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Question Number 28241 by ajfour last updated on 22/Jan/18
Commented by ajfour last updated on 22/Jan/18
Q. 28185 (alternate solution )
$${Q}.\:\mathrm{28185}\:\:\:\left({alternate}\:{solution}\:\right) \\ $$
Answered by ajfour last updated on 22/Jan/18
x^2 +y^2 −6x+6y−16=0 ⇒ (x−3)^2 +(y+3)^2 =34 slope of MN is m=4 x_M =3+(√(34))×(1/( (√(17)))) =3+(√2) y_M =−3+(√(34))×(4/( (√(17)))) =−3+4(√2) For point A we solve y=x with eq. of AM and so x_A =y_A =α is α−4(√2)+3=−(α−3−(√2)) ⇒ α=(5/( (√2))) ; hence A((5/( (√2))),(5/( (√2)))) x_N =3−(√(34))×(1/( (√(17))))=3−(√2) y_N =−3−(√(34))×(4/( (√(34))))=−3−4(√2) x_B =y_B =β which we get by solving y=x with eq. of BN: β+3+4(√2)=−(β−3+(√2)) ⇒ β=−(5/( (√2))) ; hence B(−(5/( (√2))), (5/( (√2)))) center of ellipse is midpoint of segment AB , so (0,0) . and semi major axis length =5 let us take along AB(y=x) our x axis and y=−x as new y axis (both downwards +ve) then eq. of ellipse is (x^2 /a^2 )+(y^2 /b^2 )=1 , with a=5 slope of y=4x−15 will become m=((4−tan π/4)/(1+4tan π/4)) =(3/5) point (5,5) will become (−5(√2),0) thus eq. y=4x−15 changes to y=(3/5)x+5(√2) as this is to be tangent to ellipse c^2 =a^2 m^2 +b^2 ⇒ (5(√2))^2 =(25)((3/5))^2 +b^2 b^2 =9 e^2 =1−(b^2 /a^2 ) =1−(9/(25)) e=(4/5) length of latus rectum l=((2b^2 )/a) =((2×9)/5) =((18)/5) with (a/e)=(5/(4/5))=((25)/4) eq. of directrices pass through (((25)/(4(√2))), ((25)/(4(√2)))) and (((−25)/(4(√2))), ((−25)/(4(√2)))) and are perpendicular to y=x So, eqs. are y−((25)/(4(√2)))=−(x−((25)/(4(√2)))) ⇒ x+y=±((25)/(2(√2))) .
$${x}^{\mathrm{2}} +{y}^{\mathrm{2}} −\mathrm{6}{x}+\mathrm{6}{y}−\mathrm{16}=\mathrm{0} \\ $$$$\Rightarrow\:\:\:\left({x}−\mathrm{3}\right)^{\mathrm{2}} +\left({y}+\mathrm{3}\right)^{\mathrm{2}} =\mathrm{34} \\ $$$${slope}\:{of}\:{MN}\:{is}\:{m}=\mathrm{4} \\ $$$${x}_{{M}} =\mathrm{3}+\sqrt{\mathrm{34}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{17}}}\:=\mathrm{3}+\sqrt{\mathrm{2}} \\ $$$${y}_{{M}} =−\mathrm{3}+\sqrt{\mathrm{34}}×\frac{\mathrm{4}}{\:\sqrt{\mathrm{17}}}\:=−\mathrm{3}+\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${For}\:{point}\:{A}\:{we}\:{solve}\:{y}={x}\:{with} \\ $$$${eq}.\:{of}\:{AM}\:\:{and}\:{so}\:{x}_{{A}} ={y}_{{A}} =\alpha\:{is} \\ $$$$\alpha−\mathrm{4}\sqrt{\mathrm{2}}+\mathrm{3}=−\left(\alpha−\mathrm{3}−\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\alpha=\frac{\mathrm{5}}{\:\sqrt{\mathrm{2}}}\:\:;\:{hence}\:{A}\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{2}}},\frac{\mathrm{5}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${x}_{{N}} =\mathrm{3}−\sqrt{\mathrm{34}}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{17}}}=\mathrm{3}−\sqrt{\mathrm{2}} \\ $$$${y}_{{N}} =−\mathrm{3}−\sqrt{\mathrm{34}}×\frac{\mathrm{4}}{\:\sqrt{\mathrm{34}}}=−\mathrm{3}−\mathrm{4}\sqrt{\mathrm{2}} \\ $$$${x}_{{B}} ={y}_{{B}} =\beta\:{which}\:{we}\:{get}\:{by} \\ $$$${solving}\:{y}={x}\:\:{with}\:{eq}.