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Concept 09 / 14Coordinate Geometry

Tangents and Normals to a Parabola and Its Reflection Property

Tangent at a point, in parametric form and slope form (y = mx + a/m), tangents from an external point and the intersection of tangents, the normal y = -tx + 2at + at^3 and co-normal points, chord of contact and chord with given midpoint, and the reflection property.

Figure 1
All formulas are for y2=4axy^{2} = 4ax; the TT and S1S_{1} notation works as for the circle, with S=y2−4axS = y^{2} - 4ax, S1=y12−4ax1S_{1} = y_{1}^{2} - 4ax_{1} and T=yy1−2a(x+x1)T = yy_{1} - 2a(x + x_{1}). Tangents. At a point (x1,y1)(x_{1}, y_{1}) on the curve: T=0T = 0, i.e. yy1=2a(x+x1)yy_{1} = 2a(x + x_{1}). At the parametric point (at2,2at)(at^{2}, 2at): ty=x+at2ty = x + at^{2}, with slope 1t\frac{1}{t}. Slope form: y=mx+amy = mx + \frac{a}{m} touches y2=4axy^{2} = 4ax at (am2,2am)\left(\frac{a}{m^{2}}, \frac{2a}{m}\right). The condition for y=mx+cy = mx + c to be a tangent is c=amc = \frac{a}{m}. Tangents at t1t_{1} and t2t_{2} meet at (at1t2,a(t1+t2))(at_{1}t_{2}, a(t_{1} + t_{2})): the xx-coordinate is aa times the product and the yy-coordinate aa times the sum of the parameters. So tangents at the ends of a focal chord (t1t2=−1t_{1}t_{2} = -1) meet on the directrix x=−ax = -a, and they are perpendicular (slopes 1t1,1t2\frac{1}{t_{1}}, \frac{1}{t_{2}} with product −1-1). The directrix is therefore the director circle of the parabola. Tangents from an external point (h,k)(h, k). The slopes satisfy k=mh+amk = mh + \frac{a}{m}, i.e. hm2−km+a=0hm^{2} - km + a = 0: two tangents, with m1+m2=khm_{1} + m_{2} = \frac{k}{h} and m1m2=ahm_{1}m_{2} = \frac{a}{h}. Chord of contact: T=0T = 0. Chord with midpoint (x1,y1)(x_{1}, y_{1}): T=S1T = S_{1}. Normals. At (at2,2at)(at^{2}, 2at): y=−tx+2at+at3y = -tx + 2at + at^{3}, with slope −t-t. Slope form: y=mx−2am−am3y = mx - 2am - am^{3}, the normal at (am2,−2am)(am^{2}, -2am). Through a given point (h,k)(h, k), the normal condition k=mh−2am−am3k = mh - 2am - am^{3} is a cubic in mm, so up to three normals can be drawn (co-normal points). Their slopes satisfy m1+m2+m3=0m_{1} + m_{2} + m_{3} = 0, so the sum of the corresponding yy-coordinates of the feet, −2a∑m-2a\sum m, is zero. Three real normals from a point on the axis (h,0)(h, 0) exist only if h>2ah > 2a. If the normal at t1t_{1} meets the parabola again at t2t_{2}, then t2=−t1−2t1t_{2} = -t_{1} - \frac{2}{t_{1}}. The normal chord is shortest when t12=2t_{1}^{2} = 2. Geometric properties. Reflection property: the tangent at PP makes equal angles with the focal radius SPSP and with the line through PP parallel to the axis. Hence rays parallel to the axis reflect off the parabola and pass through the focus, as the figure shows. This is why satellite dishes and car headlamps are parabolic. The foot of the perpendicular from the focus to any tangent lies on the tangent at the vertex (x=0x = 0). The tangent at PP bisects the angle between SPSP and the perpendicular from PP to the directrix. The portion of a tangent between the point of contact and the directrix subtends a right angle at the focus. The subnormal is constant: equal to 2a2a. Key Takeaways: - Tangent at tt: ty=x+at2ty = x + at^{2}; slope form y=mx+amy = mx + \frac{a}{m}. - Tangents at t1,t2t_{1}, t_{2} meet at (at1t2,a(t1+t2))(at_{1}t_{2}, a(t_{1} + t_{2})). - Tangents at focal chord ends are perpendicular and meet on the directrix. - Normal at tt: y=−tx+2at+at3y = -tx + 2at + at^{3}; slope form y=mx−2am−am3y = mx - 2am - am^{3}. - Up to 33 normals from a point; ∑m=0\sum m = 0. - Normal at t1t_{1} meets again at t2=−t1−2t1t_{2} = -t_{1} - \frac{2}{t_{1}}. - Reflection property: axis-parallel rays pass through the focus. - Pitfall: using the circle's tangency condition c2=a2(1+m2)c^{2} = a^{2}(1 + m^{2}) for a parabola. Brain-Teaser 1: Find the equation of the common tangent to y2=4xy^{2} = 4x and x2=4yx^{2} = 4y. Solution: A tangent to y2=4xy^{2} = 4x is y=mx+1my = mx + \frac{1}{m}. For it to touch x2=4yx^{2} = 4y, substitute: x2−4mx−4m=0x^{2} - 4mx - \frac{4}{m} = 0 must have a repeated root: 16m2+16m=016m^{2} + \frac{16}{m} = 0, so m3=−1m^{3} = -1 and m=−1m = -1. The common tangent is y=−x−1y = -x - 1, i.e. x+y+1=0x + y + 1 = 0. By symmetry of the two parabolas about y=xy = x, the common tangent must be perpendicular to that line, which agrees. Brain-Teaser 2: Prove that the tangents at the ends of any focal chord of y2=4axy^{2} = 4ax intersect at right angles on the directrix. Solution: The ends have parameters tt and −1t-\frac{1}{t}. The tangents have slopes 1t\frac{1}{t} and −t-t, whose product is −1-1, so they are perpendicular. They meet at (a⋅t⋅(−1t),a(t−1t))=(−a,a(t−1t))(a \cdot t \cdot (-\frac{1}{t}), a(t - \frac{1}{t})) = (-a, a(t - \frac{1}{t})), which lies on x=−ax = -a. Parameters make the proof two lines long. Brain-Teaser 3: The normal to y2=4axy^{2} = 4ax at (at2,2at)(at^{2}, 2at) meets the curve again at (aT2,2aT)(aT^{2}, 2aT). Show that T=−t−2tT = -t - \frac{2}{t}, and find the least possible value of T2T^{2}. Solution: The chord from tt to TT has slope 2t+T\frac{2}{t + T}, and the normal at tt has slope −t-t. Equating: 2t+T=−t\frac{2}{t + T} = -t, so t+T=−2tt + T = -\frac{2}{t} and T=−t−2tT = -t - \frac{2}{t}. Then T2=t2+4+4t2≥4+24=8T^{2} = t^{2} + 4 + \frac{4}{t^{2}} \ge 4 + 2\sqrt{4} = 8 by AM-GM, with equality at t2=2t^{2} = 2. So T2≥8T^{2} \ge 8: a normal chord always ends at least as far out as parameter ±22\pm 2\sqrt{2}.

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