# Inclined parabola

Please help me to solve the inclined parabola. Is it that we have to convert this entire equation in the form of axis and tangent at vertex. That would be tiresome. Any other method? For this parabola,$x^2 +y^2 +2xy -6x -2y + 3=0$,How can we find it's focus?

• See wikipedia. In this particular case it is easy to see that the equation is $-2((x-y+1)^2/2-(x-1)^2-(y-1)^2)$ so the focus is $(1,1)$. – Jan-Magnus Økland Jan 1 '18 at 10:31
• Pls help me in recognising the equation – Michael Jan 1 '18 at 10:39
• Didnt get any info about the question asked on wikipedia.can u tell how to do.pls – Michael Jan 5 '18 at 15:41
• The wikipedia-link was to show you the general form of an inclined parabola: $\frac{(ax+by+c)^2}{a^2+b^2}=(x-f_1)^2+(y-f_2)^2$, where $ax+by+c=0$ is the directrix and $(f_1,f_2)$ the focus. – Jan-Magnus Økland Jan 5 '18 at 17:28
• @jan-magnus økland .pls help me in writing the equation in general form as u told it was quite easy in that equation to recognise. But i still cant find it easy to write in general form in inclined parabola problems.pls mention the steps.plss – Michael Apr 25 at 16:43

## 1 Answer

Since you don't like the easy way (see my comment), let's do it the hard way: rotate $x^2+2xy+y^2-6x-2y+3=0$ around the origin by $\frac{\pi}{4}$:

$$x=\frac{x'-y'}{\sqrt{2}}$$ $$y=\frac{x'+y'}{\sqrt{2}}$$ substituting this and expanding the equation reduces to: $$y'+\frac{x'^2}{\sqrt{2}}-2x'+\frac{3}{2\sqrt{2}}=0$$ or $$y'-\frac1{2\sqrt{2}}=-\frac{(x'-\sqrt{2})^2}{\sqrt{2}}$$ or $$y''=-\frac{x''^2}{\sqrt{2}}=\frac1{4f}x''^2$$ which makes $f=-\frac1{2\sqrt{2}}$ and the focus for this parabola is $(x'',y'')=(0,f)$. Now we go back through the transformations and get $(x',y')=(\sqrt{2},0)$ an the inverse rotation takes this to $(x,y)=(1,1)$.