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Tangents drawn to parabola \mathrm{Y}^{2}=4 \mathrm{ax} at the point \mathrm{A} and \mathrm{B}  intersect at \mathrm{C}. If '\mathrm{S}' be the focus of the parabola then, SA, SC and SB forms

Option: 1

a A.P.


Option: 2

a G.P.


Option: 3

a H.P.


Option: 4

None of these


Answers (1)

best_answer

If \mathrm{A \equiv\left(a^{t_{1}^{2}}, 2 a^{t_{1}}\right), B \equiv\left(a^{t_{2}^{2}}, 2 a^{t_{2}}\right)},

\mathrm{\Rightarrow \quad C \equiv\left(a^{t_{1} t_{2}}, a\left(t_{1}+t_{2}\right)\right)}

Now, \mathrm{S A=a+a^{t_{1}^{2}}, S B=a+a^{t_{2}^{2}}},

\mathrm{\mathrm{SC} =\sqrt{\left(\mathrm{at}_{1} \mathrm{t}_{2}-\mathrm{a}\right)^{2}+\mathrm{a}^{2}\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}}}
\mathrm{\Rightarrow \quad \mathrm{SC} =\mathrm{a} \sqrt{\left(\mathrm{t}_{1} \mathrm{t}_{2}-1\right)^{2}+\left(\mathrm{t}_{1}+\mathrm{t}_{2}\right)^{2}} }
\mathrm{=\mathrm{a} \sqrt{\mathrm{t}_{1}^{2}+\mathrm{t}_{2}^{2}+1+\mathrm{t}_{1}^{2} \mathrm{t}_{2}^{2}} }
\mathrm{=\mathrm{a} \sqrt{\left(1+\mathrm{t}_{1}^{2}\right)\left(1+\mathrm{t}_{2}^{2}\right)} }

Clearly, \mathrm{SC}^{2}=\mathrm{SA} . \mathrm{SB}

Posted by

SANGALDEEP SINGH

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