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Find the equation of parabola if : Focus:$(2 , -3 );$ directrix :$2y-3=0$

This is the first part of the multi-part question Q1

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Toolbox:
  • General equation of a conic with focus at $F(x_1,y_1)$,directrix $lx+my+n=0$ and eccentricity 'e' is $(x-x_1)^2+(y-y_1)^2=e^2\bigg[\pm\large\frac{lx+my+n}{\sqrt{l^2+m^2}}\bigg]^2$
  • Which reduces to the form $Ax^2+Bxy+Cy^2+Dx+Ey+F=0$
  • If the eccentricity 'e' (i) is less than 1,it is an parabola.(ii) is equal to 1 then it is a parabola (iii) $e > 1\Rightarrow$it is a hyperbola.
Step 1:
Focus :(2,-3) ; directrix : $2y-3=0$
$P(x,y)$ is a point on the parabola,$F(2,-3)$ is the focus PM is $\perp$ to the directrix $2y-3=0$ $e=1$ for a parabola.
Therefore $\large\frac{FP}{PM}$$=1$
${FP}^2={PM}^2$
Step 2:
$(x-2)^2+(y+3)^2=\bigg(\pm\large\frac{2y-3}{\sqrt 4}\bigg)^2$
$x^2-4x+4+y^2+6y+9=\large\frac{4y^2-12y+9}{4}$
$4x^2-16x+4y^2+24y+52=4y^2-12y+9$
$4x^2-16x+36y+43=0$
This is the required equation.
answered Jun 13, 2013 by sreemathi.v
edited Jun 13, 2013 by sreemathi.v
 

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