Two conics `(x^(2))/(a^(2))-(y^(2))/(b^(2))=1 and x^(2)=-(a)/(b)y` intersect, if

Options

A. `0ltble(1)/(2)`
B. `0ltale(1)/(2)`
C. `a^(2)ltb^(2)`
D. `a^(2)gtb^(2)`

Answer

To solve the problem of finding the conditions under which the two conics intersect, we will follow these steps:

### Step 1: Write down the equations of the conics
The equations given are:
1. \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) (Equation of a hyperbola)
2. \(x^2 = -\frac{a}{b}y\) (Equation of a parabola)

### Step 2: Substitute the second equation into the first
We can substitute \(x^2\) from the second equation into the first equation:
\[ \frac{-\frac{a}{b}y}{a^2} - \frac{y^2}{b^2} = 1 \]
This simplifies to:
\[ -\frac{y}{ab} - \frac{y^2}{b^2} = 1 \]

### Step 3: Rearrange the equation
Multiply through by \(b^2\) to eliminate the denominators:
\[ -b y - y^2 = b^2 \]
Rearranging gives:
\[ y^2 + b y + b^2 = 0 \]

### Step 4: Analyze the quadratic equation
For the two conics to intersect, the quadratic equation must have real solutions. This occurs when the discriminant \(D\) is greater than or equal to zero:
\[ D = b^2 - 4ac \]
Here, \(a = 1\), \(b = b\), and \(c = b^2\):
\[ D = b^2 - 4(1)(b^2) = b^2 - 4b^2 = -3b^2 \]

### Step 5: Set the discriminant condition
For the discriminant to be non-negative:
\[ -3b^2 \geq 0 \]
This implies:
\[ b^2 \leq 0 \]
Since \(b^2\) is always non-negative, the only solution is:
\[ b = 0 \]

### Step 6: Conclusion about the intersection
Since \(b\) cannot be zero (as it appears in the denominator of the hyperbola equation), we conclude that the two conics do not intersect for any real values of \(a\) and \(b\) where \(b \neq 0\).

### Summary of Conditions
The conics intersect if:
- \(b\) must be greater than 0 and less than or equal to \(\frac{1}{2}\) for the hyperbola to exist.