Menu Close

Show-that-the-equation-of-tangent-to-the-ellipse-x-2-a-2-y-2-b-2-1-at-the-end-of-lactus-rectum-which-lie-in-the-1-st-quadrant-is-xe-y-a-0-merry-X-mas-and-happy-new-year-

Tinku Tara 0 Comments
Pin




Question Number 51156 by peter frank last updated on 24/Dec/18
Show that the equation of tangent to the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 at the end of lactus rectum which lie in the 1^(st) quadrant is xe+y−a=0 ∗merry X−mas and happy new year∗
$${Show}\:{that}\:{the}\:{equation} \\ $$$${of}\:{tangent}\:{to}\:{the}\:{ellipse} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:{at}\:{the}\:{end}\:{of} \\ $$$${lactus}\:{rectum}\:{which} \\ $$$${lie}\:{in}\:{the}\:\mathrm{1}^{{st}} {quadrant}\:{is} \\ $$$${xe}+{y}−{a}=\mathrm{0} \\ $$$$ \\ $$$$\ast{merry}\:{X}−{mas}\:{and}\:{happy}\:{new}\:{year}\ast \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 24/Dec/18
tangent is focus(c,0) ends of L.T are (c,±b(√(1−(c^2 /a^2 ))) ) =(c,±b(√((a^2 −c^2 )/a^2 )) )=(c,±(b^2 /a)) [c^2 =a^2 −b^2 ] tangent in 1st quadrant... ((xc)/a^2 )+(((b^2 /a)y)/b^2 )=1 ((xc)/a^2 )+(y/a)=1 ((xc)/a)+y=a xe+y=a [(c/a)=e]
$${tangent}\:{is} \\ $$$${focus}\left({c},\mathrm{0}\right)\: \\ $$$${ends}\:{of}\:{L}.{T}\:{are}\:\left({c},\pm{b}\sqrt{\mathrm{1}−\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:\right) \\ $$$$=\left({c},\pm{b}\sqrt{\frac{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }}\:\right)=\left({c},\pm\frac{{b}^{\mathrm{2}} }{{a}}\right)\:\:\:\left[{c}^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \:\right] \\ $$$${tangent}\:{in}\:\mathrm{1}{st}\:{quadrant}… \\ $$$$\frac{{xc}}{{a}^{\mathrm{2}} }+\frac{\frac{{b}^{\mathrm{2}} }{{a}}{y}}{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{xc}}{{a}^{\mathrm{2}} }+\frac{{y}}{{a}}=\mathrm{1} \\ $$$$\frac{{xc}}{{a}}+{y}={a}\:\: \\ $$$${xe}+{y}={a}\:\:\left[\frac{{c}}{{a}}={e}\right] \\ $$$$ \\ $$
Answered by peter frank last updated on 25/Dec/18
from tangent equation ((xx_1 )/a^2 )+((yy_1 )/b^2 )=1 recall (x^2 /a^2 )+(y^2 /b^2 )=1 p(ae,0) (((ae)^2 )/a^2 )+(y^2 /b^2 )=1 y^2 =b^2 (1−e^2 )....(i) from b^2 =a^2 (1−e^2 ) (b^2 /a^2 )=(1−e^2 ).....(ii) sub (ii) in (i) y^2 =(b^4 /a^2 ) y=(b^2 /a)=y_1 ((x(ae))/a^2 )+((y((b^2 /a)))/b^2 )=1 ((ae)/a)+(y/a)=1 xe+y−a=0
$${from}\:{tangent}\:{equation} \\ $$$$\frac{{xx}_{\mathrm{1}} }{{a}^{\mathrm{2}} }+\frac{{yy}_{\mathrm{1}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${recall} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${p}\left({ae},\mathrm{0}\right) \\ $$$$\frac{\left({ae}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${y}^{\mathrm{2}} ={b}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right)….\left({i}\right) \\ $$$${from} \\ $$$${b}^{\mathrm{2}} ={a}^{\mathrm{2}} \left(\mathrm{1}−{e}^{\mathrm{2}} \right) \\ $$$$\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }=\left(\mathrm{1}−{e}^{\mathrm{2}} \right)…..\left({ii}\right) \\ $$$${sub}\:\left({ii}\right)\:{in}\:\left({i}\right) \\ $$$${y}^{\mathrm{2}} =\frac{{b}^{\mathrm{4}} }{{a}^{\mathrm{2}} } \\ $$$${y}=\frac{{b}^{\mathrm{2}} }{{a}}={y}_{\mathrm{1}} \\ $$$$\frac{{x}\left({ae}\right)}{{a}^{\mathrm{2}} }+\frac{{y}\left(\frac{{b}^{\mathrm{2}} }{{a}}\right)}{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{ae}}{{a}}+\frac{{y}}{{a}}=\mathrm{1} \\ $$$${xe}+{y}−{a}=\mathrm{0} \\ $$$$ \\ $$

Pin

Leave a Reply

Your email address will not be published. Required fields are marked *