Answer to Question #234741 in Analytic Geometry for Bless

Question

Given the parabola (y-1)² = 1/2(x-3),

find;

i. The coordinates of the vertex

ii. The coordinates of the focus

iii. The equation of the directrix

iv. The equation of the axis of symmetry

v. The ends of the latus rectum

vi. The parametric equations of the parabola

vii. Sketch the parabola, clearly showing the focus, vertex, directrix, axis of symmetry, and the x-intercept.

Answer 1

Solution:

Given, $\left(y-1\right)^2=\frac{1}{2}\left(x-3\right)$

$4p\left(x-h\right)=\left(y-k\right)^2\:\\\mathrm{\:is\:the\:standard\:equation\:for\:a\:right-left\:facing\:parabola\\\:with\:vertex\:at} \left(h,\:k\right),\: \\\mathrm{and\:a\:focal\:length\:}\:|p|$

Rewriting,

$4\cdot\frac18(x-3)=(y-1)^2$

On comparing, $\left(h,\:k\right)=\left(3,\:1\right),\:p=\frac{1}{8}$

i. The coordinates of the vertex = (3,1)

ii.

$\mathrm{Parabola\:is\:symmetric\:around\:the\:x-axis\:and\:so\:the\:focus\:lies\:a\:distance\:}p \\\mathrm{\:from\:the\:center\:}\left(3,\:1\right)\mathrm{\:along\:the\:x-axis}$

$\left(3+p,\:1\right)=\left(3+\frac{1}{8},\:1\right)=\left(\frac{25}{8},\:1\right)$

So, the coordinates of the focus$=\left(\frac{25}{8},\:1\right)$

iii.

$Parabola\:is\:symmetric\:around\:the\:x-axis\:and\:so\:the\:directrix\:is\:a\:line\:parallel \\\:to\:the \:y-axis,\:a\:distance\:-p\mathrm{\:from\:the\:center\:}\left(3,\:1\right)\mathrm{\:x-coordinate}\:$

$x=3-p=3-\frac18 \\\Rightarrow x=\frac{23}8$

The equation of the directrix: $x=\frac{23}8$

iv.

$\mathrm{Parabola\:is\:of\:the\:form\:}4p\left(x-h\right)=\left(y-k\right)^2\mathrm{\:and\:is\:symmetric\:around\:the\:}x\mathrm{-axis}$

$\mathrm{Axis\:of\:symmetry\:is\:a\:line\:parallel\:to\:the\:}x\mathrm{-axis\:which\:intersects\:the\:vertex:} \\y=1$

The equation of the axis of symmetry: y=1

v. The ends of the latus rectum: $x=\frac{25}8$

vi.

$\text{The parametric equations of the parabola}\ (y - k)^2 = 4p(x - h)\ are \\ x = h + pt^2\ and\ y = k + 2pt.$

$\left(h,\:k\right)=\left(3,\:1\right),\:p=\frac{1}{8}$

So, $x=3+\frac18t^2,y=1+2(\frac18)t$

The parametric equations of the parabola: $x=3+\frac18t^2,y=1+\frac14t$

vii. Graph: