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Question Number 31113 by NECx last updated on 02/Mar/18

Two lines through the point (1,−3) are tamgent to the curve y=x^2 . Find the equation of these two lines and make a sketch to verify your results.

$${Two}\:{lines}\:{through}\:{the}\:{point}\:\left(\mathrm{1},−\mathrm{3}\right) \\ $$$${are}\:{tamgent}\:{to}\:{the}\:{curve}\:{y}={x}^{\mathrm{2}} . \\ $$$${Find}\:{the}\:{equation}\:{of}\:{these}\:{two} \\ $$$${lines}\:{and}\:{make}\:{a}\:{sketch}\:{to}\:{verify} \\ $$$${your}\:{results}. \\ $$

Answered by Tinkutara last updated on 02/Mar/18

Commented by Tinkutara last updated on 02/Mar/18

Let equation of tangent is y=mx+c It touches y=x^2 so D=0 mx+c=x^2 D=0 gives c=((−m^2 )/4) y=mx−(m^2 /4) It passes through (1,−3) −3=m−(m^2 /4) m=−2,6 Equations of line are y=−2x−1 y=6x−9

$${Let}\:{equation}\:{of}\:{tangent}\:{is}\:{y}={mx}+{c} \\ $$$${It}\:{touches}\:{y}={x}^{\mathrm{2}} \:{so}\:{D}=\mathrm{0} \\ $$$${mx}+{c}={x}^{\mathrm{2}} \\ $$$${D}=\mathrm{0}\:{gives}\:{c}=\frac{−{m}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${y}={mx}−\frac{{m}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${It}\:{passes}\:{through}\:\left(\mathrm{1},−\mathrm{3}\right) \\ $$$$−\mathrm{3}={m}−\frac{{m}^{\mathrm{2}} }{\mathrm{4}} \\ $$$${m}=−\mathrm{2},\mathrm{6} \\ $$$${Equations}\:{of}\:{line}\:{are} \\ $$$${y}=−\mathrm{2}{x}−\mathrm{1} \\ $$$${y}=\mathrm{6}{x}−\mathrm{9} \\ $$

Commented by NECx last updated on 03/Mar/18

I′m most grateful sir.Thanks.

$${I}'{m}\:{most}\:{grateful}\:{sir}.{Thanks}. \\ $$

Answered by mrW2 last updated on 02/Mar/18

eqn. of line through point (1,−3): y+3=m(x−1) or y=m(x−1)−3 intersection: y=m(x−1)−3=x^2 ⇒x^2 −mx+(m+3)=0 D=m^2 −4(m+3)=0 ⇒(m−2)^2 −16=0 ⇒m−2=±4 ⇒m=6 or −2 Eqn. of lines: y=6(x−1)−3⇒y=6x−9 y=−2(x−1)−3⇒y=−2x−1

$${eqn}.\:{of}\:{line}\:{through}\:{point}\:\left(\mathrm{1},−\mathrm{3}\right): \\ $$$${y}+\mathrm{3}={m}\left({x}−\mathrm{1}\right) \\ $$$${or}\:{y}={m}\left({x}−\mathrm{1}\right)−\mathrm{3} \\ $$$$ \\ $$$${intersection}: \\ $$$${y}={m}\left({x}−\mathrm{1}\right)−\mathrm{3}={x}^{\mathrm{2}} \\ $$$$\Rightarrow{x}^{\mathrm{2}} −{mx}+\left({m}+\mathrm{3}\right)=\mathrm{0} \\ $$$${D}={m}^{\mathrm{2}} −\mathrm{4}\left({m}+\mathrm{3}\right)=\mathrm{0} \\ $$$$\Rightarrow\left({m}−\mathrm{2}\right)^{\mathrm{2}} −\mathrm{16}=\mathrm{0} \\ $$$$\Rightarrow{m}−\mathrm{2}=\pm\mathrm{4} \\ $$$$\Rightarrow{m}=\mathrm{6}\:{or}\:−\mathrm{2} \\ $$$$ \\ $$$${Eqn}.\:{of}\:{lines}: \\ $$$${y}=\mathrm{6}\left({x}−\mathrm{1}\right)−\mathrm{3}\Rightarrow{y}=\mathrm{6}{x}−\mathrm{9} \\ $$$${y}=−\mathrm{2}\left({x}−\mathrm{1}\right)−\mathrm{3}\Rightarrow{y}=−\mathrm{2}{x}−\mathrm{1} \\ $$

Commented by NECx last updated on 03/Mar/18

Thank you so much sir.

$${Thank}\:{you}\:{so}\:{much}\:{sir}. \\ $$

Commented by NECx last updated on 02/Mar/18

and what does D represents ?

$${and}\:{what}\:{does}\:{D}\:{represents}\:\:? \\ $$

Commented by mrW2 last updated on 02/Mar/18

If the eqn. ax^2 +bx+c=0 should have only one solution, then D=b^2 −4ac must equal to zero.

$${If}\:{the}\:{eqn}.\:{ax}^{\mathrm{2}} +{bx}+{c}=\mathrm{0}\:{should} \\ $$$${have}\:{only}\:{one}\:{solution},\:{then} \\ $$$${D}={b}^{\mathrm{2}} −\mathrm{4}{ac}\:{must}\:{equal}\:{to}\:{zero}. \\ $$

Commented by NECx last updated on 02/Mar/18

why is D=0 ?

$${why}\:{is}\:{D}=\mathrm{0}\:? \\ $$

Commented by mrW2 last updated on 02/Mar/18

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