What will be the parametric equation and slope of normal to the parabola $y^{2} = 12 x$ ?

y + t x - 3 t^{3} - 6 t = 0

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y + t x + 3 t^{3} + 6 t = 0

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Slope

= - t

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Slope

= t

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Solution

The correct option is C Slope $= - t$
On comparing $y^{2} = 12 x$ with standard equation of parabola $y^{2} = 4 a x$ we get $a = 3.$
The parametric coordinates of point of contact be $(a t^{2}, 2 a t)$ i.e $(3 t^{2}, 6 t) .$
Now, the equation of normal will be $y + t x = a t^{3} + 2 a t$ i.e $y + t x = 3 t^{3} + 6 t$
Thus, from the equation of normal we found the slope of normal to be $- t .$

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