Question

Length of the shortest normal chord of the parabola $y^2=4ax$ is

• A. $2a\sqrt{27}$
• B. $9a$
• C. $a\sqrt{54}$
• D. none of these

Answer 1

The correct option is A $2a\sqrt{27}$

Let $AB$ be a normal chord where $A\equiv (a{t}^2,2at),B\equiv (a{t}_1^2,2a{t}_1)$.
We have, $t_1=-t-\frac{2}{t}$
Now, $A{B}^2=\left[a^2\right(t^2-t_1^2){]}^2+4a^2(t-t_1{)}^2$
$=a^2(t-t_1{)}^2[(t+t_1{)}^2+4]$
$=a^2{(t+t+\frac{2}{t})}^2(\frac{4}{t^2}+4)$
$=\frac{16a^2(1+t^2)}{t^4}$
$\Rightarrow \frac{d(AB{)}^2}{dt}=16a^2\left(\frac{t^4\left[3\right(1+r^2{)}^2\cdot 2t]-(1+t^2{)}^3\cdot 4t^3}{t^8}\right)$
$=32a^2(1+t^2)\left(\frac{3t^2-2-2t^2}{t^5}\right)$
$=\frac{d^2\times 32(1+t^2{)}^2}{t^5}(t^2-2)$
For $\frac{d(AB{)}^2}{dt}=0\Rightarrow t=\sqrt{2}$ for which $(AB{)}^2$ is minimum.
Thus, $A{B}_{min}=\frac{4a}{2}(1+2{)}^{3/2}=2a\sqrt{27}$ units