Question

$x-3=t^2,y=4t$ are the parametric equations of the parabola

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

Answer 1

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

$y=4t\Rightarrow \frac{y}{4}=t$
$x-3=t^2$
$\Rightarrow x-3={\left(\frac{y}{4}\right)}^2$
​​​​​​​$\Rightarrow 16(x-3)=y^2$