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**mr fantastic** (i) The point (1, 0) does not lie on the given curve. Do you mean that you need to find the equation of the tangent to $\displaystyle y^2 + 4x$ that passes through the point (1, 0)? In that case:

Let the point on the curve be $\displaystyle \left(- \frac{a^2}{4}, a \right)$. Then $\displaystyle m = - \frac{2}{a}$. But since the tangent passes through $\displaystyle \left(- \frac{a^2}{4}, a \right)$ and (1, 0) then $\displaystyle m = \frac{a}{-\frac{a^2}{4} - 1} = - \frac{4a}{a^2 + 4}$.

Therefore $\displaystyle - \frac{2}{a} = - \frac{4a}{a^2 + 4} \Rightarrow a = \pm 2$.

**Case 1: a = 2**

The point on the curve is (-1, 2) and m = -1. Therefore the equation of the tangent is $\displaystyle y - 2 = -1(x + 1) \Rightarrow y = -x + 1$.

**Case 2: a = -2**

The point on the curve is (-1, -2) and m = 1. Therefore the equation of the tangent is $\displaystyle y + 2 = 1(x + 1) \Rightarrow y = x - 1$.