Question-90773

Question Number 90773 by ajfour last updated on 26/Apr/20

Commented by ajfour last updated on 26/Apr/20

If equation of parabola is y=x^2 , find equation of AC.

$${If}\:{equation}\:{of}\:{parabola}\:{is}\:{y}={x}^{\mathrm{2}} , \\ $$$$\:{find}\:{equation}\:{of}\:{AC}. \\ $$

Answered by mr W last updated on 26/Apr/20

A(−a,a^2 ) eqn. of AC: y=a^2 +(1/(2a))(x+a) c^2 =a^2 +(1/(2a))(c+a) ...(i) (c^2 −a^2 )^2 +(c+a)^2 =c^4 ...(ii) from (i): 2ac^2 −c−a(2a^2 +1)=0 ⇒c=((1+(√(1+8a^2 (2a^2 +1))))/(4a)) from (ii): (2a^2 −1)c^2 −2ac−a^2 (1+a^2 )=0 ⇒c=((a+a^2 (√(1+2a^2 )))/(2a^2 −1)) ((1+(√(1+8a^2 (2a^2 +1))))/(4a))=((a+a^2 (√(1+2a^2 )))/(2a^2 −1)) ⇒a≈1.6159

$${A}\left(−{a},{a}^{\mathrm{2}} \right) \\ $$$${eqn}.\:{of}\:{AC}: \\ $$$${y}={a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}{a}}\left({x}+{a}\right) \\ $$$$ \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}{a}}\left({c}+{a}\right)\:\:\:…\left({i}\right) \\ $$$$\left({c}^{\mathrm{2}} −{a}^{\mathrm{2}} \right)^{\mathrm{2}} +\left({c}+{a}\right)^{\mathrm{2}} ={c}^{\mathrm{4}} \:\:\:…\left({ii}\right) \\ $$$${from}\:\left({i}\right): \\ $$$$\mathrm{2}{ac}^{\mathrm{2}} −{c}−{a}\left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right)=\mathrm{0} \\ $$$$\Rightarrow{c}=\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{8}{a}^{\mathrm{2}} \left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right)}}{\mathrm{4}{a}} \\ $$$${from}\:\left({ii}\right): \\ $$$$\left(\mathrm{2}{a}^{\mathrm{2}} −\mathrm{1}\right){c}^{\mathrm{2}} −\mathrm{2}{ac}−{a}^{\mathrm{2}} \left(\mathrm{1}+{a}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\Rightarrow{c}=\frac{{a}+{a}^{\mathrm{2}} \sqrt{\mathrm{1}+\mathrm{2}{a}^{\mathrm{2}} }}{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\frac{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{8}{a}^{\mathrm{2}} \left(\mathrm{2}{a}^{\mathrm{2}} +\mathrm{1}\right)}}{\mathrm{4}{a}}=\frac{{a}+{a}^{\mathrm{2}} \sqrt{\mathrm{1}+\mathrm{2}{a}^{\mathrm{2}} }}{\mathrm{2}{a}^{\mathrm{2}} −\mathrm{1}} \\ $$$$\Rightarrow{a}\approx\mathrm{1}.\mathrm{6159} \\ $$

Commented by mr W last updated on 26/Apr/20