Question Number 70503 by peter frank last updated on 04/Oct/19
Commented by Prithwish sen last updated on 06/Oct/19
![Let O=(x_1 ,y_1 ) ∵ It is on the circle ∴x_1 ^2 +y_1 ^2 = d^2 .....(i) Now the equation of the chord of contact PQ ≡ ((xx_1 )/a^2 ) + ((yy_1 )/b^2 ) = 1 ....(ii) Let the midpoint of P(x_2 ,y_2 ) and Q(x_3 ,y_3 ) be (h,k) i.e x_2 +x_3 = 2h.....(iii) and y_2 +y_3 = 2k .....(iv) as (ii) intersects the ellipse (x^2 /a^2 ) +(y^2 /b^2 ) = 1...(v) putting (y/b) = (b/y_1 )(1−((xx_1 )/a^2 )) from (ii) to (v) we get, (x^2 /a^2 ) +(b^2 /y_1 ^2 )(1−((xx_1 )/a^2 ))^2 = 1 ⇒ x^2 {(1/a^2 )+((b^2 x_1 ^2 )/(a^4 y_1 ^2 ))}−x.2(x_1 /y_1 ^2 ).(b^2 /a^2 ) +((b^2 /y_1 ^2 ) −1)=0 ∴ x_2 +x_3 = ((2x_1 a^2 b^2 )/((a^2 y_1 ^2 +b^2 x_1 ^2 ))) h = ((x_1 a^2 b^2 )/((a^2 y_1 ^2 +b^2 x_1 ^2 ))) similarly we can get k = ((y_1 a^2 b^2 )/((a^2 y_1 ^2 +b^2 x_1 ^2 ))) now h^2 +k^2 = ((a^4 b^4 )/((a^2 y_1 ^2 +b^2 x_1 ^2 )^2 )){x_1 ^2 +y_1 ^2 } h^2 + k^2 = d^2 .(1/(((x_1 ^2 /a^2 )+(y_1 ^2 /b^2 ))^2 )) ∵ from (i) ∴ the locus of the mipoint is x^2 +y^2 = d^2 [(x_1 ^2 /a^2 ) + (y_1 ^2 /b^2 )]^(−2) I get this . Is the given question is correct ? please check.](Q70667.png)
$$\mathrm{Let}\:\mathrm{O}=\left(\mathrm{x}_{\mathrm{1}} ,\mathrm{y}_{\mathrm{1}} \right) \\ $$$$\because\:\mathrm{It}\:\mathrm{is}\:\mathrm{on}\:\mathrm{the}\:\mathrm{circle}\: \\ $$$$\therefore\boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\mathrm{y}}_{\mathrm{1}} ^{\mathrm{2}} =\:\boldsymbol{\mathrm{d}}^{\mathrm{2}} \:.....\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\mathrm{Now}\:\mathrm{the}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{the}\:\boldsymbol{\mathrm{chord}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{contact}}\: \\ $$$$\boldsymbol{\mathrm{PQ}}\:\equiv\:\frac{\boldsymbol{\mathrm{xx}}_{\mathrm{1}} }{\boldsymbol{\mathrm{a}}^{\mathrm{2}} }\:+\:\frac{\boldsymbol{\mathrm{yy}}_{\mathrm{1}} }{\boldsymbol{\mathrm{b}}^{\mathrm{2}} }\:=\:\mathrm{1}\:....\left(\boldsymbol{\mathrm{ii}}\right) \\ $$$$\mathrm{Let}\:\mathrm{the}\:\mathrm{midpoint}\:\mathrm{of}\:\boldsymbol{\mathrm{P}}\left(\boldsymbol{\mathrm{x}}_{\mathrm{2}} ,\boldsymbol{\mathrm{y}}_{\mathrm{2}} \right)\:\mathrm{and}\:\boldsymbol{\mathrm{Q}}\left(\boldsymbol{\mathrm{x}}_{\mathrm{3}} ,\boldsymbol{\mathrm{y}}_{\mathrm{3}} \right) \\ $$$$\mathrm{be}\:\left(\boldsymbol{\mathrm{h}},\boldsymbol{\mathrm{k}}\right) \\ $$$$\boldsymbol{\mathrm{i}}.\boldsymbol{\mathrm{e}}\:\:\:\boldsymbol{\mathrm{x}}_{\mathrm{2}} +\boldsymbol{\mathrm{x}}_{\mathrm{3}} \:=\:\mathrm{2}\boldsymbol{\mathrm{h}}.....\left(\boldsymbol{\mathrm{iii}}\right) \\ $$$$\boldsymbol{\mathrm{and}}\:\boldsymbol{\mathrm{y}}_{\mathrm{2}} +\boldsymbol{\mathrm{y}}_{\mathrm{3}} \:=\:\mathrm{2}\boldsymbol{\mathrm{k}}\:.....\left(\boldsymbol{\mathrm{iv}}\right) \\ $$$$\mathrm{as}\:\left(\mathrm{ii}\right)\:\mathrm{intersects}\:\mathrm{the}\:\mathrm{ellipse}\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:+\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{b}^{\mathrm{2}} }\:=\:\mathrm{1}...\left(\mathrm{v}\right) \\ $$$$\mathrm{putting}\:\:\frac{\mathrm{y}}{\mathrm{b}}\:=\:\frac{\mathrm{b}}{\mathrm{y}_{\mathrm{1}} }\left(\mathrm{1}−\frac{\mathrm{xx}_{\mathrm{1}} }{\mathrm{a}^{\mathrm{2}} }\right)\:\:\mathrm{from}\:\:\left(\mathrm{ii}\right)\:\mathrm{to}\:\left(\mathrm{v}\right) \\ $$$$\mathrm{we}\:\mathrm{get},\: \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{a}^{\mathrm{2}} }\:+\frac{\mathrm{b}^{\mathrm{2}} }{\mathrm{y}_{\mathrm{1}} ^{\mathrm{2}} }\left(\mathrm{1}−\frac{\mathrm{xx}_{\mathrm{1}} }{\mathrm{a}^{\mathrm{2}} }\right)^{\mathrm{2}} =\:\mathrm{1} \\ $$$$\:\:\:\:\:\:\:\Rightarrow\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} \left\{\frac{\mathrm{1}}{\boldsymbol{\mathrm{a}}^{\mathrm{2}} }+\frac{\boldsymbol{\mathrm{b}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}^{\mathrm{4}} \boldsymbol{\mathrm{y}}_{\mathrm{1}} ^{\mathrm{2}} }\right\}−\boldsymbol{\mathrm{x}}.