analytic geometrygeometry
How do I find the acute angle between two tangents which can be drawn from the point $(1,1)$ to the ellipse $4x^2+9y^2=1$ ? (example)
or simply say, how do I find the angle between two tangents which can be drawn from the point $(x,y)$ to the ellipse $\frac{x^2}{a} + \frac{y^2}{b} = 1$ ?
A graphical representation would be much appreciated, thanks !
Hint: Try to use congruence. If you can prove that the two triangles are congruent then it is a kite. Now, you can prove it easily.
(Sketch.) Let $P$ and $Q$ be two points on a parabola, and let $R$ be the point where the respective tangents to the parabola intersect. Let $X$ the midpoint of $PQ$. Then $RX$ is parallel to the axis of symmetry of the parabola (proved below). Draw lines through $P$ and $Q$ parallel to $RX$, and reflect these lines in the respective tangents; the focus $F$ is the intersection of the reflected lines.
To show that $RX$ is parallel to the axis of symmetry: Drop perpendiculars from $P,Q,R$ to the directrix, meeting it at $P',Q',R'$ respectively. As you alluded to, the tangent at $P$ is the perpendicular bisector of the segment $FP'$, and likewise for $Q$ and $FQ'$. So, in $\triangle FP'Q'$, two of the perpendicular bisectors pass through $R$; therefore the third does as well. Since $RR'$ is a line through $R$ and perpendicular to $P'Q'$, it must be the perpendicular bisector, that is, $R'$ is the midpoint of $P'Q'$. By parallels, (the extension of) $RR'$ bisects $PQ$, that is, $RR'$ passes through $X$. So $RX$ is perpendicular to the directrix, as claimed.
Edit: Just for reference, here's what this looks like analytically: The direction of $RX$ is $(2,1)$; reflecting in $RP$ just means exchanging $x$ and $y$ coordinates, so the direction of $PF$ is $(1,2)$, and the line through $P$ in that direction is $2x-y=1$. Reflecting in $RQ$ means negating the $y$-coordinate, so the direction of $QF$ is $(2,-1)$, and the line through $Q$ in that direction is $x+2y=1$. The intersection of these lines is $(\frac35,\frac15)$.
Best Answer
We are going to solve this problem in the general case by considering the ellipse as a parametric curve.
The ellipse defined by the equation $$\left(\frac xa\right)^2+\left(\frac yb\right)^2=1$$ is represented parametrically as $$z:(a\cos t,b\sin t)$$ and the derivative vector of the ellipse at some parameter $t$ is $$z':(-a\sin t,b\cos t)$$ Now given a point $p=(p_x,p_y)$ and some point on the ellipse $z(t)$, the line between $p$ and $z(t)$ is a tangent to the ellipse if it is parallel to the derivative vector. In other words, the "planar cross product" $(a,b)\times(c,d)=ad-bc$ of $z(t)-p$ and $z'(t)$ has to be zero: $$(a\cos t-p_x)(b\cos t)-(b\sin t-p_y)(-a\sin t)=0$$ $$ab\cos^2t-p_xb\cos t+ab\sin^2t-p_ya\sin t=0$$ $$ab=p_xb\cos t+p_ya\sin t$$ $$\frac{p_x}a\cos t+\frac{p_y}b\sin t=1$$ This last equation has two solutions for $t$ if $p$ is outside (not on) the ellipse; call these solutions $t_1$ and $t_2$. These correspond to two points on the ellipse $z_1$ and $z_2$, and $pz_1$ and $pz_2$ are the two tangents to the ellipse from $p$. Then the angle between the tangents can be computed from their dot product: $$\theta=\cos^{-1}\frac{(z_1-p)\cdot(z_2-p)}{|z_1-p||z_2-p|}$$
For the given ellipse of $4x^2+9y^2=1$ ($a=\frac12,b=\frac13$) and $p=(1,1)$, $t=2\tan^{-1}(1\pm\frac2{\sqrt3})$. The slopes of the two tangents are $\frac49(3\pm\sqrt3)$, and the angle between them works out to be $$\cos^{-1}\frac{59}{\sqrt{5209}}$$ or approximately 35.2 degrees. Below is a picture of this ellipse and the tangents to it.