The equation of the normal drawn to the parab | vedclass

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(B) The equation of the parabola is $y^2=6x$. Comparing this with $y^2=4ax$,we get $4a=6$,so $a=1.5$. The equation of the normal to the parabola $y^2=

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$4x+y=108$

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(B) The equation of the parabola is $y^2=6x$. Comparing this with $y^2=4ax$,we get $4a=6$,so $a=1.5$. The equation of the normal to the parabola $y^2=4ax$ at the point $(x_1, y_1)$ is given by $y-y_1 = -\frac{y_1}{2a}(x-x_1)$. Substituting the values $x_1=24$,$y_1=12$,and $a=1.5$: $y-12 = -\frac{12}{2(1.5)}(x-24)$ $y-12 = -\frac{12}{3}(x-24)$ $y-12 = -4(x-24)$ $y-12 = -4x+96$ $4x+y = 108$.

The equation of the normal drawn to the parabola $y^2=6x$ at the point $(24,12)$ is

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