\mathrm{2}\frac{\boldsymbol{\mathrm{x}}_{\mathrm{1}} }{\boldsymbol{\mathrm{y}}_{\mathrm{1}} ^{\mathrm{2}} }.\frac{\boldsymbol{\mathrm{b}}^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}^{\mathrm{2}} }\:+\left(\frac{\boldsymbol{\mathrm{b}}^{\mathrm{2}} }{\boldsymbol{\mathrm{y}}_{\mathrm{1}} ^{\mathrm{2}} }\:−\mathrm{1}\right)=\mathrm{0} \\ $$$$\:\:\:\therefore\:\boldsymbol{\mathrm{x}}_{\mathrm{2}} +\boldsymbol{\mathrm{x}}_{\mathrm{3}} =\:\frac{\mathrm{2}\boldsymbol{\mathrm{x}}_{\mathrm{1}} \boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{b}}^{\mathrm{2}} }{\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} \right)} \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{h}}\:=\:\frac{\boldsymbol{\mathrm{x}}_{\mathrm{1}} \boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{b}}^{\mathrm{2}} }{\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} \right)} \\ $$$$\boldsymbol{\mathrm{similarly}}\:\boldsymbol{\mathrm{we}}\:\boldsymbol{\mathrm{can}}\:\boldsymbol{\mathrm{get}}\: \\ $$$$\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{k}}\:=\:\frac{\boldsymbol{\mathrm{y}}_{\mathrm{1}} \boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{b}}^{\mathrm{2}} }{\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} \right)} \\ $$$$\:\boldsymbol{\mathrm{now}}\: \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{h}}^{\mathrm{2}} +\boldsymbol{\mathrm{k}}^{\mathrm{2}} =\:\frac{\boldsymbol{\mathrm{a}}^{\mathrm{4}} \boldsymbol{\mathrm{b}}^{\mathrm{4}} }{\left(\boldsymbol{\mathrm{a}}^{\mathrm{2}} \boldsymbol{\mathrm{y}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\mathrm{b}}^{\mathrm{2}} \boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} \right)^{\mathrm{2}} }\left\{\boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} +\boldsymbol{\mathrm{y}}_{\mathrm{1}} ^{\mathrm{2}} \right\} \\ $$$$\:\:\:\:\:\:\:\:\:\boldsymbol{\mathrm{h}}^{\mathrm{2}} \:+\:\boldsymbol{\mathrm{k}}^{\mathrm{2}} \:=\:\boldsymbol{\mathrm{d}}^{\mathrm{2}} .\frac{\mathrm{1}}{\left(\frac{\boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}^{\mathrm{2}} }+\frac{\boldsymbol{\mathrm{y}}_{\mathrm{1}} ^{\mathrm{2}} }{\boldsymbol{\mathrm{b}}^{\mathrm{2}} }\right)^{\mathrm{2}} }\:\:\:\:\because\:\boldsymbol{\mathrm{from}}\:\left(\boldsymbol{\mathrm{i}}\right) \\ $$$$\therefore\:\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{locus}}\:\boldsymbol{\mathrm{of}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{mipoint}}\:\boldsymbol{\mathrm{is}} \\ $$$$\:\:\:\:\:\:\:\boldsymbol{\mathrm{x}}^{\mathrm{2}} +\boldsymbol{\mathrm{y}}^{\mathrm{2}} =\:\boldsymbol{\mathrm{d}}^{\mathrm{2}} \left[\frac{\boldsymbol{\mathrm{x}}_{\mathrm{1}} ^{\mathrm{2}} }{\boldsymbol{\mathrm{a}}^{\mathrm{2}} }\:+\:\frac{\boldsymbol{\mathrm{y}}_{\mathrm{1}} ^{\mathrm{2}} }{\boldsymbol{\mathrm{b}}^{\mathrm{2}} }\right]^{−\mathrm{2}} \\ $$$$\boldsymbol{\mathrm{I}}\:\boldsymbol{\mathrm{get}}\:\boldsymbol{\mathrm{this}}\:.\:\boldsymbol{\mathrm{Is}}\:\boldsymbol{\mathrm{the}}\:\boldsymbol{\mathrm{given}}\:\boldsymbol{\mathrm{question}}\:\boldsymbol{\mathrm{is}}\:\boldsymbol{\mathrm{correct}}\:? \\ $$$$\boldsymbol{\mathrm{please}}\:\boldsymbol{\mathrm{check}}. \\ $$
Commented by peter frank last updated on 06/Oct/19
$${thank}\:{very}\:{much} \\ $$
Commented by peter frank last updated on 06/Oct/19
$${thank}\:{very}\:{much} \\ $$