Tangent Quadratic 32C1Dd

1. **State the problem:** We are given a line $2y=2x+29$ and a quadratic equation $x^2 - 2ky + 32k = 0$ where $k \neq 0$. The line is tangent to the quadratic, and we need to find $k$ and the points where the quadratic crosses the y-axis. 2. **Rewrite the line equation:** Divide both sides by 2: $$y = x + \frac{29}{2}$$ 3. **Substitute $y$ from the line into the quadratic:** Replace $y$ in $x^2 - 2ky + 32k = 0$ with $x + \frac{29}{2}$: $$x^2 - 2k\left(x + \frac{29}{2}\right) + 32k = 0$$ 4. **Expand and simplify:** $$x^2 - 2kx - 29k + 32k = 0$$ $$x^2 - 2kx + 3k = 0$$ 5. **Condition for tangency:** The line is tangent to the quadratic, so the quadratic in $x$ has exactly one solution. The discriminant $\Delta$ must be zero: $$\Delta = b^2 - 4ac = 0$$ Here, $a=1$, $b=-2k$, $c=3k$. Calculate discriminant: $$(-2k)^2 - 4(1)(3k) = 4k^2 - 12k = 0$$ 6. **Solve for $k$:** $$4k^2 - 12k = 0$$ Factor out $4k$: $$4k(k - 3) = 0$$ Since $k \neq 0$, we have: $$k - 3 = 0 \implies k = 3$$ 7. **Find where the quadratic crosses the y-axis:** At the y-axis, $x=0$. Substitute $x=0$ into the quadratic: $$0^2 - 2k y + 32k = 0 \implies -2k y + 32k = 0$$ Divide both sides by $-2k$ (non-zero): $$\cancel{-2k} y + \cancel{-2k} \cdot \frac{32k}{-2k} = 0 \implies y = 16$$ 8. **Final answers:** - $k = 3$ - The quadratic crosses the y-axis at $y = 16$