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The focus of the parabola

$x^2+y^2+2xy-6x-2y+3=0$ is

what i try: Let $S(h,k)$ be focus and $P(x,y)$ be variable point on parabola and $y=mx+c$ be directrix of parabola and Let $M$ be any point on directrix

Then using definition of parabola $PS=PM$

$$\sqrt{(x-h)^2+(y-k)^2}=\bigg|\frac{mx-y+c}{\sqrt{1+m^2}}\bigg|$$

$(1+m^2)\cdot \bigg[(x-h)^2+(y-k)^2\bigg]=(mx-y+c)^2$

after simplifying

$x^2+m^2y^2+2mxy-2xh(1+m^2)+mc-2y(k(1+m^2)-c+9h^2+k^2)(1+m^2-c^2)=0$

camparing with original equation

$m^2=1$ and $2m=2$ and $2h(1+m^2+2c=6)\Rightarrow 2h+c=3\cdots (1)$

and $2k(1+m^2)-2c=2\Rightarrow 2(h^2+k^2)-c^2=3\Rightarrow 2k-c=1\cdots (2)$

and $2(h^2+k^2)-c^2=3\cdots (3)$ from $(1)$ and $(2)$ and $(3)$

we get $c=1$ and $h=1$ and $k=1$

can i solve it without heavy calculation

please explain me

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jacky
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    I get essentially a similar calculation using $$(l^2+m^2)((x-h)^2+(y-k)^2)=(lx+my+n)^2$$

    and that the axis is parallel to $x+y$ since $x^2+2xy+y^2=(x+y)^2$ and the directrix is parallel to $-x+y$. This means we can take $l^2+m^2=2$: $$2((x-h)^2+(y-k)^2)=(-x+y+n)^2$$ or $$x^2+2xy+y^2+(-4h+2n)x+(-4k-2n)y+2h^2+2k^2-n^2,$$ now $-4h+2n=-6, -4k-2n=-2, 2h^2+2k^2-n^2=3$ so $h=1,k=1,n=-1,$ giving $$2((x-1)^2+(y-1)^2)=(-x+y-1)^2$$

     – Jan-Magnus Økland Mar 02 '20 at 09:56

2 Answers2

2

Here’s a different approach that might be easier to compute.

Let $f(x,y)=x^2+y^2+2xy-6x-2y+3$. Writing the quadratic part as $(x+y)^2$, we see that the axis direction is $(1,-1)$. The parabola’s axis is therefore the polar of the point at infinity in the direction of $(1,1)$, namely $x+y-2=0$. You can also obtain this equation by expanding $$\begin{vmatrix}f_x & 1 \\ f_y & -1\end{vmatrix} = 0.$$ We can then rewrite the equation of the parabola as $$\frac12(x+y-2)^2 = \left(x-y+\frac12\right).$$ Expanding the coefficient of the parenthesized expression on the right as $$1 = \frac{4p}{\sqrt2}$$ (the factor of $\frac1{\sqrt2}$ comes from normalizing the parenthesized term) we obtain the focal length $p=\frac1{2\sqrt2}$. The parabola’s vertex is the intersection of the lines $x+y-2=0$ and $x-y+\frac12=0$, or $\left(\frac34,\frac54\right)$, and the focus is at a distance of $p$ along the axis from this point. The direction of the focus from the vertex can be determined by examining the outward normal $\nabla f(3/4,5/4)$. Alternatively, it’s the solution to the system $$\frac1{\sqrt2}\left(x-y+\frac12\right)=\frac1{2\sqrt2} \\ \frac1{\sqrt2}(x+y-2) = 0,$$ i.e., map the point $\left(\frac1{2\sqrt2},0\right)$ into the parabola’s coordinate system.

amd
  • 53,693
1

Hint:

The quadratic terms of the equation of a parabola form a perfect square. This one is

$$(x+y)^2-6x-2y+3=0.$$

We can apply the transformation $u:=x+y,v:=x-y$, which is a similarity and will preserve the parabola's shape:

$$u^2-4u-2v+3=0$$

which also writes

$$(u-2)^2=2\left(v-\frac12\right).$$

Finally, you can locate the focus of the reduced parabola $q^2=2fp$, and reverse the two transformations.

Let the directrix be $q+\dfrac f2=0$ and the focus $\left(\dfrac f2,0\right)$. The parabola has the equation $$\left(p+\frac f2\right)^2=\left(p-\frac f2\right)^2+q^2$$ or $$2fp=q^2.$$