Question

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

• A. $y^2-4x+16=0$
• B. $y^2-16x-48=0$
• C. $y^2-16x+48=0$
• D. $y^2=16x$

Answer 1

The correct option is C $y^2-16x+48=0$

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