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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?

Michael
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    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 01 '18 at 10:31
  • Pls help me in recognising the equation – Michael Jan 01 '18 at 10:39
  • Didnt get any info about the question asked on wikipedia.can u tell how to do.pls – Michael Jan 05 '18 at 15:41
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    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 05 '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 '19 at 16:43
  • Truth be told, I plotted the parabola, and the integer coordinates $(1,1)$ as focus and $y=x+1$ as directrix were the likely candidates given that the equation has integer coefficients. Then, putting that into the general form, I noticed the factor $(\pm)2$ would clear the fractions and expand it into the original equation. – Jan-Magnus Økland Apr 26 '19 at 08:15

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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 $(x,y)=(1,1)$.

Jan-Magnus Økland
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Let There be a parabola $(ax+by)^2+2gx+2fy+c=0 $

This can be written as $(ax+by)^2=-2gx-2fy-c$ Now we will add arbitrary constant in square term, $$(ax+by+\lambda)^2=xf_1(\lambda)x+f_2(\lambda)y+f_3(\lambda)-(1)$$

Now simply choose $\lambda$ such that the lines, $ax+by+\lambda$ and $xf_1(\lambda)x+f_2(\lambda)y+f_3(\lambda)$ are perpendicular.

now $xf_1(\lambda)x+f_2(\lambda)y+f_3(\lambda)=bx-ay+\mu$

we can write (1) as, $$(\frac{ax+by+\lambda}{\sqrt{a^2+b^2}})^2=4\rho(\frac{bx-ay+\mu}{\sqrt{a^2+b^2}})$$ which is of the form $Y^2=4\rho X$

Now we can do our stuff on this reduced form,

1)LATUSRECTUM:- $4\rho$=$\frac{1}{\sqrt{a^2+b^2}}$

2)AXIS:- Y=0 or ax+by+$\lambda$=0

3)EQUATION OF TANGENT AT VERTEX:- X=0 or bx-ay+$\mu$=0

4)VERTEX:- Intersection of Y=0 and X=0

5)EQUATION OF DIRECTRIX:- X+$\rho$=0

6)EQUATION OF LATUS RECTUM:- X-$\rho$=0

7)FOCUS:- By solving X-$\rho$=0 and X=0

well now we can apply all the basic parabola properties in short {^w^} .

Vishal
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