How to find the vertex of a 'non-standard' parabola? $ 9x^2-24xy+16y^2-20x-15y-60=0 $

Question

I have to find out the vertex of a parabola given by:

$$ 9x^2-24xy+16y^2-20x-15y-60=0 $$

I don't know what to do. I tried to bring it in the form:

$$ (x-a)^2 + (y-b)^2 = \dfrac {(lx+my+n)^2} {l^2+m^2} $$

but failed in doing so. Is there any other way to solve the problem? Or maybe you could help me bring the equation in the above form.

Answer 1

Think of the standard equations of a parabola you know $y=x^2$ or $y^2=4ax$ - something squared = something linear, and the squared quantity and the linear quantity represent axes at right-angles to one another.

The vertex occurs where the squared quantity is equal to zero, ie on the axis of symmetry.

Now notice that the equation can be rewritten as $$(3x-4y)^2=5(4x+3y+12)$$

Check that the axes implied by this form are perpendicular.

Answer 2

Hint: At the vertex of a parabola, the change of slope of tangent attain maximum or minimum, i.e. $y'''=0$.