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From a point (\alpha, \beta) three normals are drawn to the parabola \mathrm{y^2=4 a x} and tangents are drawn to the parabola at the feet of the normals to form a triangle then centroid of the triangle is
 

Option: 1

\mathrm{\left(\frac{3 a-\alpha}{2}, 0\right)}
 


Option: 2

\mathrm{\left(\frac{2 a-\alpha}{3}, 0\right)}
 


Option: 3

\mathrm{(a,-\beta)}
 


Option: 4

\mathrm{\left(\frac{2 a-\alpha}{3}, \frac{\beta}{2}\right)}


Answers (1)

best_answer

Equation of normal to the given parabola is \mathrm{y=m x-2 a m-a m^3}, which passes through \mathrm{(\alpha, \beta).}
\mathrm{ \therefore a m^3+(2 a-\alpha) m+\beta=0 \text {. } }

\mathrm{ Let \: m_1, m_2, m_3\: be \: its\: roots. }

\mathrm{ \therefore \quad m_1+m_2+m_3=0, m_1 m_2+m_2 m_3+m_1 m_3=\frac{2 a-\alpha}{a} }

\mathrm{ \text { and } m_1 m_2 m_3=\frac{\beta}{a} }
Now the vertices of the triangle formed by tangents are
\mathrm{ A\left(a m_1 m_2, a\left(m_1+m_2\right)\right), B\left(a m_2 m_3, a\left(m_2+m_3\right)\right), }

\mathrm{ C\left(a m_1 m_3, a\left(m_1+m_3\right)\right) }

\mathrm{ \therefore \quad \text { Centroid of } \Delta=\left(\frac{x_1+x_2+x_3}{3}, \frac{y_1+y_2+y_3}{3}\right) }

\mathrm{ \quad=\left(\frac{a\left(m_1 m_2+m_2 m_3+m_1 m_3\right)}{3}, \frac{2 a\left(m_1+m_2+m_3\right)}{3}\right) }

\mathrm{ \quad=\left(\frac{a(2 a-\alpha)}{3 a}, 0\right) }

Hence option 2 is correct.

Posted by

Ritika Kankaria

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