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New member
Feb 6, 2012
This question was asked in my exam and I could not answer it. I would like to know how it can be solved.

If $l$ and $m$ are variable real numbers such that $5l^2+6m^2-4lm+3l=0$, then a variable line $lx+my=1$ always touches a fixed parabola, whose axis is paralell to the x-axis.

(a) Find the vertex of the parabola.
(b) Find the focus of the parabola.


Staff member
Feb 24, 2012
Let the described parabola be given by:

$\displaystyle x=ay^2+by+c$

From the given line, we find:

$\displaystyle x=\frac{1-my}{l}$

Hence, we have:

$\displaystyle \frac{1-my}{l}=ay^2+by+c$

Arranging the quadratic in $\displaystyle y$ in standard form, we find:

$\displaystyle aly^2+(bl+m)y+(cl-1)=0$

We are told the line is tangent to the parabola, which means there will only be one root, and so we must have that the discriminant is zero. Equating the discriminant to zero, expanding and multiplying by a crucial number, you will find that using the given implicit relation between $\displaystyle l$ and $\displaystyle m$ you can obtain sufficient equations by equating coefficients to determine the parameters of the parabola $\displaystyle a,\,b,\,c$.

And from this, you may determine the vertex and focus of the parabola.