If two normals to the parabola y^2 = 4x passi | vedclass

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Question

(A) The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$. Here $a = 1$,so the equation is $y = mx - 2m - m^3$.

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$y = x - 3$

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(A) The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$. Here $a = 1$,so the equation is $y = mx - 2m - m^3$.Since the normal passes through $(15, 12)$,we have $12 = 15m - 2m - m^3$,which simplifies to $m^3 - 13m + 12 = 0$.Let the slopes of the three normals be $m_1, m_2, m_3$. From the cubic equation,the sum of the roots is $m_1 + m_2 + m_3 = 0$.Given the two normals $4x + y = 72$ (slope $m_1 = -4$) and $3x - y = 33$ (slope $m_2 = 3$),we have $-4 + 3 + m_3 = 0$,which gives $m_3 = 1$.The equation of the third normal with slope $m = 1$ passing through $(15, 12)$ is $y - 12 = 1(x - 15)$,which simplifies to $y = x - 3$.

If two normals to the parabola $y^2 = 4x$ passing through the point $(15, 12)$ are $4x + y = 72$ and $3x - y = 33$,find the third normal.

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