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Question: \( {\text{Find the equation of the parabola if the focus is at}}\left( { - 6, - 6} \right){\text...

Find the equation of the parabola if the focus is at(−6,−6) and the vertex is at (−2,2).  {\text{Find the equation of the parabola if the focus is at}}\left( { - 6, - 6} \right){\text{ and the vertex}} \\\ {\text{is at }}\left( { - 2,2} \right). \\\

Explanation

Solution

Let Z(x1,  y1) be the coordinates of the point of intersection of the axis and the directrix of the parabola. Then the vertex V(−2,  2)  is the mid point of the line segment joining Z(x1,  y1) and the focus S(−6,  −6). ⇒x1−62=−2  ⇒x1=2 &y1−62=2⇒y1=10 Thus the directrix meets the axis at Z(2,10). Let m1 be the slope of axis. Then, m1 = (Slope of the line joining the focus S and vertex V)=−6−2−6+2=−8−4=2 ⇒Slope of the directrix which is perpendicular to axis is  m = - 1m1=−12 ⇒equation of directrix which is passing from (2,10) is y - 10 = - 12(x−2) ⇒2y+x−22=0 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+6)2+(y+6)2=∣2y+x−2222+12∣ ⇒(x+6)2+(y+6)2=(2y+x−22)25 ⇒5x2+5y2+60x+60y+360=4y2+x2+484+4xy−44x−88y ⇒4x2+y2−4xy+104x+148y−124=0 ⇒(2x−y)2+4(26x+37y−31)=0 So, this is your required equation of parabola. NOTE: - In this particular type of questions first find the intersection  point of axis and directrix, then find out equation of directrix  then apply parabola property you will get your answer.  {\text{Let Z}}\left( {{x_1},\;{y_1}} \right){\text{ be the coordinates of the point of intersection of the axis and the directrix of the parabola}}{\text{.}} \\\ {\text{Then the vertex V}}\left( { - 2,\;2} \right)\;{\text{is the mid point of the line segment joining Z}}\left( {{x_1},\;{y_1}} \right){\text{ and the focus S}}\left( { - 6,\; - 6} \right). \\\ \Rightarrow \dfrac{{{x_1} - 6}}{2} = - 2\; \Rightarrow {x_1} = 2 \\\ \& \dfrac{{{y_1} - 6}}{2} = 2 \Rightarrow {y_1} = 10 \\\ {\text{Thus the directrix meets the axis at Z}}\left( {2,10} \right). \\\ {\text{Let }}{{\text{m}}_1}{\text{ be the slope of axis}}{\text{. Then,}} \\\ {{\text{m}}_1}{\text{ = }}\left( {{\text{Slope of the line joining the focus S and vertex V}}} \right) = \dfrac{{ - 6 - 2}}{{ - 6 + 2}} = \dfrac{{ - 8}}{{ - 4}} = 2 \\\ \Rightarrow {\text{Slope of the directrix which is perpendicular to axis is }} \\\ {\text{m = - }}\dfrac{1}{{{{\text{m}}_1}}} = - \dfrac{1}{2} \\\ \Rightarrow {\text{equation of directrix which is passing from }}\left( {2,10} \right){\text{ is}} \\\ {\text{y - 10 = - }}\frac{1}{2}\left( {x - 2} \right) \\\ \Rightarrow 2y + x - 22 = 0 \\\ {\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 + 6} \right)}^2} + {{\left( {y + 6} \right)}^2}} = \left| {\dfrac{{2y + x - 22}}{{\sqrt {{2^2} + {1^2}} }}} \right| \\\ \Rightarrow {\left( {x + 6} \right)^2} + {\left( {y + 6} \right)^2} = \dfrac{{{{\left( {2y + x - 22} \right)}^2}}}{5} \\\ \Rightarrow 5{x^2} + 5{y^2} + 60x + 60y + 360 = 4{y^2} + {x^2} + 484 + 4xy - 44x - 88y \\\ \Rightarrow 4{x^2} + {y^2} - 4xy + 104x + 148y - 124 = 0 \\\ \Rightarrow {\left( {2x - y} \right)^2} + 4\left( {26x + 37y - 31} \right) = 0 \\\ {\text{So, this is your required equation of parabola}}{\text{.}} \\\ {\text{NOTE: - In this particular type of questions first find the intersection}} \\\ {\text{ point of axis and directrix, then find out equation of directrix}} \\\ {\text{ then apply parabola property you will get your answer}}{\text{.}} \\\