How do you write the parabola $y^{2} - 10 y - 27 x + 133 = 0$ $y^2-10y-27x+133=0$ in standard form and find the vertex, focus, and directrix?

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How do you write the parabola $y^{2} - 10 y - 27 x + 133 = 0$ $y^2-10y-27x+133=0$ in standard form and find the vertex, focus, and directrix?

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Answer 1 · 2017-06-02T12:02:54

Standard form: ${(y - 5)}^{2} = 4 \cdot \frac{27}{4} (x - 4)$ $(y-5)^2 = 4*27/4(x-4)$
Vertex is at ( 4,5) , Focus is at $(\frac{43}{4}), 5$ $(43/4), 5$ . directrix is $y = - \frac{11}{7}$ $y = -11/7$

Explanation:

$y^{2} - 10 y - 27 x + 133 = 0 or y^{2} - 10 y + 25 - 25 - 27 x + 133 = 0$ $y^2-10y-27x+133=0 or y^2-10y+25-25-27x+133=0$ or
${(y - 5)}^{2} = 27 x - 108 or {(y - 5)}^{2} = 27 (x - 4)$ $(y-5)^2 =27x -108 or (y-5)^2 = 27(x-4)$ or
${(y - 5)}^{2} = 4 \cdot \frac{27}{4} (x - 4)$ $(y-5)^2 = 4*27/4(x-4)$
Comparing with standard equation ${(y - k)}^{2} = 4 \cdot a (x - h); (h, k)$ $(y-k)^2 = 4*a(x-h); (h,k)$ being vertex. we get here $h = 4, k = 5, a = \frac{27}{4}$ $h = 4 , k = 5 , a= 27/4$

Vertex is at ( 4,5) , Focus is at $((h + a), k) or (4 + \frac{27}{4}), 5 or (\frac{43}{4}, 5)$ $((h+a),k) or (4 + 27/4), 5 or (43/4,5)$ .

Vertex is equidistant from focus and directrix.

So directrix is $y = (h - a) = 4 - \frac{27}{4} or y = - \frac{11}{7}$ $y= (h-a) = 4-27/4 or y = -11/7$
graph{(y-5)^2=27(x-4) [-160, 160, -80, 80]} [Ans]

Answer 2 · 2017-06-02T12:41:05

Vertex is $(4, 5)$ $(4,5)$ and focus is $(\frac{43}{4}, 5)$ $(43/4,5)$ and axis of symmetry is $x = - \frac{11}{4}$ $x=-11/4$ .

Explanation:

Standard form of equation of parabola is $y = a x^{2} + b x + c$ $y=ax^2+bx+c$ or $x = a y^{+} b y + c$ $x=ay^+by+c$ . Asin the given equation, we have $y^{2}$ $y^2$ , we can write $y^{2} - 10 y - 27 x + 133 = 0$ $y^2-10y-27x+133=0$ as

$27 x = y^{2} - 10 y + 133$ $27x=y^2-10y+133$ and in standard form it is

$x = \frac{1}{27} y^{2} - \frac{10}{27} y + \frac{133}{27}$ $x=1/27y^2-10/27y+133/27$

Vertex form of equation of paarbola is $y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$ or $x = {(y - k)}^{2} + h$ $x=(y-k)^2+h$ , where $x = h$ $x=h$ or $y = k$ $y=k$ is axis of symmetry (focus and vertex lie on this and directrix is perpendicular to it) and vertex is $(h, k)$ $(h,k)$ .

In the equation $y = a {(x - h)}^{2} + k$ $y=a(x-h)^2+k$ , focus is $(h, k + \frac{1}{4 a})$ $(h,k+1/(4a))$ and directrix is $y = k - \frac{1}{4 a}$ $y=k-1/(4a)$ . In case equation is $x = {(y - k)}^{2} + h$ $x=(y-k)^2+h$ , focus is $(h + \frac{1}{4 a}, k)$ $(h+1/(4a),k)$ and directrix is $x = h - \frac{1}{4 a}$ $x=h-1/(4a)$ .

$27 x = y^{2} - 10 y + 133 = y^{2} - 2 \times 5 \times y + 25 - 25 + 133$ $27x=y^2-10y+133=y^2-2xx5xxy+25-25+133$

$27 x = {(y - 5)}^{2} + 108$ $27x=(y-5)^2+108$ or $x = \frac{1}{27} {(y - 5)}^{2} + 4$ $x=1/27(y-5)^2+4$

This is the vertex form of this equation and vertex is $(4, 5)$ $(4,5)$ . As we have $a = \frac{1}{27}$ $a=1/27$ , $\frac{1}{4 a} = \frac{27}{4}$ $1/(4a)=27/4$ and focus is $(4 + \frac{27}{4}, 5)$ $(4+27/4,5)$ i.e. $(\frac{43}{4}, 5)$ $(43/4,5)$ and axis of symmetry is $x = 4 - \frac{27}{4} = - \frac{11}{4}$ $x=4-27/4=-11/4$ .

graph{(y^2-10y-27x+133)(x+11/4)((x-43/4)^2+(y-5)^2-2)=0 [-45.3, 114.64, -35.3, 44.7]}

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How do you write the parabola $y^{2} - 10 y - 27 x + 133 = 0$ $y^2-10y-27x+133=0$ in standard form and find the vertex, focus, and directrix?

2 Answers

Explanation:

Explanation:

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Humanities

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How do you write the parabola y^2-10y-27x+133=0y2−10y−27x+133=0y^2-10y-27x+133=0 in standard form and find the vertex, focus, and directrix?

2 Answers

Explanation:

Explanation:

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How do you write the parabola $y^{2} - 10 y - 27 x + 133 = 0$ $y^2-10y-27x+133=0$ in standard form and find the vertex, focus, and directrix?