If the curves $x=y^{2}$ and $xy=k$ intersect at right angle, then :

Question

Accepted Answer

To determine why Option 2 ($8k^{2}=1$) is the correct answer, we need to analyze the conditions under which the curves $x = y^2$ and $xy = k$ intersect at a right angle.

### Step-by-Step Explanation:

1. **Equations of the Curves:**
   - The first curve is given by $x = y^2$.
   - The second curve is given by $xy = k$.

2. **Finding the Derivatives:**
   - For the curve $x = y^2$, differentiate with respect to $x$: 
     $$
     \frac{dy}{dx} = \frac{1}{2\sqrt{x}}
     $$
     Since $x = y^2$, we have $y = \sqrt{x}$, so $\frac{dy}{dx} = \frac{1}{2y}$.

- For the curve $xy = k$, differentiate implicitly with respect to $x$:
     $$
     y + x\frac{dy}{dx} = 0 \implies \frac{dy}{dx} = -\frac{y}{x}
     $$

3. **Condition for Perpendicular Intersection:**
   - Two curves intersect at right angles if the product of their slopes at the point of intersection is $-1$.
   - Therefore, we set:
     $$
     \left(\frac{1}{2y}\right) \left(-\frac{y}{x}\right) = -1
     $$
     Simplifying, we get:
     $$
     -\frac{1}{2x} = -1 \implies \frac{1}{2x} = 1 \implies x = \frac{1}{2}
     $$

4. **Substitute $x = \frac{1}{2}$ into $x = y^2$:**
   - We have $y^2 = \frac{1}{2}$, so $y = \pm \frac{1}{\sqrt{2}}$.

5. **Substitute $x = \frac{1}{2}$ and $y = \pm \frac{1}{\sqrt{2}}$ into $xy = k$:**
   - For $y = \frac{1}{\sqrt{2}}$, $k = \frac{1}{2} \cdot \frac{1}{\sqrt{2}} = \frac{1}{2\sqrt{2}}$.
   - For $y = -\frac{1}{\sqrt{2}}$, $k = \frac{1}{2} \cdot -\frac{1}{\sqrt{2}} = -\frac{1}{2\sqrt{2}}$.

6. **Calculate $8k^2$:**
   - For $k = \frac{1}{2\sqrt{2}}$, $k^2 = \left(\frac{1}{2\sqrt{2}}\right)^2 = \frac{1}{8}$.
   - Therefore, $8k^2 = 8 \times \frac{1}{8} = 1$.

Thus, the correct answer is $8k^2 = 1$, which corresponds to Option 2.

### Explanation for Incorrect Options:

- **Option 1 ($8k^2 = 3$):** This would imply $k^2 = \frac{3}{8}$, which does not satisfy the condition derived from the perpendicular intersection.
- **Option 3 ($8k^2 = 5$):** This would imply $k^2 = \frac{5}{8}$, which is not consistent with the derived condition.
- **Option 4 ($8k^2 = -1$):** This is not possible as $k^2$ is always non-negative, making $8k^2 = -1$ an invalid result.

Therefore, the only correct and consistent option is Option 2, $8k^2 = 1$.