Explanation
$$ \because \text{ Length of latus rectum = 4a}$$
$$ \text{ Hence Length of latus rectum of given parabola is} = \dfrac{1}{a}$$
$$\textbf{Hence, Option 'C' is correct} $$
Equation of circle having centre of origin $$(0,0)$$ and radius $$=r$$,
$$S_{1};x^{2}+y^{2}=r^{2}----(1)$$
$$\therefore f(m_{1},\dfrac{1}{m_{2}}),f(m_{2},\dfrac{1}{m_{2}}),--f(m_{4},\dfrac{1}{m_{4}})$$These points lie on $$S_{1}$$.
Let $$\displaystyle f(m, \frac{1}{m})$$ is point lie on S_{1},
$$m^{2}+\dfrac{1}{m^{2}}=r^{2}$$
$$m^{4}+1-r^{2}m^{2}=0$$
$$m^{4}-1-r^{2}m^{2}+1=0$$
$$m_{1},m_{2},m_{3}$$ and $$m_{4}$$ are roots of this equation
So, $$m_{1}m_{2}m_{3}m_{4} =1$$
$$136 (x^{2}+y^{2})=(5x+3y+7)^{2}$$
$$ \Rightarrow \displaystyle (x-0)^2+(y-0)^2= \dfrac{1}{136}(5x+3y+7)^{2}$$
$$ \Rightarrow \displaystyle (x-0)^2+(y-0)^2= \left( \frac{1}{2}\right)^2 \left(\frac{5x+3y+7}{\sqrt{34}} \right)^2$$
From the above form, we get focus of the conic has coordinates $$(0,0)$$, and directrix has equation $$5x+3y+7=0$$ and eccentricity $$e=\dfrac{1}{2}$$
Distance between focus and directrix $$=\dfrac{7}{\sqrt{34}}$$
Length of latus rectum$$=2\dfrac{b^2}{a}=2 \times e \times$$ distance bet. focus and directrix$$=2 \times \dfrac{1}{2} \times \dfrac{7}{\sqrt{34}}$$
$$= \dfrac{7}{\sqrt{34}}$$
Hence, option B.
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