Show that for all the values oft, the point P(2at,at^2) lies on thecurve x^2 = 4ay.Obtain the equation of the tangent to the curve atP. The tangent at P meet the x-axis and the y-axis at Q and Rrespectively. Find the coordinates of Q and R andobtain the (x,y) equation of the locusof T, the mid-point of QR.Show that the ratio of the distance of S(0,a) fromthe tangent at P to the distance of T from the origin is 2/t.

Show that for all the values of t, the point $P(2at,a{t}^2)$ lies on the curve $x^2=4ay$.
For point P, x = 2at and $y=a{t}^2$.If P lies on the curve $x^2=4ay$, then coordinates of P must satisfy that condition.
Since $x=2at,x^2=4a^2{t}^2=4a(a{t}^2)=4ay$. Thus P lies on this curve.
Obtain the equation of the tangent to the curve at P.
$y=(x^2)/4a$. Hence (dy/dx) = (2 x)/ 4a is theslope of the tangent at any point. At point P, x = 2at. Hence slopeof the tangent at P is (2) (2at)/4a = t. Hence the equation of thetangent would be y = tx + b. Since the tangent passes through P it must also satisfy the equation of the tangent. Hence $y=a{t}^2=t(2at)+b=2a{t}^2+b$. Hence, $b=a{t}^2-<!--\; -\; -->2a{t}^2=-<!--\; -\; -->a{t}^2$. Thus equationof the tangent will be $y=tx-<!--\; -\; -->a{t}^2$.
The tangent at P meet the x-axis and the y-axis at Q and Rrespectively. Find the coordinates of Q and R andobtain the (x,y) equation of the locusof T, the mid-point of QR.
The tangent $y=tx-<!--\; -\; -->a{t}^2$ meets the x axis at Q.At this point, y coordinate is zero. $\therefore <!--\; \therefore \; -->tx-<!--\; -\; -->a{t}^2=0\therefore <!--\; \therefore \; -->x=at$. Hence Q is(at,0)
Similarly,the tangent meets the y axis at R. At this point, x coordinate is zero. $\therefore <!--\; \therefore \; -->y=0-<!--\; -\; -->a{t}^2=-<!--\; -\; -->a{t}^2$. Hence R is $(0,-<!--\; -\; -->a{t}^2)$. The mid-point T is given by x = (at+0)/2 and $y=(0-<!--\; -\; -->a{t}^2)/2$. Thus for T x = at/2 and $y=-<!--\; -\; -->a{t}^2/2=-<!--\; -\; -->t(at/2)=-<!--\; -\; -->tx$. Hence, the locus of T is y = -tx.
Show that the ratio of the distance of S(0,a) from the tangent at P tothe distance of T from the origin is 2/t.
Let V(x,y) be a point on thetangent $y=tx-<!--\; -\; -->a{t}^2$ If SV is theshortest distance of S from the tangent then SV is perpendicular to the tangent. Hence the slope of SV should be negative reciprocal ofthe slope of the tangent. That is it must be -1/t. However, theslope of SV is (y-a)/(x-0) = (y-a)/x which must equal-1/t . $\therefore <!--\; \therefore \; -->(y-<!--\; -\; -->a)/x=-<!--\; -\; -->1/t\therefore <!--\; \therefore \; -->(y-<!--\; -\; -->a)t=-<!--\; -\; -->x\therefore <!--\; \therefore \; -->x=at-<!--\; -\; -->yt$. But since V is on the tangent $y=tx-<!--\; -\; -->a{t}^2$.
Substituting x = at -yt in this, we get $y=t(at-<!--\; -\; -->yt)-<!--\; -\; -->a{t}^2=a{t}^2-<!--\; -\; -->y{t}^2-<!--\; -\; -->a{t}^2=-<!--\; -\; -->y{t}^2.\therefore <!--\; \therefore \; -->y+y{t}^2=0$or $y(1+t^2)=0$ since $t^2$ cannot be negative,y must be zero. And $x=at-<!--\; -\; -->yt=at\therefore <!--\; \therefore \; -->V$ is (at,0). The distance SV is then ${\sqrt{at-<!--\; -\; -->0}}^2+(0-<!--\; -\; -->a{)}^2=\sqrt{a^2{t}^2}+a^2=a\sqrt{t^2+1}$ .
The distance OT is ${\sqrt{(at/2)-<!--\; -\; -->0}}^2+((-<!--\; -\; -->a{t}^2/2)-<!--\; -\; -->0{)}^2={\sqrt{at/2}}^2+(a{t}^2/2{)}^2=(at/2)\sqrt{1+t^2}$
And $SV/OT=[a\sqrt{t^2+1}]/[(at/2)\sqrt{1+t^2}]=(a)/(at/2)=2/t.$