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Question Number 157926 by mr W last updated on 29/Oct/21
if the line px+qy=r tangents the ellipse (x^2 /a^2 )+(y^2 /b^2 )=1, then 1) prove a^2 p^2 +b^2 q^2 =r^2 2) find the coordinates of the touching point.
$${if}\:{the}\:{line}\:{px}+{qy}={r}\:{tangents}\:{the} \\ $$$${ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1},\:{then}\: \\ $$$$\left.\mathrm{1}\right)\:{prove}\:\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{q}}^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \: \\ $$$$\left.\mathrm{2}\right)\:{find}\:{the}\:{coordinates}\:{of}\:{the}\: \\ $$$$\:\:\:\:\:{touching}\:{point}. \\ $$
Answered by mindispower last updated on 29/Oct/21
.... paramatric of elips{(acos(t),bsin(t))),t∈[0,2π[ M′(t)=(−asin(t),bcos(t)) M(t)∈ Line pacos(t)+qbsin(t)=r...(1) M′(t) director vector and (−q,p) line director⇒ −qbcos(t)+pasin(t)=0... (1)^2 +(2)^2 ⇔p^2 a^2 cos^2 (t)+q^2 b^2 sin^2 (t)+2paqbsin(t)cos(t) +q^2 b^2 cos^2 (t)+p^2 a^2 sin^2 (t)−2abpqsin(t)cos(t)=r^2 +0 ⇔a^2 p^2 +b^2 q^2 =r^2 { ((−qbcos(t)+pasin(t)=0....∗sin(t)..1)),((pacos(t)+qbsin(t)=r......∗cos(t)....2)) :} (1) sin(t)cos(t)≠0 ⇒(1)+(2) pa=rcos(t)⇒cos(t)=((pa)/r) M=(((a^2 p)/r),((b^2 q)/r)) sin(t)=((qb)/r) sin(t)=0⇒−qbcos(t)=0,⇒q=0 pacos(t)=r⇒cos(t)=(r/(pa))⇒∣(r/(pa))∣=1 in this case the line is x=(r/p), vertical lign M((r/p),0) cos(t)=0⇒p=0 sin(t)=(r/(qb))......The lign is y=(r/q)=^ +_− 1 vertical line M(0,(r/q))
$$….\:{paramatric}\:{of}\:{elips}\left\{\left({acos}\left({t}\right),{bsin}\left({t}\right)\right)\right),{t}\in\left[\mathrm{0},\mathrm{2}\pi\left[\right.\right. \\ $$$${M}'\left({t}\right)=\left(−{asin}\left({t}\right),{bcos}\left({t}\right)\right) \\ $$$${M}\left({t}\right)\in\:{Line}\:{pacos}\left({t}\right)+{qbsin}\left({t}\right)={r}…\left(\mathrm{1}\right) \\ $$$${M}'\left({t}\right)\:{director}\:{vector}\:{and}\:\left(−{q},{p}\right)\:{line}\:{director}\Rightarrow \\ $$$$−{qbcos}\left({t}\right)+{pasin}\left({t}\right)=\mathrm{0}… \\ $$$$\left(\mathrm{1}\right)^{\mathrm{2}} +\left(\mathrm{2}\right)^{\mathrm{2}} \:\Leftrightarrow{p}^{\mathrm{2}} {a}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({t}\right)+{q}^{\mathrm{2}} {b}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({t}\right)+\mathrm{2}{paqbsin}\left({t}\right){cos}\left({t}\right) \\ $$$$+{q}^{\mathrm{2}} {b}^{\mathrm{2}} {cos}^{\mathrm{2}} \left({t}\right)+{p}^{\mathrm{2}} {a}^{\mathrm{2}} {sin}^{\mathrm{2}} \left({t}\right)−\mathrm{2}{abpqsin}\left({t}\right){cos}\left({t}\right)={r}^{\mathrm{2}} +\mathrm{0} \\ $$$$\Leftrightarrow{a}^{\mathrm{2}} {p}^{\mathrm{2}} +{b}^{\mathrm{2}} {q}^{\mathrm{2}} ={r}^{\mathrm{2}} \\ $$$$\begin{cases}{−{qbcos}\left({t}\right)+{pasin}\left({t}\right)=\mathrm{0}….\ast{sin}\left({t}\right)..\mathrm{1}}\\{{pacos}\left({t}\right)+{qbsin}\left({t}\right)={r}……\ast{cos}\left({t}\right)….