By appropriately matching the information given in the three columns of the following table.

Columns 1, 2 and 3 contain conics, equations of tangents to the conics and points of contact, respectively.

	Column - 1	Column - 2	Column - 3
(i)	$${x^2} + {y^2} = a$$	$$my = {m^2}x + a$$	$$\left( {{a \over {{m^2}}},\,{{2a} \over m}} \right)$$
(ii)	$${x^2}{a^2}{y^2} = {a^2}]$$	$$y = mx + a\sqrt {{m^2} + 1} $$	$$\left( {{{ - ma} \over {\sqrt {{m^2} + 1} }},\,{a \over {\sqrt {{m^2} + 1} }}} \right)$$
(iii)	$${y^2} = 4ax$$	$$y = mx + \sqrt {{a^2}{m^2} - 1} $$	$$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} + 1} }},\,{1 \over {\sqrt {{a^2}{m^2} + 1} }}} \right)$$
(iv)	$${x^2} - {a^2}{y^2} = {a^2}$$	$$y = mx + \sqrt {{a^2}{m^2} + 1} $$	$$\left( {{{ - {a^2}m} \over {\sqrt {{a^2}{m^2} - 1} }},\,{{ - 1} \over {\sqrt {{a^2}{m^2} - 1} }}} \right)$$

If a tangent to a suitable conic (Column 1) is found to be y = x + 8 and its point of contact is (8, 16), then which of the following options is the only CORRECT combination?

Let $$a, r, s, t$$ be nonzero real numbers. Let $$P\,\,\left( {a{t^2},2at} \right),\,\,Q,\,\,\,R\,\,\left( {a{r^2},2ar} \right)$$ and $$S\,\,\left( {a{s^2},2as} \right)$$ be distinct points on the parabola $${y^2} = 4ax$$. Suppose that $$PQ$$ is the focal chord and lines $$QR$$ and $$PK$$ are parallel, where $$K$$ is the point $$(2a,0)$$

If $$st=1$$, then the tangent at $$P$$ and the normal at $$S$$ to the parabola meet at a point whose ordinate is

Let $$a, r, s, t$$ be nonzero real numbers. Let $$P\,\,\left( {a{t^2},2at} \right),\,\,Q,\,\,\,R\,\,\left( {a{r^2},2ar} \right)$$ and $$S\,\,\left( {a{s^2},2as} \right)$$ be distinct points on the parabola $${y^2} = 4ax$$. Suppose that $$PQ$$ is the focal chord and lines $$QR$$ and $$PK$$ are parallel, where $$K$$ is the point $$(2a,0)$$

The value of $$r$$ is

Let $$PQ$$ be a focal chord of the parabola $${y^2} = 4ax$$. The tangents to the parabola at $$P$$ and $$Q$$ meet at a point lying on the line $$y=2x+a$$, $$a>0$$.

If chord $$PQ$$ subtends an angle $$\theta $$ at the vertex of $${y^2} = 4ax$$, then tan $$\theta = $$