What is the standard equation of parabola whose vertex is $(3, 2)$ $(3,2)$ and passes through $(- 1, 14)$ $(-1,14)$ ?

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Answer 1 · 2016-05-27T12:53:56

$3 x^{2} - 18 x - 4 y + 35 = 0$ $3x^2-18x-4y+35=0$

Standard form of parabola is $y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$ where $(h, k)$ $(h,k)$ is the vertex. As vertex is $(3, 2)$ $(3,2)$ , this is

$y = a {(x - 3)}^{2} + 2$ $y=a(x-3)^2+2$ and as it passes through $(- 1, 14)$ $(-1,14)$

i.e. $14 = a {(- 1 - 3)}^{2} + 2$ $14=a(-1-3)^2+2$

or $14 = a {(- 4)}^{2} + 2$ $14=a(-4)^2+2$

or $14 = 16 a + 2$ $14=16a+2$

or $16 a = 12$ $16a=12$ and hence $a = \frac{12}{16} = \frac{3}{4}$ $a=12/16=3/4$

Hence equation of parabola is

$y = \frac{3}{4} {(x - 3)}^{2} + 2$ $y=3/4(x-3)^2+2$

or $4 y = 3 (x^{2} - 6 x + 9) + 8$ $4y=3(x^2-6x+9)+8$

or $3 x^{2} - 18 x - 4 y + 35 = 0$ $3x^2-18x-4y+35=0$

What is the standard equation of parabola whose vertex is (3,2)(3,2) and passes through (−1,14)(-1,14)?