If frac{x}{a} + frac{y}{b} = 1 is a tangent t | vedclass

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(D) Given the curve $x = Kt$ and $y = \frac{K}{t}$.Eliminating the parameter $t$,we get $t = \frac{x}{K}$,so $y = \frac{K}{x/K} = \frac{K^2}{x}$,which

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Both $(A)$ and $(C)$

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(D) Given the curve $x = Kt$ and $y = \frac{K}{t}$.Eliminating the parameter $t$,we get $t = \frac{x}{K}$,so $y = \frac{K}{x/K} = \frac{K^2}{x}$,which implies $xy = K^2$.Differentiating with respect to $x$,we get $y + x \frac{dy}{dx} = 0$,so $\frac{dy}{dx} = -\frac{y}{x} = -\frac{K^2/x}{x} = -\frac{K^2}{x^2}$.Since $K^2 > 0$ and $x^2 > 0$,the slope $\frac{dy}{dx} = -\frac{K^2}{x^2} The equation of the tangent line is given as $\frac{x}{a} + \frac{y}{b} = 1$,which can be written as $y = -\frac{b}{a}x + b$.The slope of this line is $m = -\frac{b}{a}$.Since the slope of the curve is always negative,the slope of the tangent must also be negative,so $-\frac{b}{a} 0$.This condition $\frac{b}{a} > 0$ holds if both $a$ and $b$ have the same sign.Thus,either $a > 0, b > 0$ or $a < 0, b < 0$.

If $\frac{x}{a} + \frac{y}{b} = 1$ is a tangent to the curve $x = Kt, y = \frac{K}{t}, K > 0$,then

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