$\begin{array}{1 1}(A)\;16x^2+24xy+9y^2-308x+394y-1799=0 \\(B)\;16x^2+24xy+9y^2-308x-394y-1799=0 \\(C)\;16x^2-24xy+9y^2-308x+394y-1799=0 \\(D)\;none\;of\;these \end{array}$

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Distance between focus and vertex.

$\qquad= \sqrt {(1+2)^2+(1-3)^2}=5$

Let point of intersection of direction and axis of parabola be $(x_1,y_1)$

then $\large\frac{x_1+1}{2}$$=-2$

=> $x_1=-5$

and $\large\frac{y_1-1}{2}$$=3$

=> $y_1=7$

Equation of direction is $y-7 =\large\frac{3}{4}$$(x+5)$

=>$3x-4y+43=0$

equation of the required parabola is $\bigg( \large\frac{3x-4y+43}{5}\bigg)^2$$=(x-1)^2+(y-1)^2$

$16x^2+24xy+9y^2-308x+394y-1799=0 $

Hence A is the correct answer.

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