Ellipse flash cards

1. 1.Define an ellipse using the focus-directrix property and state its eccentricity condition.

   An ellipse is the locus of a point $P$ such that its distance from a fixed point (focus) is in constant ratio $e$ ($0bounded, closed curve, unlike the parabola ($e=1$) or hyperbola ($e>1$).

   Hint: Compare with conic section eccentricity ranges

2. 2.For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ($a>b$), write the condition for the line $y=mx+c$ to be a tangent, and hence give the equation of tangent in terms of slope $m$.

   The line $y=mx+c$ touches the ellipse iff $c^2=a^2m^2+b^2$. So the tangent in slope form is $y=mx\pm\sqrt{a^2m^2+b^2}$. This is used to find tangents from an external point and to derive the director circle.

   Hint: Substitute line into ellipse, set discriminant $=0$

3. 3.What is the equation of the director circle of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, and what does it represent geometrically?

   The director circle is $x^2+y^2=a^2+b^2$. It is the locus of points from which two perpendicular tangents can be drawn to the ellipse.

   Hint: Locus of intersection of perpendicular tangents

4. 4.State the reflection (optical) property of an ellipse and its practical significance.

   A ray from one focus $S$, after reflecting off the ellipse, always passes through the other focus $S'$ — the normal at any point bisects the angle $\angle SPS'$. This property is used in whispering galleries and lithotripsy (focusing shock waves at a kidney stone placed at the second focus).

   Hint: Normal bisects the focal angle at P

5. 5.For the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ ($a>b$), state the formula for the area of the ellipse and the length of the latus rectum, and give the coordinates of the latus rectum's endpoints.

   Area of ellipse $=\pi ab$. Length of latus rectum $=\dfrac{2b^2}{a}$, with endpoints $\left(\pm ae,\; \pm\dfrac{b^2}{a}\right)$ (one pair at each focus). These follow from the ellipse being an affine (scaled) image of the auxiliary circle of radius $a$, scaled by factor $b/a$ along the minor axis.

   Hint: Ellipse = circle squashed by factor $b/a$

6. 6.If $P(a\cos\theta,b\sin\theta)$ is a point on the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ and the line $CP$ makes angle $\psi$ with the major axis, express $\tan\psi$ in terms of $\theta$. Is $\psi=\theta$ in general?

   $\tan\psi=\dfrac{y}{x}=\dfrac{b\sin\theta}{a\cos\theta}$, so $\tan\psi=\dfrac{b}{a}\tan\theta$. Since $a\neq b$, $\psi\neq\theta$ except when $\theta=0,\pi/2,\pi,\dots$; $\theta$ is the eccentric angle (angle for the corresponding point on the auxiliary circle), not the angle $CP$ makes with the axis.

   Hint: Write $\tan\psi=y/x$ using the parametric coordinates directly.

7. 7.Find the eccentric angle(s) of the extremities of the latus rectum of the ellipse $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ in terms of the eccentricity $e$.

   The right latus rectum meets the ellipse at $x=ae$. Since $x=a\cos\theta$, we need $a\cos\theta=ae\Rightarrow\cos\theta=e$. Hence $\theta=\cos^{-1}(e)$ and $\theta=-\cos^{-1}(e)$ give the right latus-rectum ends; $\theta=\pi\pm\cos^{-1}(e)$ give the left one.

   Hint: Set the $x$-coordinate of the parametric point equal to $ae$.

8. 8.$P$ and $Q$ are points on $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$ with eccentric angles $\theta$ and $\theta+\dfrac{\pi}{2}$ (so $CP,CQ$ are conjugate semi-diameters). Prove that $CP^2+CQ^2$ is independent of $\theta$.

   $CP^2=a^2\cos^2\theta+b^2\sin^2\theta$. Since $\cos(\theta+\pi/2)=-\sin\theta,\ \sin(\theta+\pi/2)=\cos\theta$, we get $CQ^2=a^2\sin^2\theta+b^2\cos^2\theta$. Adding, $CP^2+CQ^2=a^2(\cos^2\theta+\sin^2\theta)+b^2(\sin^2\theta+\cos^2\theta)=$ $a^2+b^2$, a constant.

   Hint: Substitute $\theta+\pi/2$ into the parametric coordinates and add the two squared distances.

9. 9.With $P(\theta)$ and $Q(\theta+\pi/2)$ conjugate semi-diameter points on $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$, show that $CP\cdot CQ\cdot\sin(\angle PCQ)$ is constant and find its value.

   $P=(a\cos\theta,b\sin\theta)$, $Q=(-a\sin\theta,b\cos\theta)$. Twice the area of $\triangle CPQ$ is $|x_Py_Q-x_Qy_P|=|ab\cos^2\theta+ab\sin^2\theta|=ab$. Since this area also equals $\tfrac12 CP\cdot CQ\sin(\angle PCQ)$, we get $CP\cdot CQ\cdot\sin(\angle PCQ)=ab$, independent of $\theta$.

   Hint: Compute the cross product $x_Py_Q-x_Qy_P$ to get twice the triangle's area.

10. 10.For the ellipse $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$, $P$ has eccentric angle $60^\circ$ and $Q$ (eccentric angle $150^\circ$) is such that $CP,CQ$ are conjugate semi-diameters. Find $CQ^2$.

    $CQ^2=a^2\sin^2\theta+b^2\cos^2\theta$ with $a^2=25,b^2=9,\theta=60^\circ$: $CQ^2=25\sin^2 60^\circ+9\cos^2 60^\circ=25\left(\tfrac34\right)+9\left(\tfrac14\right)=18.75+2.25=$ $21$, so $CQ=\sqrt{21}$.

    Hint: Use $CQ^2=a^2\sin^2\theta+b^2\cos^2\theta$ since $Q$'s angle is $\theta+90^\circ$.

Open the interactive deck for the other 75 cards, with self-grading so the ones you keep missing come back.