The line $y+x=2$ intersects the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ at $A,B$. A circle with diameter $AB$ is drawn to intersect...

Question

The line $y+x=2$ intersects the ellipse $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ at $A,B$. A circle with diameter $AB$ is drawn to intersect $\dfrac{x^2}{16}+\dfrac{y^2}{9}=1$ at two more points $C,D$. If the equation of the line $CD$ is $y=mx+c$, find $m$.

My thought process is to put $y=2-x$ in the ellipse and obtain two values of $x$. Accordingly, we shall get the coordinates of $A,B$. Then we can write the equation of circle. And then we can solve it with ellipse and get $C,D$.

But when I start doing it, the calculations are not very pleasant.

Is there a better way to do it?

Answer 1

Hints: No need complicated calculation. Note that lines $AB$ and $CD$ connecting the intersections of ellipse and circle are perpendicular. This is what you have to show and get the result that : $$x+y=2\Rightarrow m_1=-1$$

$$\Rightarrow m=\frac{-1}{m_1}=1$$

Second figure is for comparison, where line $y=2x+3$ intersect the ellipse. you can see that the line connecting other points of intersect of circle with ellipse is not perpendicular on the line intersection with it, that is they are perpendicular only if gradient of the line is $m=\pm1$.

Answer 2

There is no other intersection point apart of $A$ and $B$.

Answer 3

The answer is $m=1$ if you allow non-real intersection points.

Consider $x+y+c=0$ and $x^2/16+y^2/9=1.$

Substitute $y=-c-x$ into $9x^2+16y^2=9\cdot 16$ and get $9x^2+16(-c-x)^2=9\cdot 16$ or $25x^2+32cx+16c^2-144=0$

Use the quadratic formula to get $y=\frac{-9c\pm\sqrt{(9c)^2-\cdot 25\cdot (9c^2-144)}}{25}=\frac{-9c\pm 12\sqrt{25-c^2}}{25}$

The coordinates of $A, B$ are $(-c-\frac{-9c+ 12\sqrt{25-c^2}}{25},\frac{-9c+ 12\sqrt{25-c^2}}{25}),(-c-\frac{-9c- 12\sqrt{25-c^2}}{25},\frac{-9c- 12\sqrt{25-c^2}}{25})$

And find the equation of the circle on diameter $AB$ as $(x-x(A))(x-x(B))+(y-y(A))(y-y(B))=0$

$x^2+y^2+(32cx)/25+(18cy)/25+(337c^2)/625-(288(25-c^2))/625=0$

Now to find the four points of intersection of the ellipse and circle set up the system $\begin{align}9&x^2+&16&y^2-9\cdot 16&=0\\16&x^2+&16&y^2+16((32cx)/25+(18cy)/25+(337c^2)/625-(288(25-c^2))/625)&=0\end{align}$

or

$\begin{align}9&x^2+&16&y^2-9\cdot 16&=0\\7&x^2+&&+16((32cx)/25+(18cy)/25+(337c^2)/625-(288(25-c^2))/625)+9\cdot 16&=0\end{align}$

The second equation can be solved for $y:$

$y = -(175x^2+512cx+400c^2-1008)/(288c)$ and be set back into equation 1:

$(30625x^4)/(5184c^2)+(2800x^3)/(81c)-(1225x^2)/(18c^2)+(4675x^2)/54+(6400cx)/81-(1792x)/(9c)+(2500c^2)/81+196/c^2-2696/9=0$

Knowing that $A, B$ are again solutions we can factor

$((25x^2+32cx+16c^2-144)(1225x^2+5600cx+10000c^2-7056))/(5184c^2)$

and the other two give $C,D$ and we have the system:

$\begin{align}9x^2+16y^2-9\cdot 16&=0\\x^2+y^2+(32cx)/25+(18cy)/25+(337c^2)/625-(288(25-c^2))/625&=0\\1225x^2+5600cx+10000c^2-7056&=0\end{align}$

Noting that $-175/18(x^2+y^2+(32/25)cx+(18/25)cy+c^2-288/25)+175/288(9x^2+16y^2-144) +1/288(1225x^2+5600cx+10000c^2-7056)=c(7x-7y+25c)=0$

So the line through $C,D$ for varying $c$ is $7x-7y+25c=0,$ which as promised gives $m=1.$

But $1225y^2-3150cy+5625c^2-7056=0$ has discriminant $-705600(5c-7)(5c+7)$