📂Geometry

Equation of the Tangent to a Parabola

Derivation

In the Case Where the Slope is Given

Let’s first look at the case where the slope is given.

When the equation of the line tangent to the parabola $y^{ 2 }=4px$ is $y=mx+n$, the two shapes must meet at only one point, thus $$(mx+n)^{ 2 }=4px \implies m^{ 2 }x^{ 2 }+2(mn-2p)x+n^{ 2 }=0$$ by the quadratic formula, $$\frac { D }{ 4 }=m^{ 2 }n^{ 2 }-4mnp+4p^{ 2 }-m^{ 2 }n^{ 2 }=0$$ simplifying this gives $n=\frac { p }{ m }$ and substituting this into the equation of the line, the equation of the line tangent to the parabola is found as follows. $$y=mx+\frac { p }{ m }$$

In the Case Where a Point is Given

Next is the case where a point is given. However, the original rigorous proof is overly simplistic and not very helpful in the derivation, so we introduce a different derivation using differentiation, which is a bit loose.

The line $y=mx+\frac { p }{ m }$ is a tangent to the parabola $y^{ 2 }=4px$. Differentiating $y^{ 2 }=4px$ with respect to $x$ gives $2y\prime y=4p$. At point $(x_{ 1 },y_{ 1 })$ on the parabola, $y\prime =\frac { 2p }{ y_{ 1 } }$ $y\prime$ is equal to the slope of line $y=mx+\frac { p }{ m }$, hence $$y=\frac { 2p }{ y_{ 1 } }x+\frac { y_{ 1 } }{ 2p }p$$ Multiplying both sides of the above equation by $y_{ 1 }$ gives $$y_{ 1 }y=2px+\frac { y_{ 1 }^{ 2 } }{ 2 }$$ . Since point $(x_{ 1 },y_{ 1 })은$ is on the parabola, substituting $y_{ 1 }^{ 2 }=4px_{ 1 }$ into the above equation gives $$y_{ 1 }y=2px+\frac { 4px_{ 1 } }{ 2 }$$ . Therefore, the equation of the tangent line passing through point $(x_{ 1 },y_{ 1 })$ is found as follows. $$y_{ 1 }y=2p(x_{ 1 }+X)$$