If from point `P(3,3)` on the hyperbola a normal is drawn which cuts x-axis at `(9,0)` on the hyperbola `x^2/a^2-y^2/b^2=1` the value of `(a^2,e^2)` is

Options

A. `(9/2,3)`
B. `(9/2,0)`
C. `(9,3)`
D. `(3,9/2)`

Answer

To solve the problem step by step, we need to find the values of \( a^2 \) and \( e^2 \) for the hyperbola given the point \( P(3,3) \) and the normal that intersects the x-axis at \( (9,0) \).

### Step 1: Write the equation of the hyperbola
The equation of the hyperbola is given as:
\[ \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \]

### Step 2: Substitute the point \( P(3,3) \) into the hyperbola equation
Substituting \( x = 3 \) and \( y = 3 \) into the hyperbola equation:
\[ \frac{3^2}{a^2} - \frac{3^2}{b^2} = 1 \]
This simplifies to:
\[ \frac{9}{a^2} - \frac{9}{b^2} = 1 \]
Multiplying through by \( a^2b^2 \) gives:
\[ 9b^2 - 9a^2 = a^2b^2 \]
Rearranging this, we have:
\[ 9b^2 = a^2b^2 + 9a^2 \]

### Step 3: Find the slope of the tangent at point \( P(3,3) \)
The slope of the tangent to the hyperbola at any point \( (x,y) \) can be found using implicit differentiation:
\[ \frac{dy}{dx} = \frac{b^2 x}{a^2 y} \]
At point \( P(3,3) \):
\[ \frac{dy}{dx} = \frac{b^2 \cdot 3}{a^2 \cdot 3} = \frac{b^2}{a^2} \]

### Step 4: Find the slope of the normal
The slope of the normal is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{a^2}{b^2} \]

### Step 5: Write the equation of the normal line
Using the point-slope form of the line equation:
\[ y - 3 = -\frac{a^2}{b^2}(x - 3) \]
Rearranging gives:
\[ y = -\frac{a^2}{b^2}x + \left(3 + \frac{3a^2}{b^2}\right) \]

### Step 6: Find the intersection with the x-axis
To find where this normal intersects the x-axis, set \( y = 0 \):
\[ 0 = -\frac{a^2}{b^2}x + \left(3 + \frac{3a^2}{b^2}\right) \]
Solving for \( x \):
\[ \frac{a^2}{b^2}x = 3 + \frac{3a^2}{b^2} \]
\[ x = \frac{b^2(3 + \frac{3a^2}{b^2})}{a^2} = \frac{3b^2 + 3a^2}{a^2} = 9 \]
Since we know the normal intersects the x-axis at \( (9,0) \), we have:
\[ 9 = 3 + \frac{3a^2}{b^2} \]
This simplifies to:
\[ 6 = \frac{3a^2}{b^2} \]
Thus:
\[ b^2 = \frac{a^2}{2} \]

### Step 7: Substitute \( b^2 \) back into the hyperbola equation
Substituting \( b^2 = \frac{a^2}{2} \) into the equation from Step 2:
\[ 9 \cdot \frac{a^2}{2} = a^2 \cdot \frac{a^2}{2} + 9a^2 \]
This simplifies to:
\[ \frac{9a^2}{2} = \frac{a^4}{2} + 9a^2 \]
Multiplying through by 2 to eliminate the fraction:
\[ 9a^2 = a^4 + 18a^2 \]
Rearranging gives:
\[ a^4 + 9a^2 = 0 \]
Factoring out \( a^2 \):
\[ a^2(a^2 + 9) = 0 \]
Since \( a^2 \) cannot be negative, we have:
\[ a^2 = 9 \]

### Step 8: Find \( e^2 \)
Using the relationship \( b^2 = \frac{a^2}{2} \):
\[ b^2 = \frac{9}{2} \]
The eccentricity \( e \) of the hyperbola is given by:
\[ e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{\frac{9}{2}}{9} = 1 + \frac{1}{2} = \frac{3}{2} \]

### Final Result
Thus, the values of \( (a^2, e^2) \) are:
\[ (a^2, e^2) = \left(9, \frac{3}{2}\right) \]