Let the ellipse be $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

It press through $$(-3, 1)$$ so $${9 \over {{a^2}}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( i \right)$$

Also, $${b^2} = {a^2}\left( {1 - 2/5} \right)$$

$$ \Rightarrow 5{b^2} = 3{a^2}\,\,\,\,\,\,\,\,\,...\left( {ii} \right)$$

Solving $$(i)$$ and $$(ii)$$ we get $${a^2} = {{32} \over 3},{b^2} = {{32} \over 5}$$

So, the equation of the ellipse is $$3{x^2} + 5{y^2} = 32$$

The locus of perpendicular tangents is directrix

i.e., $$x=-1$$

The given ellipse is $${{{x^2}} \over 4} + {{{y^2}} \over 1} = 1$$

So $$A=(2,0)$$ and $$B = \left( {0,1} \right)$$

If $$PQRS$$ is the rectangular in which it is inscribed, then

$$P = \left( {2,1} \right).$$

Let $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$

be the ellipse circumscribing the rectangular $$PQRS$$.

Then it passes through $$P\,\,(2,1)$$

$$\therefore$$ $${4 \over {a{}^2}} + {1 \over {{b^2}}} = 1\,\,\,\,\,\,...\left( a \right)$$

Also, given that, it passes through $$(4,0)$$

$$\therefore$$ $${{16} \over {{a^2}}} + 0 = 1 \Rightarrow {a^2} = 16$$

$$ \Rightarrow {b^2} = 4/3$$ $$\left[ {\,\,} \right.$$ substituting $${{a^2} = 16\,\,}$$ in $$\left. {e{q^n}\left( a \right)\,\,} \right]$$

$$\therefore$$ The required ellipse is $${{{x^2}} \over {16}} + {{{y^2}} \over {4/3}} = 1$$

or $${x^2} + 12y{}^2 = 16$$

Vertex of a parabola is the mid point of focus and the point

where directrix meets the axis of the parabola.

Here focus is $$O\left( {0,0} \right)$$ and directrix meets the axis at $$B\left( {2,0} \right)$$

$$\therefore$$ Vertex of the parabola is $$(1,0)$$