The abscissa of a point on the curve `x y=(a+x)^2,` the normal which cuts off numerically equal intercepts from the coordinate axes, is `-1/(sqrt(2))` (b) `sqrt(2)a` (c) `a/(sqrt(2))` (d) `-sqrt(2)a`

Options

A. `-(a)/(sqrt2)`
B. `sqrt2a`
C. `(a)/(sqrt2)`
D. `-sqrt2a`

Answer

To solve the problem, we need to find the abscissa of a point on the curve \( xy = (a + x)^2 \) where the normal cuts off numerically equal intercepts from the coordinate axes.

### Step-by-Step Solution:

1. **Understand the Curve**: The given equation is \( xy = (a + x)^2 \). We can rewrite this as:
\[ y = \frac{(a + x)^2}{x} \]

2. **Differentiate the Curve**: We need to find the slope of the tangent to the curve. We will differentiate \( y \) with respect to \( x \) using the quotient rule:
\[ y' = \frac{(x \cdot 2(a + x) \cdot 1 - (a + x)^2 \cdot 1)}{x^2} \]
Simplifying this gives:
\[ y' = \frac{2x(a + x) - (a + x)^2}{x^2} = \frac{(a + x)(2x - (a + x))}{x^2} \]

3. **Find the Slope of the Normal**: The slope of the normal is the negative reciprocal of the slope of the tangent. Therefore,
\[ \text{slope of normal} = -\frac{1}{y'} \]

4. **Condition for Equal Intercepts**: The normal cuts off equal intercepts from the axes when its slope is either \( 1 \) or \( -1 \). Thus, we set:
\[ -\frac{1}{y'} = \pm 1 \]
This implies:
\[ y' = \mp 1 \]

5. **Set Up the Equation**: We can set up the equations for the two cases:
- Case 1: \( y' = 1 \)
- Case 2: \( y' = -1 \)

6. **Solve for \( x \)**: We will solve for \( x \) in both cases.

For \( y' = 1 \):
\[ \frac{(a + x)(2x - (a + x))}{x^2} = 1 \]
This simplifies to:
\[ (a + x)(2x - a - x) = x^2 \]
\[ (a + x)(x - a) = x^2 \]
Expanding gives:
\[ ax - a^2 + x^2 - ax = x^2 \]
Thus, we have:
\[ -a^2 = 0 \quad \Rightarrow \quad a = 0 \text{ (not valid)} \]

For \( y' = -1 \):
\[ \frac{(a + x)(2x - (a + x))}{x^2} = -1 \]
This simplifies to:
\[ (a + x)(2x - a - x) = -x^2 \]
\[ (a + x)(x - a) = -x^2 \]
Expanding gives:
\[ ax - a^2 + x^2 - ax = -x^2 \]
Thus:
\[ 2x^2 - a^2 = 0 \quad \Rightarrow \quad 2x^2 = a^2 \quad \Rightarrow \quad x^2 = \frac{a^2}{2} \quad \Rightarrow \quad x = \pm \frac{a}{\sqrt{2}} \]

7. **Conclusion**: The abscissa of the point on the curve where the normal cuts off equal intercepts is:
\[ x = \frac{a}{\sqrt{2}} \text{ or } x = -\frac{a}{\sqrt{2}} \]

### Final Answer:
The abscissa of the point on the curve is \( \frac{a}{\sqrt{2}} \) or \( -\frac{a}{\sqrt{2}} \).