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Home  >>  JEEMAIN and AIPMT  >>  Mathematics  >>  Class11  >>  Coordinate Geometry
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Find the equation of parabola whose focus (-4,0) & vertex (0,0)?

$\begin{array}{1 1}(a)\;y_1^2=-15x_1\\(b)\;y_1^2=-16x_1\\(c)\;y_1=-15x_1^2\\(d)\;y_1^3=-15x_1^2\end{array}$

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1 Answer

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Z is a point passes through directix and also axis of parabola.
$Z=(x,y)$
$0=\large\frac{-4+x}{2}$
$x=4$
$0=\large\frac{0+y}{2}$
$y=0$
Equation of directix $(y-0)=\large\frac{-4}{0}$$(x-4)$
$x=4$
Let $P(x_1,y_1)$ be a point on parabola
$\large\frac{SP}{PM}$$=1$
$SP=PM$
$\sqrt{x_1+4)^2+(y_1-0)^2}=\large\frac{x_1-4}{\sqrt{1^2}}$
$SP^2=PM^2$
$(x_1+4)^2+y_1^2=(x_1-4)^2$
$y_1^2=-16x_1$
Hence (b) is the correct answer.
answered Feb 5, 2014 by sreemathi.v
 
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