Day: October 18, 2022

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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.