Any point on curve
y = x 2 \displaystyle y=x^{2} y = x 2 is
P ( t , t 2 ) \displaystyle P\left ( t,t^{2} \right ) P ( t , t 2 ) $$\displaystyle
\frac{dy}{dx}=2x
e q u a t i o n o f n o r m a l a t equation of normal at e q u a t i o n o f n or ma l a t \displaystyle \left (
t,t^{2} \right )
i s is i s \displaystyle y-t^{2}=-\frac{1}{2t}\left ( x-t
\right )$$
solving with
y = x 2 \displaystyle y=x^{2} y = x 2 we get
$$\displaystyle
x^{2}-t^{2}=\frac{-1}{2t}\left ( x-t \right )\Rightarrow \left ( x-t
\right )\left ( x+t+\frac{1}{2t} \right )=0\Rightarrow
x=-t-\frac{1}{2t}$$
So normal cuts the curve again at
Q ( − t − 1 2 t , ( − t − 1 2 t ) 2 ) \displaystyle Q\left ( -t-\frac{1}{2t},\left ( -t-\frac{1}{2t} \right )^{2} \right ) Q ( − t − 2 t 1 , ( − t − 2 t 1 ) 2 ) z = P Q 2 = 4 t 2 ( 1 + 1 4 t 2 ) 3 \displaystyle z=PQ^{2}=4t^{2}\left ( 1+\frac{1}{4t^{2}} \right )^{3} z = P Q 2 = 4 t 2 ( 1 + 4 t 2 1 ) 3 $$\displaystyle
\frac{dz}{dt}=0\Rightarrow t=\pm \frac{1}{\sqrt{2}}=0,\frac{dz}{dt}$$
changes sign from negative to positive about $$\displaystyle
t=\frac{1}{\sqrt{2}}
a s w e l l a s as well as a s w e ll a s \displaystyle
t=-\frac{1}{\sqrt{2}}$$(No chord is format for t=0)
z is minimum
at
t = ± 1 2 \displaystyle t=\pm \frac{1}{\sqrt{2}} t = ± 2 1 & minimum value
of
z = P Q 2 = 3 \displaystyle z=PQ^{2}=3 z = P Q 2 = 3 shortest normal chord has
length
3 \displaystyle \sqrt{3} 3 & its equation is $$\displaystyle
x+\sqrt{2y}-\sqrt{2}=0
o r or or \displaystyle x-\sqrt{2y}+\sqrt{2}=0$$