Question
A satellite dish has a shape called a paraboloid, where each cross section is parabola. Since radio signals $($parallel to axis$)$ will bounce off the surface of the dish to the focus, the receiver should be placed at the focus. The dish is $12$ ft across, and $4.5$ ft deep at the vertex
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$i$. Name the type of curve given in the above paragraph and find the equation of curve? $(1)$
$ii.$  Find the equation of parabola whose vertex is $(3, 4)$ and focus is $(5, 4). (1)$
$iii.$ Find the equation of parabola Vertex $(0, 0)$ passing through $(2, 3)$ and axis is along $x-$ axis. and also find the length of latus rectum. $(2)$
OR
$iv$. Find focus, length of latus rectum and equation of directrix of the parabola $x^2=8 y. (2)$

Answer

$i.$ Given curve is a parabola
Equation of parabola is $x^2=4 a y$
It passes through the point $(6, 4.5)$
$\Rightarrow 36=4 \times a \times 4.5$
$\Rightarrow 36=18 a$
$\Rightarrow a=2$
Equation of parabola is $x^2=8 y$
$ii.$ Distance between focus and vertex is $= a =\sqrt{(4-4)^2+(5-3)^2}=2$
Equation of parabola is $(y-k)^2=4 a(x-h)$
where $(h, k)$ is vertex
$\Rightarrow$ Equation of parabola with vertex $(3,4) \ a=2$
$\Rightarrow(y-4)^2=8(x-3)$
$iii$. Equation of parabola with axis along $x -$ axis
$y ^2=4 ax$
which passes through $(2, 3)$
$\Rightarrow 9=4 a \times 2$
$\Rightarrow 4 a =\frac{9}{2}$
hence required equation of parabola is
$y^2=\frac{9}{2} x$
$\Rightarrow 2 y^2=9 x$
Hence length of latus rectum $= 4a = 4.5$
OR
$x^2=8 y$
$a = 2$
Focus of parabola is $(0, 2)$
length of latus rectum is $4 a=4 \times 2=8$
Equation of directrix $y + 2 = 0$

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