The tangent to the ellipse `3x^2+16y^2=12`, at the point `(1,3/4)` intersects the curves `y^2+x=0` at :

Options

A. no point
B. exactly one point
C. two distinct point
D. more than two points.

Answer

To solve the problem, we need to find the points of intersection between the tangent to the ellipse \(3x^2 + 16y^2 = 12\) at the point \((1, \frac{3}{4})\) and the curve \(y^2 + x = 0\).

### Step 1: Write the equation of the ellipse in standard form
The given equation of the ellipse is:
\[ 3x^2 + 16y^2 = 12 \]
Dividing by 12, we get:
\[ \frac{x^2}{4} + \frac{y^2}{\frac{3}{4}} = 1 \]
This shows that \(a^2 = 4\) and \(b^2 = \frac{3}{4}\).

### Step 2: Identify the values of \(a\) and \(b\)
From the standard form, we have:
- \(a = 2\)
- \(b = \frac{\sqrt{3}}{2}\)

### Step 3: Write the equation of the tangent line at the point \((1, \frac{3}{4})\)
The equation of the tangent to the ellipse at the point \((x_1, y_1)\) is given by:
\[ \frac{xx_1}{a^2} + \frac{yy_1}{b^2} = 1 \]
Substituting \(x_1 = 1\), \(y_1 = \frac{3}{4}\), \(a^2 = 4\), and \(b^2 = \frac{3}{4}\):
\[ \frac{x \cdot 1}{4} + \frac{y \cdot \frac{3}{4}}{\frac{3}{4}} = 1 \]
This simplifies to:
\[ \frac{x}{4} + y = 1 \]
Multiplying the entire equation by 4 gives:
\[ x + 4y = 4 \]

### Step 4: Substitute the tangent equation into the curve equation
The curve equation is:
\[ y^2 + x = 0 \implies x = -y^2 \]
Substituting \(x = -y^2\) into the tangent equation:
\[ -y^2 + 4y = 4 \]
Rearranging gives:
\[ y^2 - 4y + 4 = 0 \]

### Step 5: Solve the quadratic equation
Factoring the quadratic:
\[ (y - 2)^2 = 0 \]
Thus, we find:
\[ y = 2 \]
Substituting \(y = 2\) back into \(x = -y^2\):
\[ x = -2^2 = -4 \]
So the point of intersection is \((-4, 2)\).

### Conclusion
The tangent to the ellipse at the point \((1, \frac{3}{4})\) intersects the curve \(y^2 + x = 0\) at exactly one point, which is \((-4, 2)\).