The normal at $$\left(2, \frac{3}{2}\right)$$ to the ellipse $$\frac{x^2}{16} + \frac{y^2}{3} = 1$$ touches a parabola, whose equation is

The ellipse is $$\frac{x^2}{16}+\frac{y^2}{3}=1$$ and the given point on it is $$P\left(2,\frac32\right)$$.

1. Slope of the tangent at $$P$$
Differentiate the ellipse implicitly:
$$\frac{x^2}{16}+\frac{y^2}{3}=1 \;\Rightarrow\; \frac{x}{8}+\frac{2y}{3}\,\frac{dy}{dx}=0$$
$$\therefore\;\frac{dy}{dx}= -\,\frac{3x}{16y}$$

At $$P(2,\tfrac32)$$,
$$m_{\text{tangent}} = -\,\frac{3(2)}{16\left(\tfrac32\right)} = -\,\frac{6}{24} = -\frac14$$

2. Equation of the normal at $$P$$
Slope of the normal: $$m_{\text{normal}} = -\frac{1}{m_{\text{tangent}}}=4$$
Using point-slope form:
$$y-\frac32 = 4(x-2)$$
$$\Rightarrow\; y = 4x - \frac{13}{2}$$
This is the line which must be a tangent to the required parabola.

3. Slope form of a tangent to a standard parabola
• For $$y^2 = 4ax$$ (opening to the right): the tangent with slope $$m$$ is $$y = mx + \frac{a}{m}$$.
• For $$y^2 = -4ax$$ (opening to the left): the tangent with slope $$m$$ is $$y = mx - \frac{a}{m}$$.

Our line has slope $$m = 4$$ and intercept $$c = -\frac{13}{2} = -6.5$$. Comparing:

• If the parabola were $$y^2 = 4ax$$: intercept should be $$\frac{a}{m} = \frac{a}{4}$$, which is positive. Our intercept is negative, so this case is impossible.

• If the parabola is $$y^2 = -4ax$$: intercept should be $$-\frac{a}{m} = -\frac{a}{4}$$.
Set $$-\frac{a}{4} = -6.5 \Longrightarrow a = 26$$.

Hence $$4a = 104$$ and the parabola is $$y^2 = -104x$$.

4. Checking the options
Option A $$y^2 = -104x$$ matches exactly.
The other three options do not give the required intercept for $$m = 4$$.

Therefore, the parabola is represented by:

Option A which is: $$y^2 = -104x$$