If the normals to the parabola y^2 = 4ax at the points t1 & t2 intersect again on the parabola at the point ‘t3’ then t1*t2 = 2 ; t3 = – (t1 + t2) and hence prove that the line joining t1 & t2 passes through a fixed point (–2a, 0).

Question

Arun · Answer

Dear student normal to Equation at ‘t' y = – t x + 2at + at^3 t^3 + t(2a-x) – y = 0 as it has three roots t1, t2, t3 hence t1 + t2 + t3 = 0 hence t3 = – (t1 + t2)

Vikas TU · Answer

Dear student For more details and examples Please follow this link https://www.askiitians.com/iit-jee-coordinate-geometry/normal-to-a-parabola.aspx Good Luck

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If the normals to the parabola y^2 = 4ax at the points t1 & t2 intersect again on the parabola at the point ‘t3’ then t1*t2 = 2 ; t3 = – (t1 + t2) and hence prove that the line joining t1 & t2 passes through a fixed point (–2a, 0).

If the normals to the parabola y^2 = 4ax at the points t1 & t2 intersect again on the parabola at the point ‘t3’ then t1*t2 = 2 ; t3 = – (t1 + t2) and hence prove that the line joining t1 & t2 passes through a fixed point (–2a, 0).

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If the normals to the parabola y^2 = 4ax at the points t1 & t2 intersect again on the parabola at the point ‘t3’ then t1*t2 = 2 ; t3 = – (t1 + t2) and hence prove that the line joining t1 & t2 passes through a fixed point (–2a, 0).

If the normals to the parabola y^2 = 4ax at the points t1 & t2 intersect again on the parabola at the point ‘t3’ then t1*t2 = 2 ; t3 = – (t1 + t2) and hence prove that the line joining t1 & t2 passes through a fixed point (–2a, 0).

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