\:{of}\:{BN}: \\ $$$$\:\:\beta+\mathrm{3}+\mathrm{4}\sqrt{\mathrm{2}}=−\left(\beta−\mathrm{3}+\sqrt{\mathrm{2}}\right) \\ $$$$\Rightarrow\:\:\beta=−\frac{\mathrm{5}}{\:\sqrt{\mathrm{2}}}\:;\:\:{hence}\:{B}\left(−\frac{\mathrm{5}}{\:\sqrt{\mathrm{2}}},\:\frac{\mathrm{5}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${center}\:{of}\:{ellipse}\:{is}\:{midpoint}\:{of} \\ $$$${segment}\:{AB}\:,\:{so}\:\:\left(\mathrm{0},\mathrm{0}\right)\:. \\ $$$${and}\:{semi}\:{major}\:{axis}\:{length}\:=\mathrm{5} \\ $$$${let}\:{us}\:{take}\:{along}\:{AB}\left({y}={x}\right)\:{our}\:{x}\:{axis} \\ $$$${and}\:{y}=−{x}\:{as}\:{new}\:{y}\:{axis} \\ $$$$\:\left({both}\:{downwards}\:+{ve}\right) \\ $$$${then}\:{eq}.\:{of}\:{ellipse}\:{is} \\ $$$$\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1}\:\:,\:{with}\:{a}=\mathrm{5} \\ $$$${slope}\:{of}\:\:{y}=\mathrm{4}{x}−\mathrm{15}\:{will}\:{become} \\ $$$${m}=\frac{\mathrm{4}−\mathrm{tan}\:\pi/\mathrm{4}}{\mathrm{1}+\mathrm{4tan}\:\pi/\mathrm{4}}\:=\frac{\mathrm{3}}{\mathrm{5}} \\ $$$${point}\:\left(\mathrm{5},\mathrm{5}\right)\:{will}\:{become}\:\left(−\mathrm{5}\sqrt{\mathrm{2}},\mathrm{0}\right) \\ $$$${thus}\:{eq}.\:{y}=\mathrm{4}{x}−\mathrm{15}\:{changes}\:{to} \\ $$$${y}=\frac{\mathrm{3}}{\mathrm{5}}{x}+\mathrm{5}\sqrt{\mathrm{2}} \\ $$$${as}\:{this}\:{is}\:{to}\:{be}\:{tangent}\:{to}\:{ellipse} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\left(\mathrm{5}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\left(\mathrm{25}\right)\left(\frac{\mathrm{3}}{\mathrm{5}}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$${b}^{\mathrm{2}} =\mathrm{9} \\ $$$${e}^{\mathrm{2}} =\mathrm{1}−\frac{{b}^{\mathrm{2}} }{{a}^{\mathrm{2}} }\:=\mathrm{1}−\frac{\mathrm{9}}{\mathrm{25}}\: \\ $$$$\boldsymbol{{e}}=\frac{\mathrm{4}}{\mathrm{5}} \\ $$$${length}\:{of}\:{latus}\:{rectum} \\ $$$$\:\:\:\boldsymbol{{l}}=\frac{\mathrm{2}\boldsymbol{{b}}^{\mathrm{2}} }{\boldsymbol{{a}}}\:=\frac{\mathrm{2}×\mathrm{9}}{\mathrm{5}}\:=\frac{\mathrm{18}}{\mathrm{5}} \\ $$$$\boldsymbol{{with}}\:\frac{\boldsymbol{{a}}}{\boldsymbol{{e}}}=\frac{\mathrm{5}}{\mathrm{4}/\mathrm{5}}=\frac{\mathrm{25}}{\mathrm{4}} \\ $$$$\boldsymbol{{eq}}.\:\boldsymbol{{of}}\:\boldsymbol{{directrices}}\:\boldsymbol{{pass}}\:\boldsymbol{{through}} \\ $$$$\left(\frac{\mathrm{25}}{\mathrm{4}\sqrt{\mathrm{2}}},\:\frac{\mathrm{25}}{\mathrm{4}\sqrt{\mathrm{2}}}\right)\:{and}\:\left(\frac{−\mathrm{25}}{\mathrm{4}\sqrt{\mathrm{2}}},\:\frac{−\mathrm{25}}{\mathrm{4}\sqrt{\mathrm{2}}}\right) \\ $$$${and}\:{are}\:{perpendicular}\:{to}\:\boldsymbol{{y}}=\boldsymbol{{x}} \\ $$$$\boldsymbol{{S}}{o},\:{eqs}.\:{are} \\ $$$${y}−\frac{\mathrm{25}}{\mathrm{4}\sqrt{\mathrm{2}}}=−\left({x}−\frac{\mathrm{25}}{\mathrm{4}\sqrt{\mathrm{2}}}\right) \\ $$$$\Rightarrow\:\:\:\boldsymbol{{x}}+\boldsymbol{{y}}=\pm\frac{\mathrm{25}}{\mathrm{2}\sqrt{\mathrm{2}}}\:. \\ $$
Commented by ajfour last updated on 22/Jan/18
MN is tangent to this ellipse and this is given in question, we cannot amend that. from this condition we will rather find b or hence the eccenticity.