\mathrm{2}}\end{cases} \\ $$$$\left(\mathrm{1}\right)\:{sin}\left({t}\right){cos}\left({t}\right)\neq\mathrm{0} \\ $$$$\Rightarrow\left(\mathrm{1}\right)+\left(\mathrm{2}\right)\:{pa}={rcos}\left({t}\right)\Rightarrow{cos}\left({t}\right)=\frac{{pa}}{{r}} \\ $$$${M}=\left(\frac{{a}^{\mathrm{2}} {p}}{{r}},\frac{{b}^{\mathrm{2}} {q}}{{r}}\right) \\ $$$${sin}\left({t}\right)=\frac{{qb}}{{r}} \\ $$$${sin}\left({t}\right)=\mathrm{0}\Rightarrow−{qbcos}\left({t}\right)=\mathrm{0},\Rightarrow{q}=\mathrm{0} \\ $$$${pacos}\left({t}\right)={r}\Rightarrow{cos}\left({t}\right)=\frac{{r}}{{pa}}\Rightarrow\mid\frac{{r}}{{pa}}\mid=\mathrm{1} \\ $$$${in}\:{this}\:{case}\:{the}\:{line}\:{is}\:{x}=\frac{{r}}{{p}},\:{vertical}\:{lign} \\ $$$${M}\left(\frac{{r}}{{p}},\mathrm{0}\right) \\ $$$${cos}\left({t}\right)=\mathrm{0}\Rightarrow{p}=\mathrm{0} \\ $$$${sin}\left({t}\right)=\frac{{r}}{{qb}}……{The}\:{lign}\:{is}\:{y}=\frac{{r}}{{q}}\overset{} {=}\underset{−} {+}\mathrm{1}\:{vertical}\:{line} \\ $$$${M}\left(\mathrm{0},\frac{{r}}{{q}}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by mr W last updated on 30/Oct/21
thanks alot!
$${thanks}\:{alot}! \\ $$
Commented by mindispower last updated on 02/Nov/21
pleasur sir havea good day
$${pleasur}\:{sir}\:{havea}\:{good}\:{day} \\ $$
Answered by som(math1967) last updated on 30/Oct/21
px+qy=r ⇒y=((r−px)/q) (x^2 /a^2 ) +(y^2 /b^2 )=1 or. (x^2 /a^2 ) +(((r−px)^2 )/(b^2 q^2 ))=1 or. x^2 b^2 q^2 +a^2 p^2 x^2 −2a^2 pxr+r^2 a^2 =a^2 b^2 q^2 or.(b^2 q^2 +a^2 p^2 )x^2 −2a^2 pxr+a^2 r^2 −a^2 b^2 q^2 =0 ...1) px+qy=r is tangents of (x^2 /a^2 ) +(y^2 /b^2 )=1 ∴ (−2a^2 pr)^2 −4(b^2 q^2 +a^2 p^2 )(a^2 r^2 −a^2 b^2 q^2 )=0 [ for equal roots discriminant is 0] 4a^2 [a^2 p^2 r^2 −(b^2 q^2 +a^2 p^2 )(r^2 −b^2 q^2 )]=0 a^2 p^2 r^2 −b^2 q^2 r^2 +b^4 q^4 −a^2 p^2 r^2 +a^2 b^2 p^2 q^2 =0 b^2 q^2 (b^2 q^2 +a^2 p^2 −r^2 )=0 ∴ a^2 p^2 +b^2 q^2 =r^2 from equn. 1 (b^2 q^2 +a^2 p^2 )x^2 −2a^2 pxr+a^2 r^2 −a^2 b^2 q^2 =0 [(a^2 p^2 +b^2 q^2 =r^2 ] r^2 x^2 −2a^2 pxr+a^2 (r^2 −b^2 q^2 )=0 r^2 x^2 −2.rx.a^2 p+a^4 .p^2 =0 x=((a^2 p)/r) y=((r−px)/q)=((r−((a^2 p^2 )/r))/q)=((r^2 −a^2 p^2 )/(qr)) (((a^2 p)/r),((r^2 −a^2 p^2 )/(qr))) (((a^2 p)/r),((b^2 q)/r))
$${px}+{qy}={r}\:\:\:\Rightarrow{y}=\frac{{r}−{px}}{{q}} \\ $$$$\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${or}.\:\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\frac{\left({r}−{px}\right)^{\mathrm{2}} }{{b}^{\mathrm{2}} {q}^{\mathrm{2}} }=\mathrm{1} \\ $$$${or}.\:{x}^{\mathrm{2}} {b}^{\mathrm{2}} {q}^{\mathrm{2}} +{a}^{\mathrm{2}} {p}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {pxr}+{r}^{\mathrm{2}} {a}^{\mathrm{2}} ={a}^{\mathrm{2}} {b}^{\mathrm{2}} {q}^{\mathrm{2}} \\ $$$$\left.{or}.