Question Number 85298 by mr W last updated on 20/Mar/20
$${Find}\:{the}\:{symmetry}\:{axes}\:\left({if}\:{any}\right)\:{of} \\ $$$${the}\:{following}\:{curve}: \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{2}{x}−\mathrm{8}{y}−\mathrm{2}=\mathrm{0} \\ $$
Commented by jagoll last updated on 20/Mar/20
$${x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}+{y}^{\mathrm{2}} −\mathrm{8}{y}+\mathrm{16}\:=\:\mathrm{2}{xy}+\mathrm{19} \\ $$$$\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\left({y}−\mathrm{4}\right)^{\mathrm{2}} \:=\:\mathrm{2}{xy}+\mathrm{19} \\ $$$${next} \\ $$
Commented by mr W last updated on 20/Mar/20
$${any}\:{result}\:{sir}? \\ $$
Answered by MJS last updated on 21/Mar/20
$${y}={x}+\mathrm{4}\pm\sqrt{\mathrm{6}\left({x}+\mathrm{3}\right)}\:\mathrm{defined}\:\mathrm{for}\:{x}\geqslant−\mathrm{3} \\ $$$${x}={y}−\mathrm{1}\pm\sqrt{\mathrm{3}\left(\mathrm{2}{y}+\mathrm{1}\right)}\:\mathrm{defined}\:\mathrm{for}\:{y}\geqslant−\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\Rightarrow\:\mathrm{it}'\mathrm{s}\:\mathrm{a}\:\mathrm{parabola}\:\mathrm{with}\:\mathrm{direction}\:{x}^{+} \:\mathrm{and}\:{y}^{+} \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:…\mathrm{under}\:\mathrm{construction}… \\ $$
Commented by mr W last updated on 22/Mar/20
$${thank}\:{you}\:{sir}! \\ $$$${i}\:{got}\:{the}\:{symmetry}\:{axis}\:{y}={x}+\frac{\mathrm{5}}{\mathrm{2}}\:{and} \\ $$$${have}\:{checked}\:{graphically}. \\ $$
Commented by mr W last updated on 21/Mar/20
Answered by mr W last updated on 21/Mar/20
![Method 1 x^2 +y^2 −2xy+2x−8y−2=0 ⇒(x−y+1)^2 −6y−3=0 it is to see that the curve is a parabola, which has one and only one symmetry axis. the symmetry axis of a parabola is one of the lines which intersect (not tangent) the parabola at only one point. let′s say this symmetry axis is x=my+k, or y=(x/m)−(k/m). the intersection of both is the solution of { (((x−y+1)^2 −6y−3=0)),((x=my+k)) :} (my+k−y+1)^2 −6y−3=0 [(m−1)y+(k+1)]^2 −6y−3=0 (m−1)^2 y^2 +2[(m−1)(k+1)−3]y+(k+1)^2 −3=0 since it should always have only one root, the y^2 term must vanish, i.e. m=1. the eqn. is then −6y+(k+1)^2 −3=0 ⇒y=(((k+1)^2 −3)/6) ⇒x=(((k+1)^2 −3)/6)+k such that this line is parabola′s axis, the intersection point is the vertex, i.e. vertex is ((((k+1)^2 −3)/6)+k, (((k+1)^2 −3)/6)). we know at vertex the tangent of the parabola is perpendicular to the (symmetry) axis of the parabola, i.e. y′=−m=−1. from (x−y+1)^2 −6y−3=0 we get 2(x−y+1)(1−y′)−6y′=0 at the vertex: 2(k+1)(1+1)+6=0 ⇒k=−(5/2) therefore the symmetry axis is x=my+k=y−(5/2) or y=x+(5/2)](https://www.tinkutara.com/question/Q85410.png)
