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My method is to find the equation of the chord wrt to the point $P$ and the midpoint, say $O(\alpha,\beta)$ and then compare them.

Equation of the the chord wrt to $P$ is$^1$: $$T=0 \text{ where } T=12x+20y-1.$$

So the equation of the chord is: $$12x+20y-1=0\tag1$$

The equation of the chord wrt to $O$ is$^2$

$$T^{'}=S_1 \text{ where } T^{'}=3\alpha x+4\beta y-1 \text{ and } S_1=3\alpha^2+4\beta ^2-1. $$

So the equation of the chord is: $$3\alpha x+4\beta y-1-3\alpha^2-4\beta ^2+1=0\tag2$$

Now since $(1) \text{ and } (2) $ are equations of same line then they are equivalent then we have on comparing $$ O(\alpha,\beta)=O(4,5). $$

Which is wrong and absurd, but why? There seems to be no mistake in this method. The answer is $O(\frac{1}{37},\frac{5}{148}).$

The diagram of the question is here.

1: $T=0$ is true for all conics, and hence for circles as well.

2: $T=S_1$ is true for all conic curves in general, but I found this shown online for parabolas only.

Osmium
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2 Answers2

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It is appropriate to consider the chord of contact of $$3x^2+4y^2=1 \tag*{(1)} $$

from $P(4,5)$, which is $$12x+20y=1 \tag*{(2)} $$ Then we can find their intersection by solving (1) and (2).

Elimination yields $$ 3\left(\frac{1-20 y}{12}\right)^{2}+4 y^{2}=1 $$ $$y=-\frac{1}{4} \text{ or } \frac{47}{148} $$ Putting back to (2) yields the corresponding values of $$x=\frac{1}{2} \text{ and }-\frac{33}{74} $$ Therefore the mid-point of contact is $$ \left(\frac{1}{2}\left(\frac{1}{2}-\frac{33}{74}\right), \frac{1}{2}\left(-\frac{1}{4}+\frac{47}{148}\right)\right)= \left(\frac{1}{37}, \frac{5}{148}\right). $$ Diagram for reference: enter image description here

Lai
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  • Your answer is correct. But my main question was why my method is not working. I should have clarified that in the post. Thank you. – Osmium Dec 12 '21 at 09:34
  • I don’t understand your equation (2)1 – Lai Dec 12 '21 at 10:08
  • Have you checked this link? https://math.stackexchange.com/questions/3249671/equation-of-chord-of-a-parabola-whose-midpoint-is-given This answer proves this theorm for parabola but can be genralsied for all conic curves. – Osmium Dec 12 '21 at 10:30
  • O, lying inside the ellipse, has no chord of contact! O only lies on the equation (1). – Lai Dec 12 '21 at 15:21
  • By chord of contact of the ellipse from $P$, I mean the chord which passes through the point of contact of the tangents to the ellipse from $P.$ – Osmium Dec 13 '21 at 03:25
  • $O$ is the mid point of the chord. And satisfies both the equations (1) and (2). – Osmium Dec 13 '21 at 03:27
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The chord of contact obtained from $T=0$ for $(4,5)$ is $$3x \cdot 4 + 4y \cdot 5=1$$ and from $T=S_1$ for $(\alpha, \beta)$ is $$3x \cdot \alpha+4y \cdot \beta=3\alpha^2+4\beta^2$$

Their coefficients must be proportional, so $$\frac{\alpha}{4}=\frac{\beta}{5}=\frac{3\alpha^2+4\beta^2}{1}$$

From first two of these simultaneous equations, $\beta = \frac{5}{4}\alpha$. Using this with first and third, $$\frac{\alpha}{4}=3\alpha^2+4\left( \frac{5}{4}\alpha \right)^2 \Rightarrow \alpha = \frac{1}{37}$$

Hence $(\alpha, \beta) = (\frac{1}{37},\frac{5}{4}\cdot \frac{1}{37})=(\frac{1}{37},\frac{5}{148})$.

Thus the method used by comparing $T=0$ and $T=S_1$ is completely correct.

MyMolecules
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