202210181552 Solution to 1971-CE-AMATH-I-XX

Show that the straight line $x+2y+4=0$ touches the curve $y^2=4x$ .

Roughwork.

If there exists some point $(a,b)$ of intersection of the straight line and the curve, we have

$\begin{cases} a+2b+4 = 0 \\ b^2 = 4a \\ \end{cases}$

Substituting $-2b-4$ for $a$ in the second equation, we have

$\begin{aligned} & b^2 = 4(-2b-4) \\ \Leftrightarrow\quad & b^2 + 8b +16 = 0 \\ \Leftrightarrow\quad & (b+4)^2 = 0 \\ \Leftrightarrow\quad & b = -4\quad\textrm{(rep.)} \\ \Rightarrow\quad & a = -2(-4)-4= 4 \\ \end{aligned}$

$\begin{cases} a = 4 \\ b = -4 \\ \end{cases}$

Such point as $(4,-4)$ exists.

$\because$ The intersection of $\begin{cases} L: x+2y+4=0 \\ C: y^2 = 4x \\ \end{cases}$ is non-empty.

$\therefore$ The straight line touches the curve.

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