If curve `y = 1 - ax^2 and y = x^2` intersect orthogonally then the value of a is

Options

A. `(1)/(2)`
B. `(1)/(3)`
C. 2
D. 3

Answer

To solve the problem of finding the value of \( a \) such that the curves \( y = 1 - ax^2 \) and \( y = x^2 \) intersect orthogonally, we will follow these steps:

### Step 1: Set the equations equal to find points of intersection
We start by setting the two equations equal to each other:
\[ 1 - ax^2 = x^2 \]
Rearranging gives:
\[ 1 = ax^2 + x^2 \]
\[ 1 = (a + 1)x^2 \]

### Step 2: Solve for \( x^2 \)
From the equation \( 1 = (a + 1)x^2 \), we can solve for \( x^2 \):
\[ x^2 = \frac{1}{a + 1} \]

### Step 3: Find \( y \) in terms of \( a \)
Now, substituting \( x^2 \) back into either equation to find \( y \):
\[ y = x^2 = \frac{1}{a + 1} \]
Thus, the points of intersection are:
\[ \left( \pm \sqrt{\frac{1}{a + 1}}, \frac{1}{a + 1} \right) \]

### Step 4: Find the slopes of the curves at the points of intersection
Next, we need to find the derivatives (slopes) of both curves at the points of intersection.

For the first curve \( y = 1 - ax^2 \):
\[ \frac{dy}{dx} = -2ax \]

For the second curve \( y = x^2 \):
\[ \frac{dy}{dx} = 2x \]

### Step 5: Evaluate the slopes at the intersection points
At the intersection points \( x = \pm \sqrt{\frac{1}{a + 1}} \):
- For the first curve:
\[ m_1 = -2a\left(\sqrt{\frac{1}{a + 1}}\right) = -\frac{2a}{\sqrt{a + 1}} \]
- For the second curve:
\[ m_2 = 2\left(\sqrt{\frac{1}{a + 1}}\right) = \frac{2}{\sqrt{a + 1}} \]

### Step 6: Set the product of slopes equal to -1
Since the curves intersect orthogonally, we have:
\[ m_1 \cdot m_2 = -1 \]
Substituting the slopes:
\[ \left(-\frac{2a}{\sqrt{a + 1}}\right) \cdot \left(\frac{2}{\sqrt{a + 1}}\right) = -1 \]
This simplifies to:
\[ -\frac{4a}{a + 1} = -1 \]

### Step 7: Solve for \( a \)
Removing the negative sign gives:
\[ \frac{4a}{a + 1} = 1 \]
Cross-multiplying:
\[ 4a = a + 1 \]
Rearranging:
\[ 4a - a = 1 \]
\[ 3a = 1 \]
Thus:
\[ a = \frac{1}{3} \]

### Final Answer
The value of \( a \) is:
\[ \boxed{\frac{1}{3}} \]