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General Equation of Parabola

A parabolic c...

Question

A parabolic curve is described parametrically by $x - 3 = t^{2},$ $y = 4 t .$ Then equation of the parabola is

y^{2} - 4 x + 16 = 0

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y^{2} - 16 x - 48 = 0

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y^{2} - 16 x + 48 = 0

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y^{2} = 16 x

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Solution

The correct option is C $y^{2} - 16 x + 48 = 0$
Given parametric equation is
$x - 3 = t^{2}, y = 4 t$
$y = 4 t \Rightarrow \frac{y}{4} = t \dots (1)$
Putting value of $t$ from $(1)$ in $x - 3 = t^{2}$ , we get
$x - 3 = {(\frac{y}{4})}^{2}$
$\Rightarrow 16 (x - 3) = y^{2}$
$\Rightarrow y^{2} - 16 x + 48 = 0$

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Similar questions

x - 3 = t^{2},

y = 4 t .

x = t^{2} + t + 1, y = t^{2} - t + 1

x - 3 = t^{2}, y = 4 t

x = t^{2} + 2 t + 2, y = t^{2} + 2 t + 2

x = t^{2} + t + 1

y = t^{2} - t + 1

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