Find-interms-of-a-b-the-value-of-c-which-makes-the-line-y-mx-c-a-tangent-to-the-parabola-y-2-4ax-also-obtain-the-coordinate-of-the-point-of-contact-b-find-the-equation-of-tangent-x-2-4-y-2-9

Question Number 50977 by peter frank last updated on 22/Dec/18

$${Find}\:{interms}\:{of}\:\:{a},{b}\:{the} \\ $$$${value}\:{of}\:{c}\:{which}\:{makes} \\ $$$${the}\:{line}\:{y}={mx}+{c} \\ $$$${a}\:{tangent}\:{to}\:{the}\:{parabola} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{ax}.{also}\:{obtain}\:{the}\: \\ $$$${coordinate}\:{of}\:{the}\:{point}\:{of} \\ $$$${contact} \\ $$$$\left.{b}\right)\:{find}\:{the}\:{equation}\:{of}\: \\ $$$${tangent}\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{y}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1}\:{with} \\ $$$${gradient}\:\mathrm{2} \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

(mx+c)^2 =4ax m^2 x^2 +2mcx+c^2 =4ax x^2 (m^2 )+x(2mc−4a)+c^2 =0 roots are equal B^2 =4AC (2mc−4a)^2 =4m^2 c^2 4m^2 c^2 −16amc+16a^2 =4m^2 c^2 −16amc=−16a^2 c=(a/m)

$$\left({mx}+{c}\right)^{\mathrm{2}} =\mathrm{4}{ax} \\ $$$${m}^{\mathrm{2}} {x}^{\mathrm{2}} +\mathrm{2}{mcx}+{c}^{\mathrm{2}} =\mathrm{4}{ax} \\ $$$${x}^{\mathrm{2}} \left({m}^{\mathrm{2}} \right)+{x}\left(\mathrm{2}{mc}−\mathrm{4}{a}\right)+{c}^{\mathrm{2}} =\mathrm{0} \\ $$$${roots}\:{are}\:{equal}\:\:{B}^{\mathrm{2}} =\mathrm{4}{AC} \\ $$$$\left(\mathrm{2}{mc}−\mathrm{4}{a}\right)^{\mathrm{2}} =\mathrm{4}{m}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$$\mathrm{4}{m}^{\mathrm{2}} {c}^{\mathrm{2}} −\mathrm{16}{amc}+\mathrm{16}{a}^{\mathrm{2}} =\mathrm{4}{m}^{\mathrm{2}} {c}^{\mathrm{2}} \\ $$$$−\mathrm{16}{amc}=−\mathrm{16}{a}^{\mathrm{2}} \\ $$$${c}=\frac{{a}}{{m}} \\ $$

Commented by peter frank last updated on 23/Dec/18

$${thank}\:{you}\:{sir}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

b)tanngent y=mx+(√(a^2 m^2 +b^2 )) y=2x+(√(16+9)) y=2x±5

$$\left.{b}\right){tanngent}\: \\ $$$${y}={mx}+\sqrt{{a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} }\: \\ $$$${y}=\mathrm{2}{x}+\sqrt{\mathrm{16}+\mathrm{9}}\: \\ $$$${y}=\mathrm{2}{x}\pm\mathrm{5} \\ $$

Commented by peter frank last updated on 23/Dec/18

$${thank}\:{you} \\ $$

Commented by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18

$${most}\:{welcome}… \\ $$

Answered by peter frank last updated on 23/Dec/18

b) a^2 =4 b^2 =9 m=2 y=mx+c c^2 =b^2 +a^2 m^2 c=±5 y=2x±5

$$\left.{b}\right)\:{a}^{\mathrm{2}} =\mathrm{4}\:\:\:\:{b}^{\mathrm{2}} =\mathrm{9}\:\:\:\:{m}=\mathrm{2} \\ $$$${y}={mx}+{c} \\ $$$${c}^{\mathrm{2}} ={b}^{\mathrm{2}} +{a}^{\mathrm{2}} {m}^{\mathrm{2}} \\ $$$${c}=\pm\mathrm{5} \\ $$$${y}=\mathrm{2}{x}\pm\mathrm{5} \\ $$$$ \\ $$

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