Write down the equation of the tangent to the parabola $x^2 =8y$ at the point $(4p,2p^2)$ on it. Full question in description

Question

Write down the equation of the tangent to the parabola $x^2 =8y$ at the point $(4p,2p^2)$ on it. If the point $(3,1)$ is to lie on this tangent, find the values which $p$ may take. Hence deduce the equations of the tangents which may be drawn from the point $(3,1)$ to the parabola $x^2=8y$.

Answer 1

Notice, the slope of the tangent to the parabola: $x^2=8y$ $$8\frac{dy}{dx}=2x\iff \frac{dy}{dx}=\frac{x}{4}$$

Hence, the slope of tangent at the point $(4p, 2p^2)$ is $$m=\frac{4p}{4}=p$$ But the slope of the tangent passing through $(4p, 2p^2)$ & $(3, 1)$ is also determined as follows $$m=\frac{2p^2-1}{4p-3}$$ $$\implies \frac{2p^2-1}{4p-3}=p$$ $$2p^2-1=4p^2-3p\iff 2p^2-3p+1=0$$ $$(2p-1)(p-1)=0$$ $$\iff 2p-1=0\iff p=\frac{1}{2}$$ or $$\iff p-1=0\iff p=1$$ Hence, we get

$$\bbox[5pt, border:2.5pt solid #FF0000]{\color{blue}{p=\left\{ \frac{1}{2}, 1\right\}}}$$