# Let chord of contact be drawn from every point on the circle $x^2+y^2=100$ to the ellipse [CONT..]

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 $$b$$

• 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 $(x/(2/5))^2 +(y/(9/10))^2=1.$ Jun 1 '20 at 13:37
• @Jan-MagnusØkland I don’t mean to be rude, but I have no idea on what you just said Jun 1 '20 at 14:21
• I tend to give solutions cryptically in comments to give a taste. Do you want it written out as an answer? Jun 1 '20 at 14:24
• @Jan-MagnusØkland yes please Jun 1 '20 at 14:35
• @Aditya But this is the same as your previous question: find max and min distance from the center and $e=\sqrt{\text{max}^2-\text{min}^2}/\text{max}$. Jun 1 '20 at 15:31

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 $$\frac{x^2}{(\frac4{10})^2}+\frac{y^2}{(\frac9{10})^2}=1$$

• I thought it was presumed that $a>b$ for standard ellipse, but I was wrong I guess. Jun 2 '20 at 6:48
• 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? Jun 2 '20 at 14:53
• @Aditya - this is a shortcut given that the elliptical shape is known. Otherwise, the shape has be to discovered with more in involved method Jun 3 '20 at 1:44

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 $$\sqrt{1-b^2/a^2}=\sqrt{1-\frac{4/25}{81/100}}=\sqrt{65}/9.$$

• 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? Jun 1 '20 at 15:03
• Tag the question algebra-precalculus then. Jun 1 '20 at 15:04
• Thanks, I will keep that in mind. But can this be solved without calculus? Jun 1 '20 at 15:07

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 $$\frac{25}4x^2+\frac{100}{81}y^2=1.$$