Question

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

• A. a$\sqrt{27}$
• B. 3a$\sqrt{3}$
• C. 2a$\sqrt{27}$
• D. None of these

Answer 1

Answer: (C)

2a$\sqrt{27}$

Answer 2

Let AB be a normal chord, where

A ŗ (at₁², 2at₂), B ŗ (at₂², 2at₂) . We have

t₂ = –t₁ –$\frac{2}{t_{1}}$.

Now, AB² = [a²(t₁² – t₂²)]² + 4a² (t₁ – t₂)²

= a²(t₁ – t₂)² {(t₁ + t₂)² + 4}

= a² $\left( t_{1} + t_{1} + \frac{2}{t_{1}} \right)^{2}$ $\left( \frac{4}{t_{1}^{2}} + 4 \right)$

= $\frac{16a^{2}(1 + t_{1}^{2})^{3}}{t_{1}^{4}}$

Ž $\frac{d(AB^{2})}{dt_{1}}$ = 16a² $\left( \frac{t_{1}^{4}\lbrack 3(1 + t_{1}^{2})^{2}.2t_{1}\rbrack - (1 + t_{1}^{2})^{3}.4t_{1}^{3}}{t_{1}^{8}} \right)$

= $\frac{a^{2}.32(1 + t_{1}^{2})^{2}}{t_{1}^{5}}$ (t₁² – 2)

t₁ = $\sqrt{2}$ is indeed the point of minima of AB². Thus,

AB_min = $\frac{4a}{2}$ (1 + 2)^3/2 = 2a$\sqrt{27}$units.