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If the curves $a x^2+b y^2=1$ and $c x^2+d y^2=1$ intersect orthogonally, then $\frac{b-a}{d-c}=$
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The correct answer is:
$\frac{a}{c} \cdot \frac{b}{d}$
From Eqs. (i) and (ii), $(a-c) x^2+(b-d) y^2=0$
Differentiating Eq. (i) w.r.t.' $x^{\prime}$, we get
$2 a x+2 b y y^{\prime}=0 \Rightarrow y^{\prime}=-\frac{a x}{b y}$
Differentiating Eq. (ii) w.r.t' $x^{\prime}$ we get
$2 c x+2 d y y^{\prime}=0 \Rightarrow y^{\prime}=\frac{-c x}{d y}$
Curves intersect orthogonally
From Eqs. (iii) and (iv), we get
$\begin{gathered}
-\left(\frac{b-d}{a-c}\right)=-\frac{b d}{a c} \\
\Rightarrow \quad a b c-a c d=a b d-b c d \Rightarrow b c d-a c d=a b d-a b c \\
c d(b-a)=a b(d-c) \Rightarrow \frac{b-a}{d-c}=\frac{a b}{c d}
\end{gathered}$
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