Let $L$ be a point $(t,2)$ and $M$ be a point on the $y$-axis such that $LM$ has a slope $-t$. Then what is the locus of the mid point of $LM$ as $t$ varies over all real values.
note: I have used the definition of a slope of a straight line to get $\frac{2-y}{t-0}=-t$ to get the $y-$intercept as $c=2+t^2$ . Hence the equation of the line becomes $y=(-t)x+(2+t^2)$. This gives a quadratic equation of variable $t$ so using Sridharacharya’s formula we get two roots of $t$, and rearranging the equation I get $x=\frac{t^2+16}{t}$. But how can I get the locus of the mid point using the given information. Do we have to approach it in any other way?
You have found the coordinates of the intercept $M$ to be $(0, t^2+2)$, and you have the coordinates of $L(t,2)$.
So the midpoint of $LM$ is $$(x,y)=(\frac12 t, \frac 12 t^2+2)$$
Now eliminate $t=2x$ from these, and the locus of the midpoint is $$y=2x^2+2$$