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Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse $\frac{x^2}{4}+\frac{y^2}{9}=1$ such that all lines touch a standard ellipse. Find $e$ for the ellipse

Let the point $(h,k)$ lie on the given circle

The chord of the contact drawn to the given ellipse is

$$\frac{hx}{4}+\frac{ky}{9}-1=0$$

This line is coincident with the the tangent to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$

$$y=mx\pm \sqrt{a^2m^2+b^2}$$

Then comparing the two equations

$$m=\frac{-9h}{4k}$$

And $$\frac{81}{k^2}=a^2m^2+b^2$$

$$\frac{81}{k^2}=\frac{81a^2h^2}{16k^2}+b^2$$

$$(81)(16)=81a^2h^2+16k^2b^2$$

How do I proceed from here? Simply substituting $h^2=100-k^2$ doesn’t give any details for $a$ and $b$

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  • $\begingroup$ The cord of contact envelops solve([diff(h*x/4+sqrt(100-h^2)*y/9-1,h),h*x/4+sqrt(100-h^2)*y/9-1],[x,y]); [[x = h/25,y = (9*sqrt(100-h^2))/100]] which means the sought after ellipse is $2025x^2+400y^2-324=0,$ or $(x/(2/5))^2 +(y/(9/10))^2=1.$ $\endgroup$ Jun 1 '20 at 13:37
  • $\begingroup$ @Jan-MagnusØkland I don’t mean to be rude, but I have no idea on what you just said $\endgroup$
    – Aditya
    Jun 1 '20 at 14:21
  • $\begingroup$ I tend to give solutions cryptically in comments to give a taste. Do you want it written out as an answer? $\endgroup$ Jun 1 '20 at 14:24
  • $\begingroup$ @Jan-MagnusØkland yes please $\endgroup$
    – Aditya
    Jun 1 '20 at 14:35
  • $\begingroup$ @Aditya But this is the same as your previous question: find max and min distance from the center and $e=\sqrt{\text{max}^2-\text{min}^2}/\text{max}$. $\endgroup$ Jun 1 '20 at 15:31
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According to $\frac{hx}{4}+\frac{ky}{9}=1$, the horizontal and vertical lines corresponding to circular points $(h,k) = (0,10),\> (10,0)$ are $ y= \frac9{10}$, $ x = \frac4{10}$. which also corresponds to the elliptical axes $a= \frac4{10}$ and $b= \frac9{10}$. Thus, the equation of the standard ellipse is $$\frac{x^2}{(\frac4{10})^2}+\frac{y^2}{(\frac9{10})^2}=1$$

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  • $\begingroup$ I thought it was presumed that $a>b$ for standard ellipse, but I was wrong I guess. $\endgroup$
    – Aditya
    Jun 2 '20 at 6:48
  • $\begingroup$ That’s actually a really elegant way of solving it. Just out of curiosity, is this the standard approach or is there another ‘caveman’ like method of solving it? $\endgroup$
    – Aditya
    Jun 2 '20 at 14:53
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    $\begingroup$ @Aditya - this is a shortcut given that the elliptical shape is known. Otherwise, the shape has be to discovered with more in involved method $\endgroup$
    – Quanto
    Jun 3 '20 at 1:44
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What you are looking for is the curve that the chord of contact envelops. You can find (one half of) the solution ellipse by letting $k=\sqrt{100-h^2}$ and substituting into the chord of contact $\frac{hx}{4}+\frac{ky}{9}-1=0.$ The curve the family of lines $\frac{hx}{4}+\frac{\sqrt{100-h^2}y}{9}-1=0$ envelops can be found by the method in the wikipedia link above. In maxima CAS

eq1:h*x/4+sqrt(100-h^2)*y/9-1;
solve([diff(eq1,h),eq1],[x,y]);
[[x = h/25,y = (9*sqrt(100-h^2))/100]]

The answer is the parametrized curve

$$(x(h),y(h))=(\frac{h}{25}, \frac{9\sqrt{100-h^2}}{100}).$$

Implicitizing by putting $h=25x$ into $y^2=\frac{81(100-h^2)}{100^2}$ you get $2025x^2+400y^2-324=0,$ when you factor out a $5^2.$ This is $(x/(2/5))^2 +(y/(9/10))^2=1 $ in the standard form.

Now to answer the question an ellipse with semi-axes $\frac9{10},\frac25$ has eccentricity $\sqrt{1-b^2/a^2}=\sqrt{1-\frac{4/25}{81/100}}=\sqrt{65}/9.$

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  • $\begingroup$ The special condition was that the question has to be solved without differentiation, since that portion is a little beyond my scope. Can this be solved without derivatives, or is it required? $\endgroup$
    – Aditya
    Jun 1 '20 at 15:03
  • $\begingroup$ Tag the question algebra-precalculus then. $\endgroup$ Jun 1 '20 at 15:04
  • $\begingroup$ Thanks, I will keep that in mind. But can this be solved without calculus? $\endgroup$
    – Aditya
    Jun 1 '20 at 15:07
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Apply the same method that I did in this answer to a previous question of yours.

Using the usual parameterization of a circle, we get the one-parameter family of polar lines $$\frac52x\cos t+\frac{10}9y\sin t-1=0.$$ Equate coefficients with the generic line $\lambda x+\mu y+\tau=0$ and eliminate $t$ to obtain $$\frac4{25}\lambda^2+\frac{81}{100}\mu^2-\tau^2=0.$$ This conic is dual to $$\frac{25}4x^2+\frac{100}{81}y^2=1.$$

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