\left({b}^{\mathrm{2}} {q}^{\mathrm{2}} +{a}^{\mathrm{2}} {p}^{\mathrm{2}} \right){x}^{\mathrm{2}} \:−\mathrm{2}{a}^{\mathrm{2}} {pxr}+{a}^{\mathrm{2}} {r}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} {q}^{\mathrm{2}} =\mathrm{0}\:…\mathrm{1}\right) \\ $$$${px}+{qy}={r}\:{is}\:{tangents}\:{of}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$$\therefore\:\left(−\mathrm{2}{a}^{\mathrm{2}} {pr}\right)^{\mathrm{2}} −\mathrm{4}\left({b}^{\mathrm{2}} {q}^{\mathrm{2}} +{a}^{\mathrm{2}} {p}^{\mathrm{2}} \right)\left({a}^{\mathrm{2}} {r}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} {q}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\left[\:{for}\:{equal}\:{roots}\:{discriminant}\:{is}\:\mathrm{0}\right] \\ $$$$\mathrm{4}{a}^{\mathrm{2}} \left[{a}^{\mathrm{2}} {p}^{\mathrm{2}} {r}^{\mathrm{2}} −\left({b}^{\mathrm{2}} {q}^{\mathrm{2}} +{a}^{\mathrm{2}} {p}^{\mathrm{2}} \right)\left({r}^{\mathrm{2}} −{b}^{\mathrm{2}} {q}^{\mathrm{2}} \right)\right]=\mathrm{0} \\ $$$${a}^{\mathrm{2}} {p}^{\mathrm{2}} {r}^{\mathrm{2}} −{b}^{\mathrm{2}} {q}^{\mathrm{2}} {r}^{\mathrm{2}} \:+{b}^{\mathrm{4}} {q}^{\mathrm{4}} −{a}^{\mathrm{2}} {p}^{\mathrm{2}} {r}^{\mathrm{2}} +{a}^{\mathrm{2}} {b}^{\mathrm{2}} {p}^{\mathrm{2}} {q}^{\mathrm{2}} =\mathrm{0} \\ $$$${b}^{\mathrm{2}} {q}^{\mathrm{2}} \left({b}^{\mathrm{2}} {q}^{\mathrm{2}} +{a}^{\mathrm{2}} {p}^{\mathrm{2}} −{r}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$$\therefore\:\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{q}}^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \\ $$$${from}\:{equn}.\:\mathrm{1} \\ $$$$\left({b}^{\mathrm{2}} {q}^{\mathrm{2}} +{a}^{\mathrm{2}} {p}^{\mathrm{2}} \right){x}^{\mathrm{2}} \:−\mathrm{2}{a}^{\mathrm{2}} {pxr}+{a}^{\mathrm{2}} {r}^{\mathrm{2}} −{a}^{\mathrm{2}} {b}^{\mathrm{2}} {q}^{\mathrm{2}} =\mathrm{0} \\ $$$$\left[\left(\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{p}}^{\mathrm{2}} +\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{q}}^{\mathrm{2}} =\boldsymbol{{r}}^{\mathrm{2}} \right]\right. \\ $$$${r}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}{a}^{\mathrm{2}} {pxr}+{a}^{\mathrm{2}} \left({r}^{\mathrm{2}} −{b}^{\mathrm{2}} {q}^{\mathrm{2}} \right)=\mathrm{0} \\ $$$${r}^{\mathrm{2}} {x}^{\mathrm{2}} −\mathrm{2}.{rx}.{a}^{\mathrm{2}} {p}+{a}^{\mathrm{4}} .{p}^{\mathrm{2}} =\mathrm{0} \\ $$$${x}=\frac{{a}^{\mathrm{2}} {p}}{{r}} \\ $$$${y}=\frac{{r}−{px}}{{q}}=\frac{{r}−\frac{{a}^{\mathrm{2}} {p}^{\mathrm{2}} }{{r}}}{{q}}=\frac{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} {p}^{\mathrm{2}} }{{qr}} \\ $$$$\left(\frac{{a}^{\mathrm{2}} {p}}{{r}},\frac{{r}^{\mathrm{2}} −{a}^{\mathrm{2}} {p}^{\mathrm{2}} }{{qr}}\right) \\ $$$$\left(\frac{\boldsymbol{{a}}^{\mathrm{2}} \boldsymbol{{p}}}{\boldsymbol{{r}}},\frac{\boldsymbol{{b}}^{\mathrm{2}} \boldsymbol{{q}}}{\boldsymbol{{r}}}\right) \\ $$$$ \\ $$
Commented by mr W last updated on 30/Oct/21
thanks alot!
$${thanks}\:{alot}! \\ $$

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