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A chord which is normal at "t" to the parabola y^{2}=4ax\, subtends a right angle at the vertex. Then prove that t=\pm {\sqrt  {2}}\,

Let the normal at P(at^{2},2at)\, meet the conic again at Q(at_{1}^{{2}},2at_{1})\,

t_{1}=-t-{\frac  {2}{t}}\,. Let this equation be 1.

Slope of AP is {\frac  {2at}{at^{2}}}={\frac  {2}{t}}\,

Slope of AQ is {\frac  {2}{t_{1}}}\,

Given AP is perpendicular to AQ,hence

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


From the equation 1,


t^{2}=2,t=\pm {\sqrt  {2}}\,

Main page:Geometry:The Parabola