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Question Number 90989 by ajfour last updated on 27/Apr/20

Commented by ajfour last updated on 27/Apr/20

If eq. of parabola is 9y=4(x^2 −9) and AB=BC=CD, find eq. of the line AD.

$${If}\:{eq}.\:{of}\:{parabola}\:{is}\:\:\mathrm{9}{y}=\mathrm{4}\left({x}^{\mathrm{2}} −\mathrm{9}\right) \\ $$$${and}\:{AB}={BC}={CD},\:{find}\:{eq}.\:{of} \\ $$$${the}\:{line}\:{AD}. \\ $$

Answered by mr W last updated on 27/Apr/20

say eqn. of parabola is generally y=px^2 −q A(a,0) D(0,−d) eqn. of AD: (x/a)−(y/d)=1 ⇒y=d((x/a)−1) intersection of line with parabola: px^2 −q=d((x/a)−1) apx^2 −dx+a(d−q)=0 x_1 +x_2 =(d/(ap)) x_1 x_2 =((d−q)/p) we have ((x_1 +x_2 )/2)=(a/2) ⇒(d/(ap))=a ⇒d=pa^2 ...(i) x_1 −x_2 =(a/3) (x_1 −x_2 )^2 =(x_1 +x_2 )^2 −4x_1 x_2 =(a^2 /9) ⇒a^2 −((4(d−q))/p)=(a^2 /9) ⇒((2a^2 )/9)=((d−q)/p) ⇒2pa^2 =9d−9q ...(ii) 7pa^2 =9q ⇒a=3(√(q/(7p))) ⇒d=((9q)/7) ⇒eqn. y=3(√((pq)/7))x−((9q)/7)

$${say}\:{eqn}.\:{of}\:{parabola}\:{is}\:{generally} \\ $$$${y}={px}^{\mathrm{2}} −{q} \\ $$$${A}\left({a},\mathrm{0}\right) \\ $$$${D}\left(\mathrm{0},−{d}\right) \\ $$$${eqn}.\:{of}\:{AD}: \\ $$$$\frac{{x}}{{a}}−\frac{{y}}{{d}}=\mathrm{1}\:\Rightarrow{y}={d}\left(\frac{{x}}{{a}}−\mathrm{1}\right) \\ $$$${intersection}\:{of}\:{line}\:{with}\:{parabola}: \\ $$$${px}^{\mathrm{2}} −{q}={d}\left(\frac{{x}}{{a}}−\mathrm{1}\right) \\ $$$${apx}^{\mathrm{2}} −{dx}+{a}\left({d}−{q}\right)=\mathrm{0} \\ $$$${x}_{\mathrm{1}} +{x}_{\mathrm{2}} =\frac{{d}}{{ap}} \\ $$$${x}_{\mathrm{1}} {x}_{\mathrm{2}} =\frac{{d}−{q}}{{p}} \\ $$$${we}\:{have} \\ $$$$\frac{{x}_{\mathrm{1}} +{x}_{\mathrm{2}} }{\mathrm{2}}=\frac{{a}}{\mathrm{2}}\:\Rightarrow\frac{{d}}{{ap}}={a}\:\Rightarrow{d}={pa}^{\mathrm{2}} \:\:\:...\left({i}\right) \\ $$$${x}_{\mathrm{1}} −{x}_{\mathrm{2}} =\frac{{a}}{\mathrm{3}}\: \\ $$$$\left({x}_{\mathrm{1}} −{x}_{\mathrm{2}} \right)^{\mathrm{2}} =\left({x}_{\mathrm{1}} +{x}_{\mathrm{2}} \right)^{\mathrm{2}} −\mathrm{4}{x}_{\mathrm{1}} {x}_{\mathrm{2}} =\frac{{a}^{\mathrm{2}} }{\mathrm{9}} \\ $$$$\Rightarrow{a}^{\mathrm{2}} −\frac{\mathrm{4}\left({d}−{q}\right)}{{p}}=\frac{{a}^{\mathrm{2}} }{\mathrm{9}} \\ $$$$\Rightarrow\frac{\mathrm{2}{a}^{\mathrm{2}} }{\mathrm{9}}=\frac{{d}−{q}}{{p}} \\ $$$$\Rightarrow\mathrm{2}{pa}^{\mathrm{2}} =\mathrm{9}{d}−\mathrm{9}{q}\:\:\:...\left({ii}\right) \\ $$$$\mathrm{7}{pa}^{\mathrm{2}} =\mathrm{9}{q} \\ $$$$\Rightarrow{a}=\mathrm{3}\sqrt{\frac{{q}}{\mathrm{7}{p}}} \\ $$$$\Rightarrow{d}=\frac{\mathrm{9}{q}}{\mathrm{7}} \\ $$$$\Rightarrow{eqn}.\:{y}=\mathrm{3}\sqrt{\frac{{pq}}{\mathrm{7}}}{x}−\frac{\mathrm{9}{q}}{\mathrm{7}} \\ $$

Commented by ajfour last updated on 27/Apr/20

Thanks Sir, too good and great, you solved it generally!

$${Thanks}\:{Sir},\:{too}\:{good}\:{and}\:{great}, \\ $$$${you}\:{solved}\:{it}\:{generally}! \\ $$

Commented by mr W last updated on 27/Apr/20

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