The abscissa of the point on the curve sqrt{x | vedclass

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(D) Given the curve equation: $\sqrt{xy} = a + x$. Squaring both sides,we get $xy = (a + x)^2 = a^2 + x^2 + 2ax$.Solving for $y$,we have $y = \frac{a^

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Both $(A)$ and $(B)$

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(D) Given the curve equation: $\sqrt{xy} = a + x$. Squaring both sides,we get $xy = (a + x)^2 = a^2 + x^2 + 2ax$.Solving for $y$,we have $y = \frac{a^2}{x} + x + 2a$.Differentiating with respect to $x$,we get $\frac{dy}{dx} = -\frac{a^2}{x^2} + 1$.Since the tangent cuts off equal intercepts from the coordinate axes,its slope must be $\pm 1$. However,for a line to cut off equal intercepts $c$ and $c$ (where $c 
eq 0$),the slope is $-1$.Setting the derivative equal to $-1$: $-\frac{a^2}{x^2} + 1 = -1$.This simplifies to $\frac{a^2}{x^2} = 2$,which implies $x^2 = \frac{a^2}{2}$.Thus,$x = \pm \frac{a}{\sqrt{2}}$.Since $a > 0$,both values are valid abscissae.

The abscissa of the point on the curve $\sqrt{xy} = a + x$ where the tangent cuts off equal intercepts from the coordinate axes is $(a > 0)$.

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