logo

Ask Questions, Get Answers

 
X
 Search
Want to ask us a question? Click here
Browse Questions
Ad
0 votes

Find the equation of parabola if : Focus: $(-1 , 3 )$; directrix $ 2x+3y=3$.

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

Can you answer this question?
 
 

1 Answer

0 votes
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 :(-1,3) ; directrix : $2x+3y-3=0$
$P(x,y)$ is a point on the parabola,$F(-1,3)$ is the focus PM is $\perp$ to the directrix $2x+3y-3=0$, $e=1$ for a parabola.
Therefore $\large\frac{FP}{PM}$$=1$
${FP}^2={PM}^2$
Step 2:
$(x+1)^2+(y-3)^2=\bigg(\large\frac{\pm(2x+3y-31)}{4+9}\bigg)^2$
$x^2+2x+1+y^2-6y+9=\large\frac{4x^2+9y^2+9+12xy-12x-18y}{13}$
$13x^2+26x+13y-78+130=4x^2+9y^2+9+12xy-12x-18y$
$9x^2-12xy+4y^2+38x-10y+121=0$
This is the required equation.
answered Jun 13, 2013 by sreemathi.v
 

Related questions

Ask Question
student study plans
x
JEE MAIN, CBSE, NEET Mobile and Tablet App
The ultimate mobile app to help you crack your examinations
Get the Android App
...