What is the midpoint of the chord of contact form a point $P(4,5)$ to the curve $3x^2+4y^2=1?$

Question

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 $O(\frac{1}{37},\frac{5}{148}).$

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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.

Answer 1

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.

Answer 2

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: