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Question: \( {\text{The equation of parabola with focus}}\left( { - 1, - 1} \right){\text{ and directrix }...

The equation of parabola with focus(1,1) and directrix 2x3y+6=0 is ax2+2hxy+by2+2gx+2fy+c=0.Then, ab is equal to -   {\text{The equation of parabola with focus}}\left( { - 1, - 1} \right){\text{ and directrix }}2x - 3y + 6 = 0 \\\ {\text{is }}a{x^2} + 2hxy + b{y^2} + 2gx + 2fy + c = 0.{\text{Then, }}\left| {a - b} \right|{\text{ is equal to - }} \\\

Explanation

Solution

Let P(x,y) be a point on parabola. Then, Distance of P from the focus = Perpendicular distance of P from the Directrix (Parabola property) (x+1)2+(y+1)2=2x3y+622+32 (x+1)2+(y+1)2=(2x3y+6)213 13x2+13y2+26x+26y+26=4x2+9y2+3612xy+24x36y 9x2+4y2+12xy+2x+62y10=0 So, on comparing with given equation a = 9, b = 4, 2h = 12, 2g = 2, 2f = 62, c = - 10 ab=94=5=5 NOTE: - In this particular type of questions apply parabola property and solve you  get your desired answer.  {\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{.}} \\\