Mathematics - Parabola Question with Solution | TestHub

Rewrite the parabola as $(y+3{)}^2=4\left(x+\frac{9}{4}\right)$. With $X=x+\frac{9}{4},Y=y+3$, the normal at the parametric point $t$ is $Y=-tX+2t+t^3$.

For the point $(-1,4)$: $X=\frac{5}{4},Y=7$, so $7=-\frac{5}{4}t+2t+t^3$, i.e. $t^3+\frac{3}{4}t-7=0$.

Since $f^{\prime }(t)=3t^2+\frac{3}{4}>0$, this cubic is strictly increasing and has exactly one real root. Hence exactly one normal can be drawn from $(-1,4)$.