Find the equation to the normal of the curve $x=2t, y=t^2$ at the point with parameter $t$. If this normal meets the x- and y-axes at points A and B respectively, find the equation of the locus of the mid-point of AB.
I have said:
$x=2t, y= t^2$
$\frac{dy}{dx} = t$
$\rightarrow$ gradient of the normal is $\frac{-1}{t}$
$\rightarrow \frac{y-t^2}{x-2t}=\frac{-1}{t} \rightarrow x+ty=t^3 +2t$
At point A, $y=0, x=t^3 + 2t$
At point B, $x=0, y=t^2+2$
The coordinates of the mid point are $[\frac{t^3+2t}{2},\frac{t^2+2}{2}]$
But where to go from here?
You got till the point that locus of midpoint is $ \displaystyle \left [2x = t^3+2t, 2y = t^2+2\right]$
$2y = t^2+2 \implies t = \pm \sqrt{2y-2}$
$2x = t^3+2t = t \ (t^2 + 2) = (\pm \sqrt{2y-2}) \ 2y$
$x^2 = (2y-2) y^2 \implies 2y^3 - 2y^2 = x^2$
Here is a diagram of the parabola curve and locus of the midpoint of intersection of its normal lines with with x and y axes.