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Show that the locus of the point,two of the normals from which to the parabola y^{2}=4ax\, are coincident is 27ay^{2}=4(x-2a)^{3}\,

The equation to the normal at "t" to the given parabola is


Let this be 1

Let 1 pass thro' P(x_{1},y_{1})\,


at^{3}+t(2a-x_{1})-y_{1}=0\, equation 2

Since two of the three normals coincide,two roots of the equation 2 are equal then by theory of



Differentiating the bove equation,we get

3at^{2}+(2a-x_{1})=0\, have a common root. SOlving this and 2, t={\frac  {3y_{1}}{2(2a-x_{1})}}\,

Substituting this in 2,

a[{\frac  {27y_{1}^{{3}}}{8(2a-x_{1})^{3}}}]+{\frac  {3y_{1}}{2(2a-x_{1})}}\cdot (2a-x_{1})-y_{1}=0\,

{\frac  {27ay_{1}^{{2}}}{4(2a-x_{1})^{3}}}+1=0\,

Therefore,the locus of P is 27ay^{2}=4(x-2a)^{3}\,

Main Page:Geometry:The Parabola