A common tangent to the conics $$x^2 = 6y$$ and $$2x^2 - 4y^2 = 9$$ is:

We are given two conics: $$x^2 = 6y$$ (a parabola) and $$2x^2 - 4y^2 = 9$$ (a hyperbola). We need to find a common tangent from the given options. A tangent line touches a conic at exactly one point, so we will substitute each line into both conics and set the discriminant to zero for tangency.

First, rewrite the options in slope-intercept form $$y = mx + c$$ for easier substitution:

• Option A: $$x - y = \frac{3}{2}$$ → $$y = x - \frac{3}{2}$$ (slope $$m = 1$$, intercept $$c = -\frac{3}{2}$$)
• Option B: $$x + y = 1$$ → $$y = -x + 1$$ (slope $$m = -1$$, intercept $$c = 1$$)
• Option C: $$x + y = \frac{9}{2}$$ → $$y = -x + \frac{9}{2}$$ (slope $$m = -1$$, intercept $$c = \frac{9}{2}$$)
• Option D: $$x - y = 1$$ → $$y = x - 1$$ (slope $$m = 1$$, intercept $$c = -1$$)

Start with the parabola $$x^2 = 6y$$. Substitute $$y = mx + c$$:

$$x^2 = 6(mx + c)$$

$$x^2 = 6mx + 6c$$

$$x^2 - 6mx - 6c = 0$$

For tangency, the discriminant must be zero. Discriminant $$D = (-6m)^2 - 4(1)(-6c) = 36m^2 + 24c$$. Set $$D = 0$$:

$$36m^2 + 24c = 0$$

$$36m^2 = -24c$$

$$c = -\frac{36m^2}{24} = -\frac{3m^2}{2}$$

So, for the parabola, the condition is $$c = -\frac{3}{2}m^2$$.

Check each option against this condition:

• Option A: $$m = 1$$, $$c = -\frac{3}{2}$$. Compute $$-\frac{3}{2}(1)^2 = -\frac{3}{2}$$. Matches.
• Option B: $$m = -1$$, $$c = 1$$. Compute $$-\frac{3}{2}(-1)^2 = -\frac{3}{2}$$. But $$1 \neq -\frac{3}{2}$$, does not match.
• Option C: $$m = -1$$, $$c = \frac{9}{2}$$. Compute $$-\frac{3}{2}(-1)^2 = -\frac{3}{2}$$. But $$\frac{9}{2} \neq -\frac{3}{2}$$, does not match.
• Option D: $$m = 1$$, $$c = -1$$. Compute $$-\frac{3}{2}(1)^2 = -\frac{3}{2}$$. But $$-1 \neq -\frac{3}{2}$$, does not match.

Only Option A satisfies the tangency condition for the parabola.

Now, check if Option A is also tangent to the hyperbola $$2x^2 - 4y^2 = 9$$. Substitute $$y = x - \frac{3}{2}$$:

$$2x^2 - 4\left(x - \frac{3}{2}\right)^2 = 9$$

First, expand $$\left(x - \frac{3}{2}\right)^2 = x^2 - 3x + \frac{9}{4}$$:

$$2x^2 - 4\left(x^2 - 3x + \frac{9}{4}\right) = 9$$

$$2x^2 - 4x^2 + 12x - 9 = 9$$ (since $$4 \times \frac{9}{4} = 9$$)

$$-2x^2 + 12x - 9 = 9$$

Bring all terms to one side:

$$-2x^2 + 12x - 9 - 9 = 0$$

$$-2x^2 + 12x - 18 = 0$$

Multiply by $$-1$$ to simplify:

$$2x^2 - 12x + 18 = 0$$

Divide by 2:

$$x^2 - 6x + 9 = 0$$

Discriminant $$D = (-6)^2 - 4(1)(9) = 36 - 36 = 0$$. Since discriminant is zero, there is exactly one solution, so the line is tangent to the hyperbola.

Thus, Option A is tangent to both conics. For completeness, we can verify that the other options are not tangent to the hyperbola (though they already fail the parabola condition):

• Option B: Substitute $$y = -x + 1$$ into hyperbola: $$2x^2 - 4(-x + 1)^2 = 9$$. Expand $$(-x + 1)^2 = x^2 - 2x + 1$$, so $$2x^2 - 4(x^2 - 2x + 1) = 9$$ → $$2x^2 - 4x^2 + 8x - 4 = 9$$ → $$-2x^2 + 8x - 13 = 0$$. Discriminant $$8^2 - 4(-2)(-13) = 64 - 104 = -40 < 0$$, no real solution.
• Option C: Substitute $$y = -x + \frac{9}{2}$$ into hyperbola: $$2x^2 - 4\left(-x + \frac{9}{2}\right)^2 = 9$$. Expand $$\left(-x + \frac{9}{2}\right)^2 = x^2 - 9x + \frac{81}{4}$$, so $$2x^2 - 4\left(x^2 - 9x + \frac{81}{4}\right) = 9$$ → $$2x^2 - 4x^2 + 36x - 81 = 9$$ → $$-2x^2 + 36x - 90 = 0$$. Multiply by $$-1$$: $$2x^2 - 36x + 90 = 0$$ → divide by 2: $$x^2 - 18x + 45 = 0$$. Discriminant $$(-18)^2 - 4(1)(45) = 324 - 180 = 144 > 0$$, two solutions.
• Option D: Substitute $$y = x - 1$$ into hyperbola: $$2x^2 - 4(x - 1)^2 = 9$$. Expand $$(x - 1)^2 = x^2 - 2x + 1$$, so $$2x^2 - 4(x^2 - 2x + 1) = 9$$ → $$2x^2 - 4x^2 + 8x - 4 = 9$$ → $$-2x^2 + 8x - 13 = 0$$. Discriminant same as Option B: $$64 - 104 = -40 < 0$$, no real solution.

Hence, only Option A is a common tangent to both conics. So, the answer is Option A.