If the tangents to the ellipse at $$M$$ and $$N$$ meet at $$R$$ and the normal to the parabola at $$M$$ meets the $$x$$-axis at $$Q$$, then the ratio of area of the triangle $$MQR$$ to area of the quadrilateral $$M{F_1}N{F_2}$$is
Equation of tangent at M(3/2, $$\sqrt6$$) to $${{{x^2}} \over 9} + {{{y^2}} \over 8} = 1$$ is
$${3 \over 2}.{x \over 9} + \sqrt 6 .{y \over 8} = 1$$ ....... (i)
which intersect X-axis at (6, 0).
Also, equation of tangent at N(3/2, $$-$$$$\sqrt6$$) is
$${3 \over 2}.{x \over 9} - \sqrt 6 .{y \over 8} = 1$$ ....... (ii)
Eqs. (i) and (ii) intersect on X-axis at R(6, 0). ........ (iii)
Also, normal at $$M(3/2,\sqrt 6 )$$ is
$$y - \sqrt 6 = {{ - \sqrt 6 } \over 2}\left( {x - {3 \over 2}} \right)$$
On solving with y = 0, we get Q(7/2, 0) ....... (iv)
The area of MQR is
$$\left| {{1 \over 2}} \right|\left| {\matrix{ {3/2} & {\sqrt 6 } & 1 \cr 6 & 0 & 1 \cr {7/2} & 0 & 1 \cr } } \right| = \left| {{{\sqrt 6 } \over 2}\left( {6 - {7 \over 2}} \right)} \right| = {{5\sqrt 6 } \over 4}$$
The area of the quadrilateral MF1NF2 is
$$2(\Delta {m_1}{F_1}{F_2}) = 2\sqrt 6 $$
and the required ratio is
$${{5\sqrt 6 } \over {4\,.\,2\sqrt 6 }} = {5 \over 8}$$
If $$st=1$$, then the tangent at $$P$$ and the normal at $$S$$ to the parabola meet at a point whose ordinate is
The value of $$r$$ is
