A brief note on The Parabola . Read it before a test. Try each question at the end before you open its answer.
The curve Definition: P F = P M PF = PM P F = P M for a focus F F F and a directrix l l l . The eccentricity is e = 1 e = 1 e = 1 . Standard forms: y 2 = 4 a x y^2 = 4ax y 2 = 4 a x (focus ( a , 0 ) (a, 0) ( a , 0 ) , directrix x = − a x = -a x = − a ), y 2 = − 4 a x y^2 = -4ax y 2 = − 4 a x , x 2 = 4 a y x^2 = 4ay x 2 = 4 a y and x 2 = − 4 a y x^2 = -4ay x 2 = − 4 a y . Latus rectum: length 4 a 4a 4 a , ends ( a , ± 2 a ) (a, \pm 2a) ( a , ± 2 a ) for y 2 = 4 a x y^2 = 4ax y 2 = 4 a x . Focal distance: x 1 + a x_1 + a x 1 + a for a point ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) on y 2 = 4 a x y^2 = 4ax y 2 = 4 a x . Parametric form: ( a t 2 , 2 a t ) (at^2, 2at) ( a t 2 , 2 a t ) . Shifted form: ( y − k ) 2 = 4 a ( x − h ) (y - k)^2 = 4a(x - h) ( y − k ) 2 = 4 a ( x − h ) has the vertex ( h , k ) (h, k) ( h , k ) . Tangents and normals Tangent at ( x 1 , y 1 ) (x_1, y_1) ( x 1 , y 1 ) : y y 1 = 2 a ( x + x 1 ) yy_1 = 2a(x + x_1) y y 1 = 2 a ( x + x 1 ) . At t t t : t y = x + a t 2 ty = x + at^2 t y = x + a t 2 . Tangent with slope m m m : y = m x + a m y = mx + \dfrac{a}{m} y = m x + m a , touching at ( a m 2 , 2 a m ) \left(\dfrac{a}{m^2}, \dfrac{2a}{m}\right) ( m 2 a , m 2 a ) . Normal at t t t : y + t x = 2 a t + a t 3 y + tx = 2at + at^3 y + t x = 2 a t + a t 3 . With slope m m m : y = m x − 2 a m − a m 3 y = mx - 2am - am^3 y = m x − 2 am − a m 3 . Subtangent and subnormal: 2 a t 2 2at^2 2 a t 2 and 2 a 2a 2 a . Perpendicular tangents meet on the directrix. Normals through a point: up to 3 3 3 , from a m 3 + ( 2 a − h ) m + k = 0 am^3 + (2a - h)m + k = 0 a m 3 + ( 2 a − h ) m + k = 0 . Focal chords and reflection Ends: t t t and − 1 t -\dfrac{1}{t} − t 1 . Length: a ( t + 1 t ) 2 a\left(t + \dfrac{1}{t}\right)^2 a ( t + t 1 ) 2 . Harmonic property: 1 S P + 1 S Q = 1 a \dfrac{1}{SP} + \dfrac{1}{SQ} = \dfrac{1}{a} S P 1 + S Q 1 = a 1 . Tangents at the ends are perpendicular and meet on the directrix. Reflection: a ray parallel to the axis goes through the focus, since F T = F P FT = FP F T = F P . A worked example Find the tangent to y 2 = 8 x y^2 = 8x y 2 = 8 x at ( 2 , 4 ) (2, 4) ( 2 , 4 ) , and the normal there.
Check. 16 = 8 × 2 16 = 8 \times 2 16 = 8 × 2 . Here a = 2 a = 2 a = 2 . Tangent. 4 y = 4 ( x + 2 ) 4y = 4(x + 2) 4 y = 4 ( x + 2 ) , so y = x + 2 y = x + 2 y = x + 2 . Normal. The slope is − y 1 2 a = − 1 -\dfrac{y_1}{2a} = -1 − 2 a y 1 = − 1 , so y − 4 = − ( x − 2 ) y - 4 = -(x - 2) y − 4 = − ( x − 2 ) , which is x + y = 6 x + y = 6 x + y = 6 . Practice with new numbers
A program to try The program checks the tangent and the normal at a point and then finds the focal chord through the same point.
Run it. Which printed numbers show that the tangent and the normal are perpendicular?
1. Change t = 2 to t = 3. What is the length of the focal chord now?
Mistakes to avoid a a a read as the coefficient. For y 2 = 12 x y^2 = 12x y 2 = 12 x the value is a = 3 a = 3 a = 3 . Tangent slope. It is 2 a y 1 \dfrac{2a}{y_1} y 1 2 a or 1 t \dfrac{1}{t} t 1 , not y 1 2 a \dfrac{y_1}{2a} 2 a y 1 . Focal chord. The product t 1 t 2 t_1t_2 t 1 t 2 is − 1 -1 − 1 . Directrix. It is on the other side of the vertex from the focus. Check yourself Find the focus and the directrix of y 2 = 20 x y^2 = 20x y 2 = 20 x . Stuck? Show a hint Show the answerFind the tangent to y 2 = 4 x y^2 = 4x y 2 = 4 x with slope 2 2 2 . Stuck? Show a hint Show the answerFind the length of the focal chord of y 2 = 4 x y^2 = 4x y 2 = 4 x that has one end at t = 3 t = 3 t = 3 . Stuck? Show a hint Show the answerWhere do perpendicular tangents to a parabola meet? Stuck? Show a hint Show the answerWhat is the subnormal of y 2 = 12 x y^2 = 12x y 2 = 12 x ? Stuck? Show a hint Show the answerSources NCERT, Mathematics for Class XI, Chapter 10 (Conic Sections), the section on the parabola. OpenStax, Precalculus 2e , Chapter 12 (Analytic Geometry), the section on the parabola.