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  • Length of the shortest normal chord of the parabola is<div class=

Length of the shortest normal chord of the parabola \mathrm{y^{2}=4 a x} is

Option: 1

a \sqrt{27}


Option: 2

3 a \sqrt{3}


Option: 3

2 \mathrm{a} \sqrt{27}


Option: 4

None of these


Answers (1)

best_answer

Let AB be a normal chord where \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{\text { We have, } t_{2}=t_{1}-\frac{2}{t_{1}} }.

\mathrm{A B^{2}=\left[a^{2}\left(t_{1}^{2}-t_{2}^{2}\right)\right]^{2}+4 a^{2}\left(t_{1}-t_{2}\right)^{2} }
          \mathrm{=a^{2}\left(t_{1}-t_{2}\right)^{2}\left[\left(t_{1}+t_{2}\right)^{2}+4\right]}
          \mathrm{=a^{2}\left(t_{1}+t_{1}+\frac{2}{t_{1}}\right)^{2}\left(\frac{4}{t_{1}^{2}}+4\right)}
          \mathrm{=\frac{16 a^{2}\left(1+t_{1}^{2}\right)^{3}}{t_{1}^{4}}}
\mathrm{\Rightarrow \frac{d\left(A B^{2}\right)}{d t_{1}} }
\mathrm{=16 \mathrm{a}^{2}\left(\frac{\mathrm{t}_{1}^{4}\left[3\left(1+\mathrm{t}_{1}^{2}\right)^{2} 2 \mathrm{t}_{1}\right]-\left(1+\mathrm{t}_{1}^{2}\right)^{3} \cdot 4 \mathrm{t}_{1}^{3}}{\mathrm{t}_{1}^{8}}\right) }
\mathrm{ =\frac{a^{2} \cdot 32\left(1+t_{1}^{2}\right)^{2}}{t_{1}^{5}}\left(t_{1}^{2}-2\right) }
\mathrm{t_{1}=\sqrt{2} }  is indeed the point of minima of \mathrm{A B^{2}}.

Thus, \mathrm{\mathrm{AB}_{\text {mini }}=\frac{4 \mathrm{a}}{2}(1+2)^{3 / 2}=2 \mathrm{a} \sqrt{27}} units.

Posted by

Suraj Bhandari

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