The condition for the curves ax^2 + by^2 = 1 | vedclass

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(A) Let the curves be $C_1: ax^2 + by^2 = 1$ and $C_2: a'x^2 + b'y^2 = 1$.Differentiating $C_1$ with respect to $x$,we get $2ax + 2by \frac{dy}{dx} =

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$\frac{1}{a} - \frac{1}{a'} = \frac{1}{b} - \frac{1}{b'}$

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(A) Let the curves be $C_1: ax^2 + by^2 = 1$ and $C_2: a'x^2 + b'y^2 = 1$.Differentiating $C_1$ with respect to $x$,we get $2ax + 2by \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{ax}{by}$.Differentiating $C_2$ with respect to $x$,we get $2a'x + 2b'y \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{a'x}{b'y}$.For orthogonal intersection,the product of the slopes at the point of intersection $(x, y)$ must be $-1$:$(-\frac{ax}{by}) 	imes (-\frac{a'x}{b'y}) = -1$$\frac{aa'x^2}{bb'y^2} = -1$Subtracting the two original equations: $(a - a')x^2 + (b - b')y^2 = 0$,which gives $\frac{x^2}{y^2} = -\frac{b - b'}{a - a'} = \frac{b' - b}{a - a'}$.Substituting this into the product condition:$\frac{aa'}{bb'} 	imes \frac{b' - b}{a - a'} = -1$$aa'(b' - b) = -bb'(a - a')$$aa'b' - aa'b = -bba' + bb'a$$aa'b' - bb'a = aa'b - bba'$Dividing by $aa'bb'$:$\frac{1}{b} - \frac{1}{b'} = \frac{1}{a'} - \frac{1}{a}$Rearranging gives $\frac{1}{a} - \frac{1}{a'} = \frac{1}{b'} - \frac{1}{b}$,which is equivalent to $\frac{1}{a} - \frac{1}{a'} = \frac{1}{b} - \frac{1}{b'}$.

Column-$I$	Column-$II$
$A$. Circle	$P$. Locus of point $(h, k)$ such that the line $hx + ky = 1$ touches the circle $x^2 + y^2 = 4$
$B$. Parabola	$Q$. Point $z$ in the complex plane satisfies $\|z + 2\| - \|z - 2\| = \pm 3$
$C$. Hyperbola	$R$. Eccentricity of the conic lies in the interval $[1, \infty)$
	$S$. Point $z$ in the complex plane satisfies $Re(z + 1)^2 = \|z\|^2 + 1$

The condition for the curves $ax^2 + by^2 = 1$ and $a'x^2 + b'y^2 = 1$ to intersect each other orthogonally is

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