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Question Number 51921 by peter frank last updated on 01/Jan/19
A normal chord to an ellipse (x^2 /a^2 )+(y^2 /b^2 )=1 make an angle of 45^° with the axis.prove that the square of its length is equal to ((32a^4 b^4 )/((a^2 +b^2 )^3 ))
$${A}\:{normal}\:{chord}\:{to}\:{an}\: \\ $$$${ellipse}\:\frac{{x}^{\mathrm{2}} }{{a}^{\mathrm{2}} }+\frac{{y}^{\mathrm{2}} }{{b}^{\mathrm{2}} }=\mathrm{1} \\ $$$${make}\:{an}\:{angle}\:{of}\:\mathrm{45}^{°} \\ $$$${with}\:{the}\:{axis}.{prove} \\ $$$${that}\:{the}\:{square}\:{of}\:{its}\: \\ $$$${length}\:{is}\:{equal}\:{to} \\ $$$$\frac{\mathrm{32}{a}^{\mathrm{4}} {b}^{\mathrm{4}} }{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{3}} }\: \\ $$
Answered by ajfour last updated on 01/Jan/19
let x_P = acos θ , y_P =bsin θ −(dx/dy)∣_P = ((asin θ)/(bcos θ)) = (a/b)tan θ = tan 45°=1 ⇒ tan θ = (b/a) ⇒ acos θ = (a^2 /( (√(a^2 +b^2 )))) = (a^2 /s) (say) bsin θ = (b^2 /s) eq. of normal let this normal intersects ellipse aot the other point Q(h,k) let PQ = l h = acos θ−(l/( (√2))) , k = bcos θ−(l/( (√2))) or Q = ((a^2 /s)−(l/( (√2))) , (b^2 /s)−(l/( (√2)))) as Q is on ellipse, b^2 ((a^2 /s)−(l/( (√2))))^2 +a^2 ((b^2 /s)−(l/( (√2))))^2 = a^2 b^2 ⇒ b^2 (a^2 (√2)−sl)^2 +a^2 (b^2 (√2)−sl)^2 = 2a^2 b^2 s^2 ⇒ 2a^2 b^2 (a^2 +b^2 )+s^2 l^2 (a^2 +b^2 ) −2(√2)sa^2 b^2 l −2a^2 b^2 s^2 = 0 ⇒ since s^2 = a^2 +b^2 , we get l^2 −((2(√2)a^2 b^2 )/s^3 )l = 0 ⇒ l^2 = ((8a^4 b^4 )/((a^2 +b^2 )^3 )) .
$${let}\:\:\:{x}_{{P}} =\:{a}\mathrm{cos}\:\theta\:\:,\:{y}_{{P}} ={b}\mathrm{sin}\:\theta \\ $$$$\:\:−\frac{{dx}}{{dy}}\mid_{{P}} \:=\:\frac{{a}\mathrm{sin}\:\theta}{{b}\mathrm{cos}\:\theta}\:=\:\frac{{a}}{{b}}\mathrm{tan}\:\theta\:=\:\mathrm{tan}\:\mathrm{45}°=\mathrm{1} \\ $$$$\Rightarrow\:\mathrm{tan}\:\theta\:=\:\frac{{b}}{{a}} \\ $$$$\Rightarrow\:\:{a}\mathrm{cos}\:\theta\:=\:\frac{{a}^{\mathrm{2}} }{\:\sqrt{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }}\:=\:\frac{{a}^{\mathrm{2}} }{{s}}\:\:\:\left({say}\right) \\ $$$$\:\:\:\:\:\:\:{b}\mathrm{sin}\:\theta\:=\:\frac{{b}^{\mathrm{2}} }{{s}} \\ $$$${eq}.\:{of}\:{normal}\: \\ $$$${let}\:{this}\:{normal}\:{intersects}\:{ellipse} \\ $$$${aot}\:{the}\:{other}\:{point}\:{Q}\left({h},{k}\right) \\ $$$${let}\:\:{PQ}\:=\:{l} \\ $$$$\:\:{h}\:=\:{a}\mathrm{cos}\:\theta−\frac{{l}}{\:\sqrt{\mathrm{2}}}\:\:,\:{k}\:=\:{b}\mathrm{cos}\:\theta−\frac{{l}}{\:\sqrt{\mathrm{2}}} \\ $$$${or}\:\:\:\:{Q}\:=\:\left(\frac{{a}^{\mathrm{2}} }{{s}}−\frac{{l}}{\:\sqrt{\mathrm{2}}}\:,\:\frac{{b}^{\mathrm{2}} }{{s}}−\frac{{l}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$${as}\:{Q}\:{is}\:{on}\:{ellipse}, \\ $$$$\:\:\:\:\:\:{b}^{\mathrm{2}} \left(\frac{{a}^{\mathrm{2}} }{{s}}−\frac{{l}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \left(\frac{{b}^{\mathrm{2}} }{{s}}−\frac{{l}}{\:\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} =\:{a}^{\mathrm{2}} {b}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:{b}^{\mathrm{2}} \left({a}^{\mathrm{2}} \sqrt{\mathrm{2}}−{sl}\right)^{\mathrm{2}} +{a}^{\mathrm{2}} \left({b}^{\mathrm{2}} \sqrt{\mathrm{2}}−{sl}\right)^{\mathrm{2}} =\:\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {s}^{\mathrm{2}} \\ $$$$\Rightarrow\:\:\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)+{s}^{\mathrm{2}} {l}^{\mathrm{2}} \left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right) \\ $$$$\:\:\:\:\:\:\:\:\:\:−\mathrm{2}\sqrt{\mathrm{2}}{sa}^{\mathrm{2}} {b}^{\mathrm{2}} {l}\:−\mathrm{2}{a}^{\mathrm{2}} {b}^{\mathrm{2}} {s}^{\mathrm{2}} \:=\:\mathrm{0} \\ $$$$\Rightarrow\:{since}\:\:{s}^{\mathrm{2}} \:=\:{a}^{\mathrm{2}} +{b}^{\mathrm{2}} \:\:,\:{we}\:{get} \\ $$$$\:\:{l}^{\mathrm{2}} −\frac{\mathrm{2}\sqrt{\mathrm{2}}{a}^{\mathrm{2}} {b}^{\mathrm{2}} }{{s}^{\mathrm{3}} }{l}\:=\:\mathrm{0} \\ $$$$\Rightarrow\:\:\:{l}^{\mathrm{2}} \:=\:\frac{\mathrm{8}{a}^{\mathrm{4}} {b}^{\mathrm{4}} }{\left({a}^{\mathrm{2}} +{b}^{\mathrm{2}} \right)^{\mathrm{3}} }\:\:. \\ $$
Commented by peter frank last updated on 01/Jan/19
thank sir
$${thank}\:{sir}\: \\ $$$$ \\ $$

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