A point on the parabola whose focus and verte | vedclass

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(D) The general equation of a parabola opening to the right is $(y - k)^2 = 4a(x - h)$,where $(h, k)$ is the vertex.Given vertex $(h, k) = (1, -2)$ an

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$(10, 1)$

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(D) The general equation of a parabola opening to the right is $(y - k)^2 = 4a(x - h)$,where $(h, k)$ is the vertex.Given vertex $(h, k) = (1, -2)$ and focus $(h + a, k) = \left(\frac{5}{4}, -2ight)$.Calculating $a$: $h + a = \frac{5}{4}$ $\Rightarrow 1 + a = \frac{5}{4}$ $\Rightarrow a = \frac{1}{4}$.Substituting the values into the equation:$(y - (-2))^2 = 4 	imes \frac{1}{4}(x - 1)$$(y + 2)^2 = (x - 1)$.Now,check the given options:For option $(D)$,substitute $(10, 1)$ into the equation:$(1 + 2)^2 = 3^2 = 9$$(10 - 1) = 9$.Since $9 = 9$,the point $(10, 1)$ lies on the parabola.

$A$ point on the parabola whose focus and vertex are respectively at $\left(\frac{5}{4}, -2\right)$ and $(1, -2)$ is

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