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Maths-

General

Easy

Question

Length of the shortest normal chord of the parabola y² = 4x is-

a $\sqrt{27}$
3a $\sqrt{3}$
2a $\sqrt{27}$
None of these

The correct answer is: 2a $square root of 27$

Let AB be a normal chord, where
A ≡ (at₁², 2at₂), B ≡ (at₂², 2at₂) . We have t₂ = –t₁ – $fraction numerator 2 over denominator t subscript 1 end subscript end fraction$ .
Now, AB² = [a²(t₁² – t₂²)]² + 4a² (t₁ – t₂)²
= a²(t₁ – t₂)² {(t₁ + t₂)² + 4}
= a² $open parentheses t subscript 1 end subscript plus t subscript 1 end subscript plus fraction numerator 2 over denominator t subscript 1 end subscript end fraction close parentheses to the power of 2 end exponent$ $open parentheses fraction numerator 4 over denominator t subscript 1 end subscript superscript 2 end superscript end fraction plus 4 close parentheses$
= $fraction numerator 16 a to the power of 2 end exponent left parenthesis 1 plus t subscript 1 end subscript superscript 2 end superscript right parenthesis to the power of 3 end exponent over denominator t subscript 1 end subscript superscript 4 end superscript end fraction$
$rightwards double arrow fraction numerator d left parenthesis A B squared right parenthesis over denominator d t subscript 1 end fraction$ = 16a² $open parentheses fraction numerator t subscript 1 end subscript superscript 4 end superscript left square bracket 3 left parenthesis 1 plus t subscript 1 end subscript superscript 2 end superscript right parenthesis to the power of 2 end exponent.2 t subscript 1 end subscript right square bracket minus left parenthesis 1 plus t subscript 1 end subscript superscript 2 end superscript right parenthesis to the power of 3 end exponent.4 t subscript 1 end subscript superscript 3 end superscript over denominator t subscript 1 end subscript superscript 8 end superscript end fraction close parentheses$
= $fraction numerator a to the power of 2 end exponent.32 left parenthesis 1 plus t subscript 1 end subscript superscript 2 end superscript right parenthesis to the power of 2 end exponent over denominator t subscript 1 end subscript superscript 5 end superscript end fraction$ (t₁² – 2)
t₁ = $square root of 2$ is indeed the point of minima of AB². Thus,
AB_min = $fraction numerator 4 a over denominator 2 end fraction$ (1 + 2)^3/2 = 2a $square root of 27$ units.

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