Geo5.1.41

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If the normals at the points t_{1},t_{2}\, on the parabola y^{2}=4ax\, meet on the parabola, prove that t_{1}t_{2}=2\,

The equations of the normal at t1,t2 are

y+t_{1}x=2at_{1}+at_{1}^{{3}},y+t_{2}x=2at_{2}+at_{2}^{{3}}\,

Let the normal at t1 and t2 meet at t3 on the parabola

Therefore (at_{3}^{{2}},2at_{3})\, lies on the first two equations,hence

2at_{3}+t_{1}(at_{3}^{{2}})=2at_{1}+at_{1}^{{3}}\,

2at_{3}+t_{2}(at_{3}^{{2}})=2at_{2}+at_{3}^{{2}}\,.Let these two equations be 3 and 4.

From 3,

2(t_{3}-t_{1})=t_{1}(t_{1}^{{2}}-t_{3}^{{2}})\, (Since t1 not equal to t3).

2=-t_{1}(t_{1}+t_{3}),t_{3}={\frac  {-2}{t_{1}}}-t_{1}\,

From 4,

t_{3}={\frac  {-2}{t_{2}}}-t_{2}\,

Therefore,

{\frac  {-2}{t_{1}}}-t_{1}={\frac  {-2}{t_{2}}}-t_{2}\,

t_{1}-t_{2}={\frac  {2(t_{1}-t_{2})}{t_{1}t_{2}}}\,

t_{1}t_{2}=2\, (Since t1 si not equal to t2).


Main Page:Geometry:The Parabola