Find the equation of the normal to the parabo | vedclass

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(B) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.The equation of a normal to the parabola

Vedclass · Accepted Answer

$4x + 16y = 33$

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(B) The equation of the parabola is $y^2 = 16x$. Comparing this with $y^2 = 4ax$,we get $4a = 16$,so $a = 4$.The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is given by $y = mx - 2am - am^3$.Given $m = -1/4$ and $a = 4$,we substitute these values into the formula:$y = (-1/4)x - 2(4)(-1/4) - 4(-1/4)^3$$y = -x/4 + 2 - 4(-1/64)$$y = -x/4 + 2 + 1/16$Multiply the entire equation by $16$ to clear the denominators:$16y = -4x + 32 + 1$$16y = -4x + 33$$4x + 16y = 33$

Find the equation of the normal to the parabola $y^2 = 16x$ having a slope of $-1/4$.

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