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# Equation of the parabola if its vertex and focus are $(-2,3)$ and $(1,-1)$ respectively is

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

$\qquad= \sqrt {(1+2)^2+(1-3)^2}=5$
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$
$16x^2+24xy+9y^2-308x+394y-1799=0$