Menu Close

Prove-that-the-expression-ax-2-2hxy-by-2-2gx-2fy-c-0-can-be-resolved-into-two-linear-rational-factors-if-abc-2fgh-af-2-bg-2-ch-2-0-

Tinku Tara 0 Comments
Pin




Question Number 20297 by Tinkutara last updated on 25/Aug/17
Prove that the expression ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 can be resolved into two linear rational factors if Δ = abc + 2fgh − af^2 − bg^2 − ch^2 = 0
$$\mathrm{Prove}\:\mathrm{that}\:\mathrm{the}\:\mathrm{expression}\:{ax}^{\mathrm{2}} \:+\:\mathrm{2}{hxy} \\ $$$$+\:{by}^{\mathrm{2}} \:+\:\mathrm{2}{gx}\:+\:\mathrm{2}{fy}\:+\:{c}\:=\:\mathrm{0}\:\mathrm{can}\:\mathrm{be} \\ $$$$\mathrm{resolved}\:\mathrm{into}\:\mathrm{two}\:\mathrm{linear}\:\mathrm{rational}\:\mathrm{factors} \\ $$$$\mathrm{if}\:\Delta\:=\:{abc}\:+\:\mathrm{2}{fgh}\:−\:{af}^{\mathrm{2}} \:−\:{bg}^{\mathrm{2}} \:−\:{ch}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$
Answered by ajfour last updated on 25/Aug/17
let a(x−x_0 )^2 +b(y−y_0 )^2 + 2h(x−x_0 )(y−y_0 )=ax^2 +by^2 +2hxy 2gx+2fy+c=0 comparing coefficients of x and y, and constant term we get: ax_0 +hy_0 =−g .....(i) by_0 +hx_0 =−f ......(ii) ax_0 ^2 +by_0 ^2 +2hx_0 y_0 =c (iii) using (i) and (ii) in (iii): ⇒ x_0 (−g−hx_0 )+y_0 (−f−hx_0 ) +2hx_0 y_0 =c or gx_0 +fy_0 =−c ....(iv) (i) , (ii), and (iv) form a system of three linear equations in two unknowns, has unique solution only if determinant ((a,h,g),(h,b,f),(g,f,c))=0 first row we get from coefficients of x, y, and constant term of equation (i), second row→(ii) third row → (iv) ⇒a(bc−f^2 )−h(ch−gf)+g(hf−bg)=0 ⇒ abc+2fgh−af^2 −ch^2 −bg^2 =0 .
$${let}\:{a}\left({x}−{x}_{\mathrm{0}} \right)^{\mathrm{2}} +{b}\left({y}−{y}_{\mathrm{0}} \right)^{\mathrm{2}} + \\ $$$$\:\mathrm{2}{h}\left({x}−{x}_{\mathrm{0}} \right)\left({y}−{y}_{\mathrm{0}} \right)={ax}^{\mathrm{2}} +{by}^{\mathrm{2}} +\mathrm{2}{hxy} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\mathrm{2}{gx}+\mathrm{2}{fy}+{c}=\mathrm{0} \\ $$$${comparing}\:{coefficients}\:{of}\:{x}\:{and} \\ $$$${y},\:{and}\:{constant}\:{term}\:{we}\:{get}: \\ $$$$\:\:{ax}_{\mathrm{0}} +{hy}_{\mathrm{0}} =−{g}\:\:\:\:\:…..\left({i}\right) \\ $$$$\:\:{by}_{\mathrm{0}} +{hx}_{\mathrm{0}} =−{f}\:\:\:\:\:……\left({ii}\right) \\ $$$${ax}_{\mathrm{0}} ^{\mathrm{2}} +{by}_{\mathrm{0}} ^{\mathrm{2}} +\mathrm{2}{hx}_{\mathrm{0}} {y}_{\mathrm{0}} ={c}\:\:\:\left({iii}\right) \\ $$$${using}\:\left({i}\right)\:{and}\:\left({ii}\right)\:{in}\:\left({iii}\right): \\ $$$$\Rightarrow\:{x}_{\mathrm{0}} \left(−{g}−{hx}_{\mathrm{0}} \right)+{y}_{\mathrm{0}} \left(−{f}−{hx}_{\mathrm{0}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:+\mathrm{2}{hx}_{\mathrm{0}} {y}_{\mathrm{0}} ={c} \\ $$$${or}\:\:\:\:\:{gx}_{\mathrm{0}} +{fy}_{\mathrm{0}} =−{c}\:\:\:\:\:….\left({iv}\right) \\ $$$$\left({i}\right)\:,\:\left({ii}\right),\:{and}\:\left({iv}\right)\:{form}\:{a}\:{system} \\ $$$${of}\:{three}\:{linear}\:{equations}\:{in}\:{two} \\ $$$${unknowns},\:{has}\:{unique}\:{solution}\:{only}\:{if} \\ $$$$\begin{vmatrix}{{a}}&{{h}}&{{g}}\\{{h}}&{{b}}&{{f}}\\{{g}}&{{f}}&{{c}}\end{vmatrix}=\mathrm{0} \\ $$$${first}\:{row}\:{we}\:{get}\:{from}\:{coefficients} \\ $$$${of}\:{x},\:{y},\:{and}\:{constant}\:{term}\:{of} \\ $$$${equation}\:\left({i}\right),\:\:{second}\:{row}\rightarrow\left({ii}\right) \\ $$$${third}\:{row}\:\rightarrow\:\left({iv}\right) \\ $$$$\Rightarrow{a}\left({bc}−{f}^{\mathrm{2}} \right)−{h}\left({ch}−{gf}\right)+{g}\left({hf}−{bg}\right)=\mathrm{0} \\ $$$$\Rightarrow\:\:{abc}+\mathrm{2}{fgh}−{af}^{\mathrm{2}} −{ch}^{\mathrm{2}} −{bg}^{\mathrm{2}} =\mathrm{0}\:. \\ $$
Commented by Tinkutara last updated on 25/Aug/17
Thank you very much Sir!
$$\mathrm{Thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much}\:\mathrm{Sir}! \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *