Calculate the distance between intersection points of tangents to a parabola

Question

• Question

  Tangent lines $T_1$ and $T_2$ are drawn at two points $P_1$ and $P_2$ on the parabola $y=x^2$ and they intersect at a point $P$. Another tangent line $T$ is drawn at a point between $P_1$ and $P_2$; it intersects $T_1$ at $Q_1$ and $T_2$ at $Q_2$. Show that $$\frac{|PQ_1|}{|PP_1|} + \frac{|PQ_2|}{|PP_2|} = 1$$

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• My attempt at the question

  I include a possible scenario of the graph for convenience'(I hope) sake:

  The outer two tangents are tangents $T_1$ and $T_2$, and the inner tangent is tangent $T$.

  • Since points $P_1$ and $P_2$ are points on the parabola, I can give them coordinates as follows, $$\tag 1 P_1(P_{1x}, P^2_{1x})$$ and $$\tag 2 P_2(P_{2x}, P^2_{2x})$$

• Using $y\prime = 2x$, I calculate the equations for the tangents $T_1$ and $T_2$ respectively, they are,

  $$T_1 = y = 2P_{1x}(x - P_{1x}) + P^2_{1x}$$

  and

  $$T_2 = y = 2P_{2x}(x - P_{2x}) + P^2_{2x}$$

  • By setting $T_1 = T_2$ and then solving for $x$ I show that the two tangents intersect at a point $x = \frac{P_{1x} + P_{2x}}{2}$, which in words is the two tangents to the parabola is halfway between points $P_1$ and $P_2$.

    Then substituting $x = \frac{P_{1x} + P_{2x}}{2}$ into any of the tangent line equations I get the $y$ coordinate of the tangent lines' intersection, which is $y = P_{1x}\cdot P_{2x}$

    Now I have the coordinates for point $P$, that is $$\tag 3 P\Big(\frac{P_{1x} + P_{2x}}{2}, P_{1x}\cdot P_{2x}\Big)$$

  • To get coordinates for points $Q_1$ and $Q_2$ I will substitute $Q_{1x}$ in to the equation of tangent $T_1$ and substitute $Q_{2x}$ in to the equation of tangent $T_2$.

    That yields the following for coordinates: $$\tag 4 Q_1(Q_{1x}, \,\,2P_{1x}Q_{1x} - P_{1x}^2)$$ $$\tag 5 Q_2(Q_{2x}, \,\,2P_{2x}Q_{2x} - P_{2x}^2)$$

  • Since I have all the points necessary to calculate $\frac{|PQ_1|}{|PP_1|} + \frac{|PQ_2|}{|PP_2|}$, I feel inclined to apply the distance formula. Doing so yielded the following:

  $$\tag 6 |PQ_1| = \frac{\sqrt{(4P_{1x}^2 + 1)(P_{1x} + P_{2x} - 2Q_{1x})^2}}{2}$$

  $$\tag 7 |PP_1| = \frac{\sqrt{(4P_{1x}^2 + 1)(P_{1x} - P_{2x})^2}}{2}$$

  $$\tag 8 |PQ_2| = \frac{\sqrt{(4P_{2x}^2 + 1)(P_{1x} + P_{2x} - 2Q_{2x})^2}}{2}$$

  $$\tag 9 |PP_2| = \frac{\sqrt{(4P_{1x}^2 + 1)(P_{1x} - P_{2x})^2}}{2}$$

  • Now I calculate $\frac{|PQ_1|}{|PP_1|} + \frac{|PQ_2|}{|PP_2|}$ using the above: $$\frac{|PQ_1|}{|PP_1|} + \frac{|PQ_2|}{|PP_2|} = \frac{\frac{\sqrt{(4P_{1x}^2 + 1)(P_{1x} + P_{2x} - 2Q_{1x})^2}}{2}}{\frac{\sqrt{(4P_{1x}^2 + 1)(P_{1x} - P_{2x})^2}}{2}} + \frac{\frac{\sqrt{(4P_{2x}^2 + 1)(P_{1x} + P_{2x} - 2Q_{2x})^2}}{2}}{\frac{\sqrt{(4P_{2x}^2 + 1)(P_{1x} - P_{2x})^2}}{2}}$$ $$\tag {10} =\frac{\sqrt{(P_{1x} + P_{2x} - 2Q_{1x})^2} + \sqrt{(P_{1x} + P_{2x} - 2Q_{2x})^2}}{\sqrt{(P_{1x} - P_{2x})^2}}$$

  • I can't seem to find a way to show that $(10)$ is equal to $1$. I have, however tested a few instances and it held up, for what it's worth. But for now, I'm at a loss as to how to proceed.

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Any hints, suggestions, or alternative approaches?

Answer 1

Suppose that the third tangent is drawn at a point $A$ with coordinates $A(a, a^2)$. Then its tangent intersects $T_{i}$ at

$$ Q_{i}\left(\frac{P_{ix} + a}{2}, P_{ix}a \right) $$

using your equation $(3)$. In other words,

$$ 2(Q_{1x} - Q_{2x})=P_{1x}-P_{2x} $$

Therefore, in the argument of the square root in the numerator of the first term of (10),

\begin{align*} P_{1x} + P_{2x} - 2Q_{1x} &= P_{1x} + P_{2x} - 2[(P_{1x} - P_{2x})/2 + Q_{2x}] \\ &= 2(P_{2x} - Q_{2x}) \end{align*}

In general, $\sqrt{x^2} = |x|$ so we need to figure out the relative size of all of this. Assume without loss of generality that $P_{2x} > P_{1x}$ so the denominator of $(10)$ is $P_{2x} - P_{1x}$. Clearly $P_{2x} > Q_{2x}$ so the first term in the numerator is $2(P_{2x} - Q_{2x})$. Finally,

\begin{align*} P &< Q_{2x} \\ (P_{1x} + P_{2x})/2 &< Q_{2x} \\ P_{1x} + P_{2x} - 2Q_{2x} &< 0 \end{align*}

so the second term in the numerator is $2Q_{2x} - P_{1x} - P_{2x}.$ Putting this all together:

\begin{align*} \frac{|P Q_{1}|}{|P P_{1}|} + \frac{|P Q_{2}|}{|P P_{2}|} &= \frac{2(P_{2x} - Q_{2x})+ (2Q_{2x} - P_{1x} - P_{2x})}{P_{2x} - P_{1x}} \\ &= \frac{(2P_{2x} - P_{2x}) - P_{1x} + (2Q_{2x} - 2Q_{2x})}{P_{2x} - P_{1x}} \\ &= \frac{P_{2x} - P_{1x}}{P_{2x} - P_{1x}} \\ &= 1 \end{align*}