$${MN}\:{is}\:{tangent}\:{to}\:{this}\:{ellipse} \\ $$$${and}\:{this}\:{is}\:{given}\:{in}\:{question},\:{we} \\ $$$${cannot}\:{amend}\:{that}. \\ $$$${from}\:{this}\:{condition}\:{we}\:{will} \\ $$$${rather}\:{find}\:\boldsymbol{{b}}\:{or}\:{hence}\:{the} \\ $$$${eccenticity}. \\ $$
Commented by mrW2 last updated on 22/Jan/18
I assumed that the tangents from M and N should at ends of both axes of the ellipse. This is much more complicated than what is asked. I had to read the question more carefully.
$${I}\:{assumed}\:{that}\:{the}\:{tangents}\:{from}\:{M} \\ $$$${and}\:{N}\:{should}\:{at}\:{ends}\:{of}\:{both}\:{axes}\:{of} \\ $$$${the}\:{ellipse}.\:{This}\:{is}\:{much}\:{more} \\ $$$${complicated}\:{than}\:{what}\:{is}\:{asked}. \\ $$$${I}\:{had}\:{to}\:{read}\:{the}\:{question}\:{more} \\ $$$${carefully}. \\ $$
Commented by ajfour last updated on 22/Jan/18
must be , let me see if i can follow along that interpretation..
$${must}\:{be}\:,\:{let}\:{me}\:{see}\:{if}\:{i}\:{can} \\ $$$${follow}\:{along}\:{that}\:{interpretation}.. \\ $$
Commented by mrW2 last updated on 23/Jan/18
Is there an easy way to get c^2 =a^2 m^2 +b^2 ? I know this is true. But I can derive it only through a lengthy way: (...)x^2 +(...)x+(...)=0 Δ=(...)^2 −4(...)(...)=0 ⇒c^2 =a^2 m^2 +b^2
$${Is}\:{there}\:{an}\:{easy}\:{way}\:{to}\:{get} \\ $$$${c}^{\mathrm{2}} ={a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \:? \\ $$$${I}\:{know}\:{this}\:{is}\:{true}.\:{But}\:{I}\:{can}\:{derive} \\ $$$${it}\:{only}\:{through}\:{a}\:{lengthy}\:{way}: \\ $$$$\left(…\right){x}^{\mathrm{2}} +\left(…\right){x}+\left(…\right)=\mathrm{0} \\ $$$$\Delta=\left(…\right)^{\mathrm{2}} −\mathrm{4}\left(…\right)\left(…\right)=\mathrm{0} \\ $$$$\Rightarrow{c}^{\mathrm{2}} ={a}^{\mathrm{2}} {m}^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$
Commented by mrW2 last updated on 23/Jan/18
I got following: rotated with angle −(π/4) eq. y=4x−15 changes to ((x+y)/( (√2)))=((4(x−y))/( (√2)))−15 ⇒5y=3x−15(√2) ⇒y=(3/5)x−3(√2) (but you got y=(3/5)x+5(√2)) (−3(√2))^2 =((3/5)×5)^2 +b^2 ⇒b^2 =9 (you got also b^2 =9)
$${I}\:{got}\:{following}: \\ $$$${rotated}\:{with}\:{angle}\:−\frac{\pi}{\mathrm{4}}\:{eq}.\:{y}=\mathrm{4}{x}−\mathrm{15}\:{changes}\:{to} \\ $$$$\frac{{x}+{y}}{\:\sqrt{\mathrm{2}}}=\frac{\mathrm{4}\left({x}−{y}\right)}{\:\sqrt{\mathrm{2}}}−\mathrm{15} \\ $$$$\Rightarrow\mathrm{5}{y}=\mathrm{3}{x}−\mathrm{15}\sqrt{\mathrm{2}} \\ $$$$\Rightarrow{y}=\frac{\mathrm{3}}{\mathrm{5}}{x}−\mathrm{3}\sqrt{\mathrm{2}}\:\:\:\left({but}\:{you}\:{got}\:{y}=\frac{\mathrm{3}}{\mathrm{5}}{x}+\mathrm{5}\sqrt{\mathrm{2}}\right) \\ $$$$\left(−\mathrm{3}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} =\left(\frac{\mathrm{3}}{\mathrm{5}}×\mathrm{5}\right)^{\mathrm{2}} +{b}^{\mathrm{2}} \\ $$$$\Rightarrow{b}^{\mathrm{2}} =\mathrm{9}\:\left({you}\:{got}\:{also}\:{b}^{\mathrm{2}} =\mathrm{9}\right) \\ $$

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