Question and Answers Forum

All Questions      Topic List

Coordinate Geometry Questions

Previous in All Question      Next in All Question      

Previous in Coordinate Geometry      Next in Coordinate Geometry      

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

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

Terms of Service

Privacy Policy

Contact: info@tinkutara.com