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Question-46813

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Question Number 46813 by peter frank last updated on 31/Oct/18
Commented by peter frank last updated on 02/Nov/18
help please
$$\mathrm{help}\:\mathrm{please} \\ $$
Answered by ajfour last updated on 04/Nov/18
eq. of asymptotes: tan θ=(y/x)=±(b/a) let eq. of line PQR : y=m(x−h)+k h=asec φ , k=btan φ m= tan α for x_Q : ((bx_Q )/a) = m(x_Q −h)+k ⇒ x_Q = ((k−mh)/((b/a)−m)) r_1 =(h−x_Q )sec α = (((((bh)/a)−k)/((b/a)−m)))sec α similarly r_2 =(h−x_R )sec α = (((((bh)/a)+k)/((b/a)+m)))sec α r_1 r_2 =[(((((bh)/a))^2 −k^2 )/(((b/a))^2 −m^2 ))]sec^2 α =[(((h^2 /a^2 )−(k^2 /b^2 ))/((1/a^2 )−(m^2 /b^2 )))](1+m^2 ) but P (h,k) lies on hyperbola so (h^2 /a^2 )−(k^2 /b^2 ) = 1 , hence ⇒ r_1 r_2 = ((a^2 b^2 (1+m^2 ))/(b^2 −a^2 m^2 )) .
$${eq}.\:{of}\:{asymptotes}: \\ $$$$\:\:\:\:\mathrm{tan}\:\theta=\frac{{y}}{{x}}=\pm\frac{{b}}{{a}} \\ $$$${let}\:\:{eq}.\:{of}\:{line}\:{PQR}\:: \\ $$$$\:\:{y}={m}\left({x}−{h}\right)+{k} \\ $$$$\:\:\:\:\:{h}={a}\mathrm{sec}\:\phi\:\:,\:{k}={b}\mathrm{tan}\:\phi \\ $$$$\:\:\:\:\:{m}=\:\mathrm{tan}\:\alpha \\ $$$${for}\:\:{x}_{{Q}} \:: \\ $$$$\:\:\frac{{bx}_{{Q}} }{{a}}\:=\:{m}\left({x}_{{Q}} −{h}\right)+{k} \\ $$$$\Rightarrow\:\:{x}_{{Q}} \:=\:\frac{{k}−{mh}}{\frac{{b}}{{a}}−{m}} \\ $$$${r}_{\mathrm{1}} =\left({h}−{x}_{{Q}} \right)\mathrm{sec}\:\alpha \\ $$$$\:\:\:\:\:=\:\left(\frac{\frac{{bh}}{{a}}−{k}}{\frac{{b}}{{a}}−{m}}\right)\mathrm{sec}\:\alpha \\ $$$${similarly} \\ $$$${r}_{\mathrm{2}} =\left({h}−{x}_{{R}} \right)\mathrm{sec}\:\alpha \\ $$$$\:\:\:\:=\:\left(\frac{\frac{{bh}}{{a}}+{k}}{\frac{{b}}{{a}}+{m}}\right)\mathrm{sec}\:\alpha \\ $$$${r}_{\mathrm{1}} {r}_{\mathrm{2}} =\left[\frac{\left(\frac{{bh}}{{a}}\right)^{\mathrm{2}} −{k}^{\mathrm{2}} }{\left(\frac{{b}}{{a}}\right)^{\mathrm{2}} −{m}^{\mathrm{2}} }\right]\mathrm{sec}\:^{\mathrm{2}} \alpha \\ $$$$\:\:\:\:\:=\left[\frac{\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}{\frac{\mathrm{1}}{{a}^{\mathrm{2}} }−\frac{{m}^{\mathrm{2}} }{{b}^{\mathrm{2}} }}\right]\left(\mathrm{1}+{m}^{\mathrm{2}} \right) \\ $$$${but}\:{P}\:\left({h},{k}\right)\:{lies}\:{on}\:{hyperbola} \\ $$$${so}\:\:\:\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\:=\:\mathrm{1}\:,\:{hence} \\ $$$$\Rightarrow\:\:{r}_{\mathrm{1}} {r}_{\mathrm{2}} \:=\:\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)}{{b}^{\mathrm{2}} −{a}^{\mathrm{2}} {m}^{\mathrm{2}} }\:. \\ $$
Commented by peter frank last updated on 04/Nov/18
thank you sir
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$
Answered by MrW3 last updated on 04/Nov/18
let point P be (h,k) (h^2 /a^2 )−(k^2 /b^2 )=1 line through point P with slope m: y−k=m(x−h) eqn. of asymptote 1: y=−(b/a)x y_q =−(b/a)x_q y_q −k=m(x_q −h) ⇒k=−(b/a)x_q −m(x_q −h) ⇒((b/a)+m)x_q =mh−k ⇒x_q =((a(mh−k))/(ma+b)) ⇒y_q =−((b(mh−k))/(ma+b)) eqn. of asymptote 2: y=(b/a)x y_r =(b/a)x_r y_r −k=m(x_r −h) ⇒k=(b/a)x_r −m(x_r −h) ⇒((b/a)−m)x_r =−(mh−k) ⇒x_r =((a(mh−k))/(am−b)) ⇒y_r =((b(mh−k))/(am−b)) QP=(√((x_q −h)^2 +(y_q −k)^2 ))=(√((((a(mh−k))/(ma+b))−h)^2 +(−((b(mh−k))/(ma+b))−k)^2 )) ⇒QP=∣(((bh+ak)(√((1+m^2 ))))/(ma+b))∣ RP=(√((x_r −h)^2 +(y_r −k)^2 ))=(√((((a(mh−k))/(am−b))−h)^2 +(((b(mh−k))/(am−b))−k)^2 )) ⇒RP=∣(((bh−ak)(√((1+m^2 ))))/(ma−b))∣ QP×RP=∣(((bh+ak)(√((1+m^2 ))))/(ma+b))×(((bh−ak)(√((1+m^2 ))))/(ma−b))∣ =∣(((b^2 h^2 −a^2 k^2 )(1+m^2 ))/(m^2 a^2 −b^2 ))∣ =∣((((h^2 /a^2 )−(k^2 /b^2 ))a^2 b^2 (1+m^2 ))/(m^2 a^2 −b^2 ))∣ =((a^2 b^2 (1+m^2 ))/(∣m^2 a^2 −b^2 ∣))
