Prove that the tangents to the curve y = (1+3x²)/(3+x²)

Prove that the tangents to the curve $y = \dfrac{1+3x^{2}}{3+x^{2}}$ drawn at the points where $y=1$ intersect at the origin.

Step 1: The points where $y=1$

\[ \frac{1+3x^{2}}{3+x^{2}} = 1 \;\Longrightarrow\; 1+3x^{2}=3+x^{2} \;\Longrightarrow\; x^{2}=1 \;\Longrightarrow\; x=\pm1. \tag{1} \]

The two points are $P_{1}=(1,1)$ and $P_{2}=(-1,1)$.

Step 2: Derivative

\[ y' = \frac{6x(3+x^{2}) - (1+3x^{2})\cdot2x}{(3+x^{2})^{2}} = \frac{x\bigl[6(3+x^{2})-2(1+3x^{2})\bigr]}{(3+x^{2})^{2}} = \frac{16x}{(3+x^{2})^{2}}. \tag{2} \]

Step 3: Slopes at the two points

\[ k_{1}=y'(1)=\frac{16}{16}=1, \qquad k_{2}=y'(-1)=\frac{-16}{16}=-1. \tag{3} \]

Step 4: The two tangents

\[ \text{At } P_{1}\ (\text{slope }1)\colon\ y-1=x-1 \;\Longrightarrow\; y=x. \tag{4} \]

\[ \text{At } P_{2}\ (\text{slope }-1)\colon\ y-1=-(x+1) \;\Longrightarrow\; y=-x. \tag{5} \]

Step 5: Intersection

Setting (4) equal to (5): $x=-x \Rightarrow x=0,\ y=0$. The two tangents meet at the origin.

Step 6: Verification

\[P_1 \text{ on the curve: } \tfrac{1+3}{3+1}=1\ \checkmark; \qquad P_2 \text{ on the curve: } \tfrac{1+3}{3+1}=1\ \checkmark\]

\[\text{Origin on both tangents: } y(0)=0 \text{ in (4) and (5)}\ \checkmark\]