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If the points \mathrm{\left(a u^2, 2 a u\right)} and \mathrm{\left(a v^2, 2 a v\right)} are the extremities of a focal chord of the parabola \mathrm{y^2=4 a x, \quad} then

Option: 1

\mathrm{u v-1=0}


Option: 2

\mathrm{u v+1=0}


Option: 3

\mathrm{u+v=0}


Option: 4

\mathrm{u-v=0}


Answers (1)

best_answer

Equation of focal chord for the parabola \mathrm{y^2=4 a x} passes through the point \mathrm{\left(a u^2, 2 a u\right)} and \mathrm{\left(a v^2, 2 a v\right)}

\mathrm{\Rightarrow \quad y-2 a u=\frac{2 a v-2 a u}{a v^2-a u^2}\left(x-a u^2\right) \Rightarrow y-2 a u=\frac{2 a(v-u)}{a(v-u)(v+u)}\left(x-a u^2\right) \Rightarrow y-2 a u=\frac{2}{v+u}\left(x-a u^2\right)}It this is focal chord, so it would passes through focus (a, 0)

\mathrm{\Rightarrow \quad 0-2 a u=\frac{2}{v+u}\left(a-a u^2\right) \Rightarrow-u v-u^2=1-u^2, \therefore u v+1=0}

Trick: Given points \mathrm{\left(a u^2, 2 a u\right)} and \mathrm{\left(a v^2, 2 a v\right)}, then \mathrm{t_1=u} and \mathrm{t_2=v,} we know that \mathrm{t_1 t_2=-1}. Hence \mathrm{u v+1=0}

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Pankaj

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