What will be the parametric equation and slope of normal to the parabola $x^{2} = - 8 y$ ?

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

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

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Slope

= \frac{1}{t}

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Slope

= t

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Solution

The correct option is C Slope $= \frac{1}{t}$
On comparing $x^{2} = - 8 y$ with standard equation of parabola $x^{2} = - 4 a y$ we get $a = 2.$
The parametric coordinates of point of contact be $(2 a t, - a t^{2})$ i.e $(4 t, - 2 t^{2}) .$
Now, the equation of normal will be $x - t y = a t^{3} + 2 a t$ i.e $x - t y = 2 t^{3} + 4 t$
Thus, from the equation of normal we found the slope of normal to be $\frac{1}{t} .$

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