If the curves frac{x^2}{a^2} + frac{y^2}{b^2} | vedclass

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(C) Let the curves be $C_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and $C_2: x^2 - y^2 = c^2$.Differentiating $C_1$ with respect to $x$: $\frac{2x}{a^

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$a^2 - b^2 = 2c^2$

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(C) Let the curves be $C_1: \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ and $C_2: x^2 - y^2 = c^2$.Differentiating $C_1$ with respect to $x$: $\frac{2x}{a^2} + \frac{2yy'}{b^2} = 0 \implies y'_1 = -\frac{b^2x}{a^2y}$.Differentiating $C_2$ with respect to $x$: $2x - 2yy'_2 = 0 \implies y'_2 = \frac{x}{y}$.Since the curves intersect at right angles,the product of their slopes at the point of intersection $(x, y)$ is $-1$:$y'_1 	imes y'_2 = -1$$(-\frac{b^2x}{a^2y}) 	imes (\frac{x}{y}) = -1$$\frac{b^2x^2}{a^2y^2} = 1 \implies b^2x^2 = a^2y^2$.From $C_2$,$y^2 = x^2 - c^2$. Substituting this into the equation above:$b^2x^2 = a^2(x^2 - c^2)$$b^2x^2 = a^2x^2 - a^2c^2$$x^2(a^2 - b^2) = a^2c^2 \implies x^2 = \frac{a^2c^2}{a^2 - b^2}$.From $C_1$,$\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Substituting $y^2 = x^2 - c^2$:$\frac{x^2}{a^2} + \frac{x^2 - c^2}{b^2} = 1$$x^2(\frac{1}{a^2} + \frac{1}{b^2}) = 1 + \frac{c^2}{b^2}$$x^2(\frac{a^2 + b^2}{a^2b^2}) = \frac{b^2 + c^2}{b^2}$$x^2 = \frac{a^2(b^2 + c^2)}{a^2 + b^2}$.Equating the two expressions for $x^2$:$\frac{a^2c^2}{a^2 - b^2} = \frac{a^2(b^2 + c^2)}{a^2 + b^2}$$c^2(a^2 + b^2) = (b^2 + c^2)(a^2 - b^2)$$a^2c^2 + b^2c^2 = a^2b^2 - b^4 + a^2c^2 - b^2c^2$$2b^2c^2 = a^2b^2 - b^4$Dividing by $b^2$ $(b 
eq 0)$:$2c^2 = a^2 - b^2$.

If the curves $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ $(a > b)$ and $x^2 - y^2 = c^2$ intersect at right angles,then:

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