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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^2 =4ax.also obtain the   coordinate of the point of  contact  b) find the equation of   tangent (x^2 /4)+(y^2 /9)=1 with  gradient 2
$${Find}\:{interms}\:{of}\:\:{a},{b}\:{the} \\ $$$${value}\:{of}\:{c}\:{which}\:{makes} \\ $$$${the}\:{line}\:{y}={mx}+{c} \\ $$$${a}\:{tangent}\:{to}\:{the}\:{parabola} \\ $$$${y}^{\mathrm{2}} =\mathrm{4}{ax}.{also}\:{obtain}\:{the}\: \\ $$$${coordinate}\:{of}\:{the}\:{point}\:{of} \\ $$$${contact} \\ $$$$\left.{b}\right)\:{find}\:{the}\:{equation}\:{of}\: \\ $$$${tangent}\:\frac{{x}^{\mathrm{2}} }{\mathrm{4}}+\frac{{y}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1}\:{with} \\ $$$${gradient}\:\mathrm{2} \\ $$
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.
$${thank}\:{you}\:{sir}. \\ $$
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
$${thank}\:{you} \\ $$
Commented by tanmay.chaudhury50@gmail.com last updated on 23/Dec/18
most welcome...
$${most}\:{welcome}… \\ $$
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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