202212131509 Solution to 1985-CE-AMATH-II-6

Find the equations of the two tangents drawn from the point $(-1,0)$ to the parabola $y^2=4x$ .

Roughwork.

$\begin{aligned} y^2 & = 4x \\ 2y\,\mathrm{d}y & = 4\,\mathrm{d}x \\ \frac{\mathrm{d}y}{\mathrm{d}x} & = \frac{2}{y} \\ \end{aligned}$

Let $(a_1,b_1)$ and $(a_2,b_2)$ be two points at which the two tangents touches the parabola. Then,

$\displaystyle{\frac{2}{b_i} =\frac{\mathrm{d}y}{\mathrm{d}x}\bigg|_{(a_i,b_i)} =\frac{b_i-(0)}{a_i-(-1)}}$

By solving two equations in two unknowns:

$\begin{cases} b_i^2 = 2a_i+2 \\ b_i^2 = 4a_i \\ \end{cases} \Longrightarrow\quad \begin{pmatrix} a_i\\b_i \end{pmatrix}_{i=1,2} = \Bigg\{ \begin{pmatrix} 1\\ 2 \end{pmatrix} , \begin{pmatrix} 1 \\ -2 \end{pmatrix} \Bigg\}$

This problem is not to be attempted.

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