Geo5.1.42

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Prove that the locus of the point of intersection of two perpendicular normals to the parabola y^{2}=4ax\, is the parabola y^{2}=a(x-3a)\,

Let P(x_{1},y_{1})\, be the ppoint of intersection of the normals at A(t_{1}),B(t_{2})\,

Equation of the normal at t is y+tx=2at+at^{3}\,

If it passes through (x1,y1)

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

If t1,t2,t3 are the roots of the above equation, t_{1}t_{2}t_{3}={\frac  {y_{1}}{a}}\,.

Let this be equation 2

Slope of the normal at t1=-t1.

Slope of the normal at t2=-t2.

Given PA is perpendicular to PB (-t_{1})(-t_{2})=-1,t_{1}t_{2}=1\,.equation 3

From 2 and 3

t_{3}={\frac  {-y}{a}}\,

Since the normal at t3 also passes thro' P(x_{1},y_{1})\,

y_{1}+x_{1}t_{3}=2at_{3}+at_{3}^{{3}}\,. equation 4.

y_{1}+x_{1}({\frac  {-y_{1}}{a}})=2a({\frac  {-y_{1}}{a}}+a({\frac  {-y_{1}^{{3}}}{a^{3}}}\,

1-{\frac  {x_{1}}{a}}=-2-{\frac  {y_{1}^{{2}}}{a^{2}}}\,

y_{1}^{{2}}=a(x_{1}-3a)\,

Hence the locus of P is

y^{2}=a(x-3a)\,


Main Page:Geometry:The Parabola