A) \[\frac{1}{a}+\frac{1}{a'}=\frac{1}{b}+\frac{1}{b'}\]
B) \[\frac{1}{a}+\frac{1}{b}=\frac{1}{a'}+\frac{1}{b'}\]
C) \[\frac{1}{a}-\frac{1}{b}=\frac{1}{a'}-\frac{1}{b'}\]
D) None of these
Correct Answer: C
Solution :
Given, curve is \[a{{x}^{2}}+b{{y}^{2}}=1\] ... (i) and \[a'{{x}^{2}}+b'{{y}^{2}}=1\] ...(ii) On differentiating w.r.t.\[x,\]we get \[2ax+2by\frac{dy}{dx}=0\] and \[2a'x+2b'y\frac{dy}{dx}=0\] \[\Rightarrow \] \[{{m}_{1}}=\frac{dy}{dx}=-\frac{ax}{by}\] ?.(iii) and \[{{m}_{2}}=\frac{dy}{dx}=-\frac{a'x}{b'y}\] ?.(iv) Since, curves intersect each other orthogonally. \[\therefore \] \[{{m}_{1}}{{m}_{2}}=-1\] \[\Rightarrow \] \[\left( -\frac{ax}{by} \right)\left( -\frac{a'x}{b'y} \right)=-1\] \[\Rightarrow \] \[\frac{aa'}{bb'}.\frac{{{x}^{2}}}{{{y}^{2}}}=-1\] ?.(v) On subtracting Eq. (ii) from Eq. (i), we get \[{{x}^{2}}(a-a')+{{y}^{2}}(b-b')=0\] \[\Rightarrow \] \[\frac{{{x}^{2}}}{{{y}^{2}}}=-\left( \frac{b-b'}{a-a'} \right)\] On putting this value in'Eq. (v), we get \[-\frac{aa'}{bb'}\left( \frac{b-b'}{a-a'} \right)=-1\] \[\Rightarrow \] \[\frac{b-b'}{bb'}=\frac{a-a'}{aa'}\] \[\Rightarrow \] \[\frac{1}{b'}-\frac{1}{b}=\frac{1}{a'}-\frac{1}{a}\] \[\Rightarrow \] \[\frac{1}{a}-\frac{1}{b}=\frac{1}{a'}-\frac{1}{b'}\]
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