1. **State the problem:** Find the possible values of $a$ such that the tangent to the curve $y = x^2 - 3$ at the point $(a,b)$ passes through the point $(0,-12)$. 2. **Recall the formula for the tangent line:** The slope of the tangent line to the curve at $x = a$ is given by the derivative: $$y' = \frac{dy}{dx} = 2x$$ So the slope at $x = a$ is: $$m = 2a$$ 3. **Find the coordinates of the point of tangency:** Since the point lies on the curve, $$b = a^2 - 3$$ 4. **Equation of the tangent line at $(a,b)$:** Using point-slope form: $$y - b = m(x - a)$$ Substitute $m = 2a$ and $b = a^2 - 3$: $$y - (a^2 - 3) = 2a(x - a)$$ 5. **The tangent passes through $(0,-12)$:** Substitute $x=0$, $y=-12$: $$-12 - (a^2 - 3) = 2a(0 - a)$$ Simplify left side: $$-12 - a^2 + 3 = -2a^2$$ $$-9 - a^2 = -2a^2$$ 6. **Solve for $a$:** Bring all terms to one side: $$-9 - a^2 + 2a^2 = 0$$ $$-9 + a^2 = 0$$ $$a^2 = 9$$ 7. **Find possible values of $a$:** $$a = \pm 3$$ **Final answer:** The possible values of $a$ are $3$ and $-3$.