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Prove-that-length-of-lactus-rectum-when-directrix-are-parallel-to-y-axis-is-y-1-y-2-2b-2-a-

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Question Number 51614 by peter frank last updated on 29/Dec/18
Prove that length of lactus rectum when directrix are parallel to y−axis is y_1 −y_2 =((2b^2 )/a)
$${Prove}\:{that}\:{length}\:{of} \\ $$$${lactus}\:{rectum}\:{when}\: \\ $$$${directrix}\:{are}\:{parallel}\:{to} \\ $$$${y}−{axis}\:{is}\: \\ $$$${y}_{\mathrm{1}} −{y}_{\mathrm{2}} =\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}} \\ $$
Answered by tanmay.chaudhury50@gmail.com last updated on 29/Dec/18
for ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 ditectrix is ∥ to yaxis focus(c,0) and (−c,0) so eqn of st line ∥to yaxis passing through (c,0) is x=c solve x=c and (x^2 /a^2 )+(y^2 /b^2 )=1 (y^2 /b^2 )=1−(c^2 /a^2 ) so y=±(b/a)(√(a^2 −c^2 )) now diztance between (c,(b/a)(√(a^2 −c^2 )) )and(c,((−b)/a)(√(a^2 −c^2 )) ) is=((2b)/a)(√(a^2 −c^2 )) [for ellipse c^2 =a^2 −b^2 ] =((2b)/a)×b=((2b^2 )/a) for hyperbola (x^2 /a^2 )−(y^2 /b^2 )=1 directrix is ∥ to yaxis focus(c,0) (−c,0) eqn of st line passing through(c,0) and ∥ to yaxis is x=c solve x=c and (x^2 /a^2 )−(y^2 /b^2 )=1 (y^2 /b^2 )=(c^2 /a^2 )−1 y=±(b/a)(√(c^2 −a^2 )) y=±(b/a)×b [for hyperbola c^2 =a^2 +b^2 ] y=±(b^2 /a) so latus rectum is ((2b^2 )/a)
$${for}\:{ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:{ditectrix}\:{is}\:\parallel\:{to}\:{yaxis} \\ $$$${focus}\left({c},\mathrm{0}\right)\:{and}\:\left(−{c},\mathrm{0}\right)\:\:{so}\:{eqn}\:{of}\:{st}\:{line}\:\parallel{to}\:{yaxis} \\ $$$${passing}\:{through}\:\left({c},\mathrm{0}\right)\:{is}\:{x}={c} \\ $$$${solve}\:{x}={c}\:\:{and}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}−\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:\:{so}\:{y}=\pm\frac{{b}}{{a}}\sqrt{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }\: \\ $$$${now}\:{diztance}\:{between}\:\left({c},\frac{{b}}{{a}}\sqrt{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }\:\right){and}\left({c},\frac{−{b}}{{a}}\sqrt{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }\:\right) \\ $$$${is}=\frac{\mathrm{2}{b}}{{a}}\sqrt{{a}^{\mathrm{2}} −{c}^{\mathrm{2}} }\:\:\:\left[{for}\:{ellipse}\:{c}^{\mathrm{2}} ={a}^{\mathrm{2}} −{b}^{\mathrm{2}} \right] \\ $$$$=\frac{\mathrm{2}{b}}{{a}}×{b}=\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}} \\ $$$${for}\:{hyperbola}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${directrix}\:{is}\:\parallel\:{to}\:{yaxis} \\ $$$${focus}\left({c},\mathrm{0}\right)\:\left(−{c},\mathrm{0}\right) \\ $$$${eqn}\:{of}\:{st}\:{line}\:{passing}\:{through}\left({c},\mathrm{0}\right)\:{and}\:\parallel\:{to}\:{yaxis} \\ $$$${is}\:{x}={c} \\ $$$${solve}\:{x}={c}\:{and}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\frac{{c}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\mathrm{1} \\ $$$${y}=\pm\frac{{b}}{{a}}\sqrt{{c}^{\mathrm{2}} −{a}^{\mathrm{2}} }\: \\ $$$${y}=\pm\frac{{b}}{{a}}×{b}\:\:\left[{for}\:{hyperbola}\:{c}^{\mathrm{2}} ={a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right] \\ $$$${y}=\pm\frac{{b}^{\mathrm{2}} }{{a}} \\ $$$${so}\:{latus}\:{rectum}\:{is}\:\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}} \\ $$$$ \\ $$
Commented by peter frank last updated on 29/Dec/18
very nice sir....thanks
$${very}\:{nice}\:{sir}….{thanks} \\ $$
Answered by peter frank last updated on 29/Dec/18
(((x−h)^2 )/a^2 )−(((y−k)^2 )/b^2 )=1 x=h+ae(parallel to y−axis) (((h+ae−h)^2 )/a^2 )−(((y−k)^2 )/b^2 )=1 e^2 −1=(((y−k)^2 )/b^2 ) y−k=(√(b^2 (e^2 −1))) y−k=±b(√((e^2 −1))) y=k±b(√((e^2 −1))) y_(1 ) =k−b(√(e^2 −1)) ....(i) y_(2 ) =k+b(√(e^2 −1)) ....(ii) y_(2 ) −y_1 =−2b(√(e^2 −1)) y_(1 ) −y_2 =2b(√(e^2 −1)) from b^2 =a^2 (e^2 −1) y_(1 ) −y_2 =2b×(b/a) y_(1 ) −y_2 =((2b^2 )/a)
$$\frac{\left({x}−{h}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\left({y}−{k}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${x}={h}+{ae}\left({parallel}\:{to}\:{y}−{axis}\right) \\ $$$$\frac{\left({h}+{ae}−{h}\right)^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{\left({y}−{k}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${e}^{\mathrm{2}} −\mathrm{1}=\frac{\left({y}−{k}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} } \\ $$$${y}−{k}=\sqrt{{b}^{\mathrm{2}} \left({e}^{\mathrm{2}} −\mathrm{1}\right)}\:\: \\ $$$${y}−{k}=\pm{b}\sqrt{\left({e}^{\mathrm{2}} −\mathrm{1}\right)}\:\: \\ $$$${y}={k}\pm{b}\sqrt{\left({e}^{\mathrm{2}} −\mathrm{1}\right)}\:\: \\ $$$${y}_{\mathrm{1}\:} ={k}−{b}\sqrt{{e}^{\mathrm{2}} −\mathrm{1}}\:….\left({i}\right) \\ $$$${y}_{\mathrm{2}\:} ={k}+{b}\sqrt{{e}^{\mathrm{2}} −\mathrm{1}}\:….\left({ii}\right) \\ $$$${y}_{\mathrm{2}\:} −{y}_{\mathrm{1}} =−\mathrm{2}{b}\sqrt{{e}^{\mathrm{2}} −\mathrm{1}}\: \\ $$$${y}_{\mathrm{1}\:} −{y}_{\mathrm{2}} =\mathrm{2}{b}\sqrt{{e}^{\mathrm{2}} −\mathrm{1}}\: \\ $$$${from} \\ $$$${b}^{\mathrm{2}} ={a}^{\mathrm{2}} \left({e}^{\mathrm{2}} −\mathrm{1}\right) \\ $$$${y}_{\mathrm{1}\:} −{y}_{\mathrm{2}} =\mathrm{2}{b}×\frac{{b}}{{a}} \\ $$$${y}_{\mathrm{1}\:} −{y}_{\mathrm{2}} =\frac{\mathrm{2}{b}^{\mathrm{2}} }{{a}} \\ $$$$ \\ $$

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