$${let}\:{point}\:{P}\:{be}\:\left({h},{k}\right) \\ $$$$\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${line}\:{through}\:{point}\:{P}\:{with}\:{slope}\:{m}: \\ $$$${y}−{k}={m}\left({x}−{h}\right) \\ $$$$ \\ $$$${eqn}.\:{of}\:{asymptote}\:\mathrm{1}: \\ $$$${y}=−\frac{{b}}{{a}}{x} \\ $$$${y}_{{q}} =−\frac{{b}}{{a}}{x}_{{q}} \\ $$$${y}_{{q}} −{k}={m}\left({x}_{{q}} −{h}\right) \\ $$$$\Rightarrow{k}=−\frac{{b}}{{a}}{x}_{{q}} −{m}\left({x}_{{q}} −{h}\right) \\ $$$$\Rightarrow\left(\frac{{b}}{{a}}+{m}\right){x}_{{q}} ={mh}−{k} \\ $$$$\Rightarrow{x}_{{q}} =\frac{{a}\left({mh}−{k}\right)}{{ma}+{b}} \\ $$$$\Rightarrow{y}_{{q}} =−\frac{{b}\left({mh}−{k}\right)}{{ma}+{b}} \\ $$$$ \\ $$$${eqn}.\:{of}\:{asymptote}\:\mathrm{2}: \\ $$$${y}=\frac{{b}}{{a}}{x} \\ $$$${y}_{{r}} =\frac{{b}}{{a}}{x}_{{r}} \\ $$$${y}_{{r}} −{k}={m}\left({x}_{{r}} −{h}\right) \\ $$$$\Rightarrow{k}=\frac{{b}}{{a}}{x}_{{r}} −{m}\left({x}_{{r}} −{h}\right) \\ $$$$\Rightarrow\left(\frac{{b}}{{a}}−{m}\right){x}_{{r}} =−\left({mh}−{k}\right) \\ $$$$\Rightarrow{x}_{{r}} =\frac{{a}\left({mh}−{k}\right)}{{am}−{b}} \\ $$$$\Rightarrow{y}_{{r}} =\frac{{b}\left({mh}−{k}\right)}{{am}−{b}} \\ $$$${QP}=\sqrt{\left({x}_{{q}} −{h}\right)^{\mathrm{2}} +\left({y}_{{q}} −{k}\right)^{\mathrm{2}} }=\sqrt{\left(\frac{{a}\left({mh}−{k}\right)}{{ma}+{b}}−{h}\right)^{\mathrm{2}} +\left(−\frac{{b}\left({mh}−{k}\right)}{{ma}+{b}}−{k}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{QP}=\mid\frac{\left({bh}+{ak}\right)\sqrt{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)}}{{ma}+{b}}\mid \\ $$$${RP}=\sqrt{\left({x}_{{r}} −{h}\right)^{\mathrm{2}} +\left({y}_{{r}} −{k}\right)^{\mathrm{2}} }=\sqrt{\left(\frac{{a}\left({mh}−{k}\right)}{{am}−{b}}−{h}\right)^{\mathrm{2}} +\left(\frac{{b}\left({mh}−{k}\right)}{{am}−{b}}−{k}\right)^{\mathrm{2}} } \\ $$$$\Rightarrow{RP}=\mid\frac{\left({bh}−{ak}\right)\sqrt{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)}}{{ma}−{b}}\mid \\ $$$${QP}×{RP}=\mid\frac{\left({bh}+{ak}\right)\sqrt{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)}}{{ma}+{b}}×\frac{\left({bh}−{ak}\right)\sqrt{\left(\mathrm{1}+{m}^{\mathrm{2}} \right)}}{{ma}−{b}}\mid \\ $$$$=\mid\frac{\left({b}^{\mathrm{2}} {h}^{\mathrm{2}} −{a}^{\mathrm{2}} {k}^{\mathrm{2}} \right)\left(\mathrm{1}+{m}^{\mathrm{2}} \right)}{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\mid \\ $$$$=\mid\frac{\left(\frac{{h}^{\mathrm{2}} }{{a}^{\mathrm{2}} }−\frac{{k}^{\mathrm{2}} }{{b}^{\mathrm{2}} }\right){a}^{\mathrm{2}} {b}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)}{{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} }\mid \\ $$$$=\frac{{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left(\mathrm{1}+{m}^{\mathrm{2}} \right)}{\mid{m}^{\mathrm{2}} {a}^{\mathrm{2}} −{b}^{\mathrm{2}} \mid} \\ $$
Commented by peter frank last updated on 04/Nov/18
mrW3.thank you
$$\mathrm{mrW3}.\mathrm{thank}\:\mathrm{you} \\ $$

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