Answer
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$
{\text{Let P}}\left( {x,y} \right){\text{ be a point on parabola}}{\text{. Then,}} \\
{\text{Distance of P from the focus = Perpendicular distance of P from the Directrix }}\left( {{\text{Parabola property}}} \right) \\
\Rightarrow \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {y + 1} \right)}^2}} = \left| {\dfrac{{2x - 3y + 6}}{{\sqrt {{2^2} + {3^2}} }}} \right| \\
\Rightarrow {\left( {x + 1} \right)^2} + {\left( {y + 1} \right)^2} = \dfrac{{{{\left( {2x - 3y + 6} \right)}^2}}}{{13}} \\
\Rightarrow 13{x^2} + 13{y^2} + 26x + 26y + 26 = 4{x^2} + 9{y^2} + 36 - 12xy + 24x - 36y \\
\Rightarrow 9{x^2} + 4{y^2} + 12xy + 2x + 62y - 10 = 0 \\
{\text{So, on comparing with given equation}} \\
{\text{a = 9, b = 4, 2h = 12, 2g = 2, 2f = 62, c = - 10}} \\
\Rightarrow \left| {a - b} \right| = \left| {9 - 4} \right| = \left| 5 \right| = 5 \\
{\text{NOTE: - In this particular type of questions apply parabola property and solve you}} \\
{\text{ get your desired answer}}{\text{.}} \\
$
{\text{Let P}}\left( {x,y} \right){\text{ be a point on parabola}}{\text{. Then,}} \\
{\text{Distance of P from the focus = Perpendicular distance of P from the Directrix }}\left( {{\text{Parabola property}}} \right) \\
\Rightarrow \sqrt {{{\left( {x + 1} \right)}^2} + {{\left( {y + 1} \right)}^2}} = \left| {\dfrac{{2x - 3y + 6}}{{\sqrt {{2^2} + {3^2}} }}} \right| \\
\Rightarrow {\left( {x + 1} \right)^2} + {\left( {y + 1} \right)^2} = \dfrac{{{{\left( {2x - 3y + 6} \right)}^2}}}{{13}} \\
\Rightarrow 13{x^2} + 13{y^2} + 26x + 26y + 26 = 4{x^2} + 9{y^2} + 36 - 12xy + 24x - 36y \\
\Rightarrow 9{x^2} + 4{y^2} + 12xy + 2x + 62y - 10 = 0 \\
{\text{So, on comparing with given equation}} \\
{\text{a = 9, b = 4, 2h = 12, 2g = 2, 2f = 62, c = - 10}} \\
\Rightarrow \left| {a - b} \right| = \left| {9 - 4} \right| = \left| 5 \right| = 5 \\
{\text{NOTE: - In this particular type of questions apply parabola property and solve you}} \\
{\text{ get your desired answer}}{\text{.}} \\
$
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