$5 y^2+2 x-21 y+20=0$

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(B) The general equation of a parabola with its axis parallel to the $X$-axis is given by $x = Ay^2 + By + C$. Since the parabola passes through the points $(-2, 1)$,$(1, 2)$,and $(-1, 3)$,we substitute these coordinates into the general equation: For $(-2, 1)$: $-2 = A(1)^2 + B(1) + C \Rightarrow A + B + C = -2 \quad (i)$ For $(1, 2)$: $1 = A(2)^2 + B(2) + C \Rightarrow 4A + 2B + C = 1 \quad (ii)$ For $(-1, 3)$: $-1 = A(3)^2 + B(3) + C \Rightarrow 9A + 3B + C = -1 \quad (iii)$ Subtracting $(i)$ from $(ii)$: $(4A + 2B + C) - (A + B + C) = 1 - (-2) \Rightarrow 3A + B = 3 \quad (iv)$ Subtracting $(ii)$ from $(iii)$: $(9A + 3B + C) - (4A + 2B + C) = -1 - 1 \Rightarrow 5A + B = -2 \quad (v)$ Subtracting $(iv)$ from $(v)$: $(5A + B) - (3A + B) = -2 - 3$ $\Rightarrow 2A = -5$ $\Rightarrow A = -\frac{5}{2}$ Substituting $A = -\frac{5}{2}$ into $(iv)$: $3(-\frac{5}{2}) + B = 3 \Rightarrow B = 3 + \frac{15}{2} = \frac{21}{2}$ Substituting $A$ and $B$ into $(i)$: $-\frac{5}{2} + \frac{21}{2} + C = -2$ $\Rightarrow \frac{16}{2} + C = -2$ $\Rightarrow 8 + C = -2$ $\Rightarrow C = -10$ Thus,the equation is $x = -\frac{5}{2}y^2 + \frac{21}{2}y - 10$. Multiplying by $2$: $2x = -5y^2 + 21y - 20$. Rearranging gives $5y^2 + 2x - 21y + 20 = 0$.

The equation of the parabola whose axis is parallel to the $X$-axis and which passes through the points $(-2, 1)$,$(1, 2)$,and $(-1, 3)$ is

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