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Question Number 84531 by jagoll last updated on 14/Mar/20
find for equation of image ellipse (x^2 /9) + (y^2 /8) = 1 if reflected with line x + y = −4
$$\mathrm{find}\:\mathrm{for}\:\mathrm{equation}\:\mathrm{of}\:\mathrm{image}\:\mathrm{ellipse} \\ $$$$\frac{\mathrm{x}^{\mathrm{2}} }{\mathrm{9}}\:+\:\frac{\mathrm{y}^{\mathrm{2}} }{\mathrm{8}}\:=\:\mathrm{1}\:\mathrm{if}\:\mathrm{reflected}\:\mathrm{with}\:\mathrm{line} \\ $$$$\mathrm{x}\:+\:\mathrm{y}\:=\:−\mathrm{4} \\ $$
Commented by jagoll last updated on 14/Mar/20
help me
$$\mathrm{help}\:\mathrm{me} \\ $$
Commented by mr W last updated on 14/Mar/20
i misread the question as x+y=4, but the method is the same.
$${i}\:{misread}\:{the}\:{question}\:{as}\:{x}+{y}=\mathrm{4},\:{but} \\ $$$${the}\:{method}\:{is}\:{the}\:{same}. \\ $$
Answered by mr W last updated on 14/Mar/20
image of point (u,v) in the line ax+by+c=0 is point (x,y) with ((x−u)/a)=((y−v)/b)=−((2(au+bv+c))/(a^2 +b^2 )) or x=u−((2a(au+bv+c))/(a^2 +b^2 )) y=v−((2b(au+bv+c))/(a^2 +b^2 )) in our case: the line is x+y=4, i.e. a=b=1, c=−4 ⇒x=4−v ⇒y=4−u from (x^2 /9)+(y^2 /8)=1 we get its image ⇒(((4−v)^2 )/9)+(((4−u)^2 )/8)=1 i.e. the eqn. of the image ellipse is (((4−y)^2 )/9)+(((4−x)^2 )/8)=1
$${image}\:{of}\:{point}\:\left({u},{v}\right)\:{in}\:{the}\:{line} \\ $$$${ax}+{by}+{c}=\mathrm{0}\:{is}\:{point}\:\left({x},{y}\right)\:{with} \\ $$$$\frac{{x}−{u}}{{a}}=\frac{{y}−{v}}{{b}}=−\frac{\mathrm{2}\left({au}+{bv}+{c}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${or} \\ $$$${x}={u}−\frac{\mathrm{2}{a}\left({au}+{bv}+{c}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$${y}={v}−\frac{\mathrm{2}{b}\left({au}+{bv}+{c}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} } \\ $$$$ \\ $$$${in}\:{our}\:{case}:\:{the}\:{line}\:{is}\:{x}+{y}=\mathrm{4}, \\ $$$${i}.{e}.\:{a}={b}=\mathrm{1},\:{c}=−\mathrm{4} \\ $$$$\Rightarrow{x}=\mathrm{4}−{v} \\ $$$$\Rightarrow{y}=\mathrm{4}−{u} \\ $$$${from}\:\frac{{x}^{\mathrm{2}} }{\mathrm{9}}+\frac{{y}^{\mathrm{2}} }{\mathrm{8}}=\mathrm{1}\:{we}\:{get}\:{its}\:{image} \\ $$$$\Rightarrow\frac{\left(\mathrm{4}−{v}\right)^{\mathrm{2}} }{\mathrm{9}}+\frac{\left(\mathrm{4}−{u}\right)^{\mathrm{2}} }{\mathrm{8}}=\mathrm{1} \\ $$$${i}.{e}.\:{the}\:{eqn}.\:{of}\:{the}\:{image}\:{ellipse}\:{is} \\ $$$$\frac{\left(\mathrm{4}−{y}\right)^{\mathrm{2}} }{\mathrm{9}}+\frac{\left(\mathrm{4}−{x}\right)^{\mathrm{2}} }{\mathrm{8}}=\mathrm{1} \\ $$
Commented by mr W last updated on 14/Mar/20
