Problem :

Find the points on the parabola $y^2-2y-4x=0$ whose focal length is 6 .

Solution : The given equation $y^2-2y-4x=0$ can be written as :

$ (y-1)^2=4x+1$ $\Rightarrow (y-1)^2=4(x+\frac{1}{4})$

Therefore we can say that vertex of this parabola is $(-\frac{1}{4}, 1)$ Now how to proceed further please suggest... thanks.

  • 1
    $\begingroup$ What's the definition of focal length. $\endgroup$ – Mhenni Benghorbal Dec 11 '13 at 6:25


I believe its focal distance, not focal length, as the focal length of $\displaystyle(y-k)^2=4a(x-h)$ is $a$

and the focus of $\displaystyle(y-k)^2=4a(x-h)$ will be $(h+a,k)$ as the vertex is $(h,k)$

The parametric equation of the parabola $(-\frac14+t^2,2t+1)$


Find the focus of the parabola $(p_1,p_2)$ and pick up a point $(x,y)$ which lies on the parabola and then use the distance formula between two points as

$$ d=\sqrt{(x-p_1)^2+(y-p_2)^2}=6. $$

Then use the equation of the parabola $y^2-2y-4x=0$ and solve for $x$ or $y$ (you should choose the one which makes the above equation easy to solve) and substitute back in the above equation and finish the problem.

Note: This is what I understood by "they have the same focal distance". Make sure of it.


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