If normal at P on the curve `y^2 - 3x^2 + y + 10 = 0` passes through the point `(0, 3/2)` ,then slope of tangent at P is n. The value of |n| is equal to

Explanation

To find the slope of the tangent at point P on the curve given by the equation \( y^2 - 3x^2 + y + 10 = 0 \), we can follow these steps:

### Step 1: Differentiate the curve implicitly
We start with the equation of the curve:
\[ y^2 - 3x^2 + y + 10 = 0 \]
Differentiating both sides with respect to \( x \):
\[ \frac{d}{dx}(y^2) - \frac{d}{dx}(3x^2) + \frac{d}{dx}(y) + \frac{d}{dx}(10) = 0 \]
This gives:
\[ 2y \frac{dy}{dx} - 6x + \frac{dy}{dx} = 0 \]
Combining terms:
\[ (2y + 1) \frac{dy}{dx} = 6x \]
Thus, we can express the slope of the tangent \( \frac{dy}{dx} \):
\[ \frac{dy}{dx} = \frac{6x}{2y + 1} \]

### Step 2: Find the slope of the normal
The slope of the normal line at point P, which we denote as \( (h, k) \), is the negative reciprocal of the slope of the tangent:
\[ \text{slope of normal} = -\frac{1}{\frac{dy}{dx}} = -\frac{2y + 1}{6x} \]
At point \( P(h, k) \):
\[ \text{slope of normal at } P = -\frac{2k + 1}{6h} \]

### Step 3: Use the condition that the normal passes through the point \( (0, \frac{3}{2}) \)
The normal line at point P can be expressed using the point-slope form:
\[ y - k = \text{slope of normal} \cdot (x - h) \]
Substituting the point \( (0, \frac{3}{2}) \):
\[ \frac{3}{2} - k = -\frac{2k + 1}{6h} \cdot (0 - h) \]
This simplifies to:
\[ \frac{3}{2} - k = \frac{h(2k + 1)}{6h} = \frac{2k + 1}{6} \]
Multiplying through by 6 to eliminate the fraction:
\[ 6\left(\frac{3}{2} - k\right) = 2k + 1 \]
This results in:
\[ 9 - 6k = 2k + 1 \]
Rearranging gives:
\[ 9 - 1 = 8k \implies 8 = 8k \implies k = 1 \]

### Step 4: Substitute \( k \) back into the curve equation to find \( h \)
Substituting \( k = 1 \) into the original curve equation:
\[ 1^2 - 3h^2 + 1 + 10 = 0 \]
This simplifies to:
\[ 1 - 3h^2 + 1 + 10 = 0 \implies -3h^2 + 12 = 0 \implies 3h^2 = 12 \implies h^2 = 4 \implies h = \pm 2 \]

### Step 5: Find the slope of the tangent at point P
Now we have two possible points \( P(2, 1) \) and \( P(-2, 1) \). We can find the slope of the tangent using either point:
Using \( P(2, 1) \):
\[ \frac{dy}{dx} = \frac{6(2)}{2(1) + 1} = \frac{12}{3} = 4 \]
Using \( P(-2, 1) \):
\[ \frac{dy}{dx} = \frac{6(-2)}{2(1) + 1} = \frac{-12}{3} = -4 \]

### Step 6: Find the modulus of the slope of the tangent
The modulus of the slope \( n \) is:
\[ |n| = |4| = 4 \]

### Final Answer
Thus, the value of \( |n| \) is \( \boxed{4} \).