Commented by mr W last updated on 14/Mar/20
further examples: image of (x^2 /(25))+(y^2 /9)=1 is (((4−y)^2 )/(25))+(((4−x)^2 )/9)=1 image of y=x^2 −4 is 4−x=(4−y)^2 −4 or x=8−(4−y)^2 image of y=sin x is 4−x=sin (4−y) or x=4−sin (4−y)
$${further}\:{examples}: \\ $$$${image}\:{of}\:\frac{{x}^{\mathrm{2}} }{\mathrm{25}}+\frac{{y}^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1}\:{is} \\ $$$$\frac{\left(\mathrm{4}−{y}\right)^{\mathrm{2}} }{\mathrm{25}}+\frac{\left(\mathrm{4}−{x}\right)^{\mathrm{2}} }{\mathrm{9}}=\mathrm{1} \\ $$$$ \\ $$$${image}\:{of}\:{y}={x}^{\mathrm{2}} −\mathrm{4}\:{is} \\ $$$$\mathrm{4}−{x}=\left(\mathrm{4}−{y}\right)^{\mathrm{2}} −\mathrm{4}\:{or}\:{x}=\mathrm{8}−\left(\mathrm{4}−{y}\right)^{\mathrm{2}} \\ $$$$ \\ $$$${image}\:{of}\:{y}={sin}\:{x}\:{is} \\ $$$$\mathrm{4}−{x}=\mathrm{sin}\:\left(\mathrm{4}−{y}\right)\:{or}\:{x}=\mathrm{4}−\mathrm{sin}\:\left(\mathrm{4}−{y}\right) \\ $$
Commented by mr W last updated on 14/Mar/20
Commented by mr W last updated on 14/Mar/20
Commented by mr W last updated on 14/Mar/20
Commented by jagoll last updated on 14/Mar/20
yes sir. i got result (((x−4)^2 )/9) + (((y−4)^2 )/8) = 1 i will learn your method sir. thank you very much
$$\mathrm{yes}\:\mathrm{sir}.\:\mathrm{i}\:\mathrm{got}\:\mathrm{result}\:\frac{\left(\mathrm{x}−\mathrm{4}\right)^{\mathrm{2}} }{\mathrm{9}}\:+\:\frac{\left(\mathrm{y}−\mathrm{4}\right)^{\mathrm{2}} }{\mathrm{8}}\:=\:\mathrm{1} \\ $$$$\mathrm{i}\:\mathrm{will}\:\mathrm{learn}\:\mathrm{your}\:\mathrm{method}\:\mathrm{sir}. \\ $$$$\mathrm{thank}\:\mathrm{you}\:\mathrm{very}\:\mathrm{much} \\ $$
Commented by jagoll last updated on 14/Mar/20
sir x=4−v, where your got 4 sir
$$\mathrm{sir}\:{x}=\mathrm{4}−{v},\:\mathrm{where}\:\mathrm{your}\:\mathrm{got}\:\mathrm{4}\:\mathrm{sir} \\ $$
Commented by mr W last updated on 14/Mar/20
eqn. of reflection line: x+y−4=0 with a=b=1, c=−4 x=u−((2a(au+bv+c))/(a^2 +b^2 ))=u−((2(u+v−4))/2)=4−v y=v−((2b(au+bv+c))/(a^2 +b^2 ))=v−((2(u+v−4))/2)=4−u
$${eqn}.\:{of}\:{reflection}\:{line}:\:{x}+{y}−\mathrm{4}=\mathrm{0} \\ $$$${with}\:{a}={b}=\mathrm{1},\:{c}=−\mathrm{4} \\ $$$${x}={u}−\frac{\mathrm{2}{a}\left({au}+{bv}+{c}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={u}−\frac{\mathrm{2}\left({u}+{v}−\mathrm{4}\right)}{\mathrm{2}}=\mathrm{4}−{v} \\ $$$${y}={v}−\frac{\mathrm{2}{b}\left({au}+{bv}+{c}\right)}{{a}^{\mathrm{2}} +{b}^{\mathrm{2}} }={v}−\frac{\mathrm{2}\left({u}+{v}−\mathrm{4}\right)}{\mathrm{2}}=\mathrm{4}−{u} \\ $$

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