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

Options

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

Answer

To solve the problem step by step, we will find the equation of the tangent to the ellipse at the given point and then determine where this tangent intersects the line \( y + x = 0 \).

### Step 1: Write the equation of the ellipse in standard form.
The given equation of the ellipse is:
\[ 3x^2 + 16y^2 = 12 \]
Dividing the entire equation by 12 gives:
\[ \frac{x^2}{4} + \frac{y^2}{\frac{3}{4}} = 1 \]
This shows that the ellipse is centered at the origin with semi-major axis \( a = 2 \) and semi-minor axis \( b = \frac{\sqrt{3}}{2} \).

### Step 2: Verify that the point lies on the ellipse.
We need to check if the point \( (1, \frac{3}{4}) \) lies on the ellipse:
\[ 3(1)^2 + 16\left(\frac{3}{4}\right)^2 = 3 + 16 \cdot \frac{9}{16} = 3 + 9 = 12 \]
Since the left-hand side equals the right-hand side, the point lies on the ellipse.

### Step 3: Find the slope of the tangent line at the point.
To find the slope of the tangent line, we differentiate the equation of the ellipse implicitly:
\[ \frac{d}{dx}(3x^2) + \frac{d}{dx}(16y^2) = \frac{d}{dx}(12) \]
This gives:
\[ 6x + 32y \frac{dy}{dx} = 0 \]
Solving for \( \frac{dy}{dx} \):
\[ 32y \frac{dy}{dx} = -6x \implies \frac{dy}{dx} = -\frac{6x}{32y} = -\frac{3x}{16y} \]
At the point \( (1, \frac{3}{4}) \):
\[ \frac{dy}{dx} = -\frac{3(1)}{16 \cdot \frac{3}{4}} = -\frac{3}{12} = -\frac{1}{4} \]

### Step 4: Write the equation of the tangent line.
Using the point-slope form of the line equation:
\[ y - y_1 = m(x - x_1) \]
Substituting \( m = -\frac{1}{4} \) and the point \( (1, \frac{3}{4}) \):
\[ y - \frac{3}{4} = -\frac{1}{4}(x - 1) \]
Simplifying this gives:
\[ y - \frac{3}{4} = -\frac{1}{4}x + \frac{1}{4} \]
\[ y = -\frac{1}{4}x + 1 \]

### Step 5: Find the intersection of the tangent line with the line \( y + x = 0 \).
The equation of the line \( y + x = 0 \) can be rewritten as:
\[ y = -x \]
Setting the two equations equal to each other:
\[ -\frac{1}{4}x + 1 = -x \]
Solving for \( x \):
\[ -\frac{1}{4}x + x = 1 \implies \frac{3}{4}x = 1 \implies x = \frac{4}{3} \]
Substituting back to find \( y \):
\[ y = -\frac{4}{3} \]

### Final Answer:
The tangent to the ellipse at the point \( (1, \frac{3}{4}) \) intersects the line \( y + x = 0 \) at the point:
\[ \left(\frac{4}{3}, -\frac{4}{3}\right) \]