Length of the chord of an ellipse whose midpoint is given

Question

Find the length of the chord of the ellipse $$\frac{x^2}{25}+\frac{y^2}{16}=1\tag1$$where mid point is $(\frac{1}{2},\frac{2}{3})$

My attempt: I know the equation of the chord of an ellipse whose mid-point is given is $T=S_1$,where T is $\left(\frac{x.x_1}{25}+\frac{y.y_1}{16} - 1\right)$ and $S_1$ is $\left(\frac{x_1^2}{25}+\frac{y_1^2}{16} - 1\right)$. Using this I get the equation of the chord as $$\frac{x}{50}+\frac{y}{24}=\frac{17}{450}\tag2$$ Now by solving (1) & (2), I should get the point of intersections and eventually the length of the chord. I want to know if there is a better way to solve this question.

Answer 1

Here’s a hybrid geometric/analytic approach:

Besant’s Proposition XX states that if $PCp$ and $DCd$ are conjugate diameters of an ellipse and $QV$ is an ordinate of $Pp$ (i.e., is parallel to $Dd$), then $$QV^2:PV\cdot Vp::CD^2:CP^2.$$

For our problem $V=\left(\frac12,\frac23\right)$ and $C$ is the origin. The chord length we seek is $2\cdot QV$.

The ellipse has semi-axis lengths $5$ and $4$. We can find the points $D$, $P$ and $p$ by mapping to a unit circle and then transforming the corresponding points back to our ellipse. We have $$V' = \left(\frac12\cdot\frac15,\frac23\cdot\frac14\right) = \left(\frac1{10},\frac1{6}\right).$$ Normalizing this gives $P'=\left({3\over\sqrt{34}},{5\over\sqrt{34}}\right)$ which becomes $P=\left(5\cdot{3\over\sqrt{34}},4\cdot{5\over\sqrt{34}}\right)=\left({15\over\sqrt{34}},{20\over\sqrt{34}}\right)$, and $p=-P$. We also have $D'=\left(-{5\over\sqrt{34}},{3\over\sqrt{34}}\right)$, so $D=\left(-{25\over\sqrt{34}},{12\over\sqrt{34}}\right)$. Computing $QV$ is then a matter of a few applications of the formula for the distance between two points and a bit of arithmetic. When all is said and done, this calculation yields a value of approximately $9.3302$ for the chord length.

Then again, we can construct the chord directly from $V'$ and $D'$ and then compute its length. The end points of the unit circle chord bisected by $V'$ are, by the Pythagorean theorem, $$V'\pm\sqrt{1-(OV')^2}\cdot\overrightarrow{O{D'}}$$ with chord length $2\sqrt{1-(OV')^2}\cdot OD'$. Affine transformations preserve segment length ratios of parallel lines, therefore the corresponding ellipse chord length is $2\sqrt{1-(OV')^2}\cdot OD$. Evaluating this expression is a bit less work than the one at top, and deriving it didn’t require ferreting out an obscure, albeit easy to prove analytically, property of ellipses.

Incidentally, an Internet search turns up an almost identical problem in various places, though with the point $\left(\frac12,\frac25\right)$ instead. That $y$-value produces much “nicer” numbers in the calculations. The chord length ends up being exactly $\frac75\sqrt{41}$.

Answer 2

Line passing through midpoint $(\tfrac{1}{2}, \tfrac{2}{3})$ is $$p(r)=\langle\tfrac{1}{2} + r \cos t, \tfrac{2}{3} + r \sin t\rangle \tag 1$$ where $r$ is the distance along the line and $t$ is the angle of slope of line.

Using your equation of chord, we can calculate $\cos t$ and $\sin t$:

$$\frac{x}{50}+\frac{y}{24}=\frac{17}{450} \tag 2$$

$\tan t =\frac{-24}{50} = \frac{-12}{25}$, so $\sin t= \frac{12}{\sqrt{25^2+12^2}}$ and $\cos t = \frac{-25}{\sqrt{25^2 + 12^2}}$

Substitute $\cos t$ and $\sin t$ in equation $1$, substitute $p(r)$ in the ellipse equation and solve for $r$. This gives a quadratic in $r$: $$r^2 = \dfrac{332977}{15300}$$

So the length of chord is $2r$, ie

$$2r = \frac{1}{15}\sqrt{\dfrac{332977}{17}} \approx 9.3302$$

Answer 3

Hint:

WLOG A$(5\cos t,4\sin t)$ and

B$\left(1-5\cos t,\dfrac43-4\sin t\right)$

Now use the fact that B lies on the given ellipse to find $t$