$${Method}\:\mathrm{1} \\ $$$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{2}{xy}+\mathrm{2}{x}−\mathrm{8}{y}−\mathrm{2}=\mathrm{0} \\ $$$$\Rightarrow\left({x}−{y}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{6}{y}−\mathrm{3}=\mathrm{0} \\ $$$${it}\:{is}\:{to}\:{see}\:{that}\:{the}\:{curve}\:{is}\:{a}\:{parabola}, \\ $$$${which}\:{has}\:{one}\:{and}\:{only}\:{one}\:{symmetry} \\ $$$${axis}. \\ $$$$ \\ $$$${the}\:{symmetry}\:{axis}\:{of}\:{a}\:{parabola}\:{is} \\ $$$${one}\:{of}\:{the}\:{lines}\:{which}\:{intersect} \\ $$$$\left({not}\:{tangent}\right)\:{the}\:{parabola}\:{at}\:{only}\:{one} \\ $$$${point}. \\ $$$${let}'{s}\:{say}\:{this}\:{symmetry}\:{axis}\:{is} \\ $$$${x}={my}+{k},\:{or}\:{y}=\frac{{x}}{{m}}−\frac{{k}}{{m}}. \\ $$$$ \\ $$$${the}\:{intersection}\:{of}\:{both}\:{is}\:{the}\: \\ $$$${solution}\:{of} \\ $$$$\begin{cases}{\left({x}−{y}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{6}{y}−\mathrm{3}=\mathrm{0}}\\{{x}={my}+{k}}\end{cases} \\ $$$$\left({my}+{k}−{y}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{6}{y}−\mathrm{3}=\mathrm{0} \\ $$$$\left[\left({m}−\mathrm{1}\right){y}+\left({k}+\mathrm{1}\right)\right]^{\mathrm{2}} −\mathrm{6}{y}−\mathrm{3}=\mathrm{0} \\ $$$$\left({m}−\mathrm{1}\right)^{\mathrm{2}} {y}^{\mathrm{2}} +\mathrm{2}\left[\left({m}−\mathrm{1}\right)\left({k}+\mathrm{1}\right)−\mathrm{3}\right]{y}+\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$${since}\:{it}\:{should}\:{always}\:{have}\:{only}\:{one} \\ $$$${root},\:{the}\:{y}^{\mathrm{2}} \:{term}\:{must}\:{vanish},\:{i}.{e}. \\ $$$${m}=\mathrm{1}.\:{the}\:{eqn}.\:{is}\:{then} \\ $$$$−\mathrm{6}{y}+\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}=\mathrm{0} \\ $$$$\Rightarrow{y}=\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}}{\mathrm{6}} \\ $$$$\Rightarrow{x}=\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}}{\mathrm{6}}+{k} \\ $$$${such}\:{that}\:{this}\:{line}\:{is}\:{parabola}'{s}\:{axis}, \\ $$$${the}\:{intersection}\:{point}\:{is}\:{the}\:{vertex}, \\ $$$${i}.{e}.\:{vertex}\:{is}\:\left(\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}}{\mathrm{6}}+{k},\:\frac{\left({k}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{3}}{\mathrm{6}}\right). \\ $$$$ \\ $$$${we}\:{know}\:{at}\:{vertex}\:{the}\:{tangent}\:{of}\:{the} \\ $$$${parabola}\:{is}\:{perpendicular}\:{to}\:{the} \\ $$$$\left({symmetry}\right)\:{axis}\:{of}\:{the}\:{parabola}, \\ $$$${i}.{e}.\:{y}'=−{m}=−\mathrm{1}. \\ $$$${from}\:\left({x}−{y}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{6}{y}−\mathrm{3}=\mathrm{0}\:{we}\:{get} \\ $$$$\mathrm{2}\left({x}−{y}+\mathrm{1}\right)\left(\mathrm{1}−{y}'\right)−\mathrm{6}{y}'=\mathrm{0} \\ $$$${at}\:{the}\:{vertex}: \\ $$$$\mathrm{2}\left({k}+\mathrm{1}\right)\left(\mathrm{1}+\mathrm{1}\right)+\mathrm{6}=\mathrm{0} \\ $$$$\Rightarrow{k}=−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${therefore}\:{the}\:{symmetry}\:{axis}\:{is} \\ $$$${x}={my}+{k}={y}−\frac{\mathrm{5}}{\mathrm{2}} \\ $$$${or}\:{y}={x}+\frac{\mathrm{5}}{\mathrm{2}} \\ $$