Find the equation of a parabola, whose vertex is at $(- 3, 2)$ $(-3,2)$ and passes through $(4, 7)$ $(4,7)$ ?

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Answer 1 · 2017-04-05T12:45:15

$5 x^{2} + 30 x - 49 y + 143 = 0$ $5x^2+30x-49y+143=0$ or $7 y^{2} - 25 x - 28 y - 47 = 0$ $7y^2-25x-28y-47=0$

There could be two type of parabolas with vertex as $(- 3, 2)$ $(-3,2)$

A $(y - 2) = a {(x + 3)}^{2}$ $(y-2)=a(x+3)^2$ If this passes through $(4, 7)$ $(4,7)$ , we have

$(7 - 2) = a {(4 + 3)}^{2}$ $(7-2)=a(4+3)^2$ i.e. $a = \frac{5}{49}$ $a=5/49$ and equation of parabola is

$(y - 2) = \frac{5}{49} {(x + 3)}^{2}$ $(y-2)=5/49(x+3)^2$ i.e. $49 y - 98 = 5 x^{2} + 30 x + 45$ $49y-98=5x^2+30x+45$ or

$5 x^{2} + 30 x - 49 y + 143 = 0$ $5x^2+30x-49y+143=0$

B $(x + 3) = a {(y - 2)}^{2}$ $(x+3)=a(y-2)^2$ If this passes through $(4, 7)$ $(4,7)$ , we have

$(4 + 3) = a {(7 - 2)}^{2}$ $(4+3)=a(7-2)^2$ i.e. $a = \frac{7}{25}$ $a=7/25$ and equation of parabola is

$(x + 3) = \frac{7}{25} {(y - 2)}^{2}$ $(x+3)=7/25(y-2)^2$ i.e. $25 x + 75 = 7 y^{2} - 28 y + 28$ $25x+75=7y^2-28y+28$ or

$7 y^{2} - 25 x - 28 y - 47 = 0$ $7y^2-25x-28y-47=0$

graph{(5x^2+30x-49y+143)(7y^2-25x-28y-47)=0 [-11, 9, -1.36, 8.64]}

Find the equation of a parabola, whose vertex is at (−3,2)(-3,2) and passes through (4,7)(4,7)?