If the curves ax^2+by^2=1 and cx^2+dy^2=1 int | vedclass

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(A) Let the curves be $ax^2+by^2=1$ $(i)$ and $cx^2+dy^2=1$ (ii). Subtracting (ii) from $(i)$,we get $(a-c)x^2 + (b-d)y^2 = 0$,which implies $x^2 = -\

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$\frac{a}{c} \cdot \frac{b}{d}$

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(A) Let the curves be $ax^2+by^2=1$ $(i)$ and $cx^2+dy^2=1$ (ii). Subtracting (ii) from $(i)$,we get $(a-c)x^2 + (b-d)y^2 = 0$,which implies $x^2 = -\frac{b-d}{a-c} y^2 = \frac{d-b}{a-c} y^2$. Differentiating $(i)$ with respect to $x$,we get $2ax + 2byy' = 0$,so $y'_1 = -\frac{ax}{by}$. Differentiating (ii) with respect to $x$,we get $2cx + 2dyy' = 0$,so $y'_2 = -\frac{cx}{dy}$. Since the curves intersect orthogonally,the product of their slopes at the point of intersection $(x, y)$ is $-1$: $y'_1 \cdot y'_2 = -1 \Rightarrow \left(-\frac{ax}{by}\right) \left(-\frac{cx}{dy}\right) = -1 \Rightarrow \frac{acx^2}{bdy^2} = -1$. Substituting $x^2 = \frac{d-b}{a-c} y^2$ into the equation: $\frac{ac}{bd} \left(\frac{d-b}{a-c}\right) = -1 \Rightarrow ac(d-b) = -bd(a-c) \Rightarrow acd - abc = -abd + bcd$. Rearranging terms: $acd + abd = abc + bcd \Rightarrow ad(c+b) = bc(a+d)$. Actually,using the condition $y'_1 y'_2 = -1$ at intersection point $(x, y)$: $\frac{ac x^2}{bd y^2} = -1 \Rightarrow \frac{x^2}{y^2} = -\frac{bd}{ac}$. From $(a-c)x^2 + (b-d)y^2 = 0$,we have $\frac{x^2}{y^2} = -\frac{b-d}{a-c} = \frac{d-b}{a-c}$. Equating the two expressions for $\frac{x^2}{y^2}$: $\frac{d-b}{a-c} = -\frac{bd}{ac} \Rightarrow ac(d-b) = -bd(a-c) \Rightarrow acd - abc = -abd + bcd$. $abd - abc = bcd - acd \Rightarrow ab(d-c) = cd(b-a)$. Therefore,$\frac{b-a}{d-c} = \frac{ab}{cd}$.

List-$I$	List-$II$
$A$. The equation of the polar of $(4, 3)$ with respect to $C$	$I$. $y+5=0$
$B$. The equation of the tangent at $(9, -5)$ on $C$	$II$. $x=1$
$C$. The equation of the normal at $(-7, -5)$ on $C$	$III$. $3x+8y=27$
$D$. The equation of the diameter passing through $(1, -5)$ and $(1, 3)$	$IV$. $x=9$

If the curves $ax^2+by^2=1$ and $cx^2+dy^2=1$ intersect orthogonally,then $\frac{b-a}{d-c}=$

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