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	$(p)$ The locus of the point $(h, k)$ for which the line $h x+k y=1$ touches the circle $x^2+y^2=4$
$(B)$ Parabola	$(q)$ Points $z$ in the complex plane satisfying $\|z+2\|-\|z-2\|= \pm 3$
$(C)$ Ellipse	$(r)$ Points of the conic have parametric representation $x=\sqrt{3}\left(\frac{1-t^2}{1+t^2}\right), y=\frac{2 t}{1+t^2}$
$(D)$ Hyperbola	$(s)$ The eccentricity of the conic lies in the interval $1 \leq x < \infty$
	$(t)$ Points $z$ in the complex plane satisfying $\operatorname{Re}(z+1)^2=\|z\|^2+1$

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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