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Sum

If the normal at the point ‘t_{1}‘ on the parabola y^{2} = 4ax meets the parabola again at the point ‘t_{2}‘, then prove that t_{2} = `- ("t"_1 + 2/"t"_1)`

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#### Solution

Equation of normal to y^{2} = 4at’ t’ is y + xt = 2at + at^{3}.

So equation of normal at ‘t_{1}’ is y + xt_{1} = 2at_{1} + at_{1}^{3}

The normal meets the parabola y^{2} = 4ax at ‘t_{2}’

(ie.,) at (at_{2}^{2}, 2at_{2})

⇒ 2at_{2} + at_{1}t_{2}^{2} = 2at_{1} + at_{1}^{3}

So 2a(t_{2} – t_{1}) = at_{1}^{3} – at_{1}t_{2}^{2}

⇒ 2a(t_{2} – t_{1}) = at_{1}(t_{1}^{2} – t_{2}^{2})

⇒ 2(t_{2} – t_{1}) = t_{1}(t_{1} + t_{2})(t_{1} – t_{2})

⇒ 2= – t_{1}(t_{1} + t_{2})

⇒ t_{1} + t_{2} = `(-2)/"t"_1`

⇒ t_{2} = `- "t"_1 - 2/"t"_1`

= `- ("t"_1 + 2/"t"_1)`

Concept: Tangents and Normals to Conics

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