Conic Section: Parabola In this note, we will discuss:
• Parabola,
• Some terminology,
• Standard Equation of parabola and its different forms,
• Equation of parabola with its axis parallel to x-axis,
• General Equation of parabola,
• A point and a parabola,
• Condition that a line is tangent to a parabola,
• Equation of tangent and normal to the parabola y² = 4ax at a point P(x1,y1)

Parabola:

A conic section is called a parabola if its eccentricity is unity, i.e. e = 1

OR,

A parabola is a locus of a point which moves in a plane such that the ratio of the distances from a fixed point and a fixed line in always unity.

i.e. it is at equidistance from a fixed point and from a fixed straight line.

Terminology

Focus : the fixed point

Axis of parabola : the perpendicular line which passes through the focus and vertex of the parabola

Focal distance : the distance between any point on parabola and the focus

Focal chord : any chord of the parabola passing through focus

Latus rectum : the focal chord perpendicular to the axis of parabola

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Standard Equation of parabola:

y² = 4ax Let S(a,0) is focus of the parabola whose axis is x-axis and vertex at V(0,0). The equation of directrix is x + a = 0

Let P(x,y) be any point on the parabola. Join PS and draw PM perpendicular on the directrix of the parabola.

For parabola,

PS = PM

or,        PS² = PM²

or, or,        x² + y² - 2ax + a² = x² + 2ax + a²

or,        y² = 4ax           is the required equation.

-        Latus rectum:

Let L = (a,y’)

y’² = 4a.a  y’ = ± 2a

L(a,2a) and L’(a,-2a) are the points of the latus rectum on parabola and length of latus rectum is 4a.

Different forms of standard equation

i.    y² = - 4ax ii.    x² = 4ay iii.    x² = - 4ay Equation of parabola with its axis parallel to x-axis:

(y - k)² = 4a(x - h) Let V(h,k) is vertex of parabola of which axis is parallel to x-axis.

Let S(h + a, k) is focus of parabola then, the equation of directrix is

x - (h - a) = 0

Let P(x,y) is any point on the parabola. Join PS and draw PM perpendicular on the directrix.

For parabola,

PS = PM

or,       PS² = PM²

or, or,        x² - 2x(h + a) + (h + a)² + (y - k)² = x² - 2x(h - a) + (h – a)²

or,        (y - k)² = 2x[h + a - h + a] + (h - a)² - (h + a)²

or,        (y - k)² = 4ax – 4ah

or,        (y - k)² = 4a(x - h)

Similarly, the equation of parabola of which axis is parallel to y-axis is

(x - h)² = 4a(y - k) General equation of parabola

(a’² + b’²){(x - h)² + (y - k)²} = (a’x + b’y + c’)² Let S(h,k) be the focus of parabola of which equation of directrix is

a’x + b’y + c’ = 0

Let P(x,y) be any point on the parabola. Join PS and draw PM perpendicular on directrix

For parabola,

PS = PM

or,        PS² = PM²

or, or,       (a’² + b’²){(x - h)² + (y - k)²} = (a’x + b’y + c’)²  is the required equation.

A point and a parabola: Let P(x1,y1) be any point. Draw PM perpendicular to the axis of parabola.

OM = x1, PM = y1 ___

Let PM meet the parabola at point Q. Then,

QM² = 4ax1___ⓑ

The point P lies outside or on or inside the parabola according as

PM² > QM²

or,       PM² = QM²

or,        PM² < QM²

i.e.

y1² > 4ax1

or,        y1² = 4ax1

or,       y1² < 4ax1

Condition that a line is tangent to a parabola: Let y = mx + c ___

be a line and

y² = 4ax ___

be a parabola.
Solving
and , we have

(mx + c)² = 4ax

or,        m²x² + 2x(mc - 2a) + c² = 0 ___

This gives two values of x and corresponding two values of y.

The line will be a tangent to the parabola if discriminant of equation is zero i.e.

4(mc - 2a) ² - 4m²c² = 0

or,        4a² - 4mca = 0

or,        4a(a - mc) = 0

4a ≠ 0

a – mc = 0

or,        c = a/m            is the required condition.

The line

y = mx + a/m   is a tangent to the parabola y² = 4ax.

Equation of tangent and normal to the parabola y² = 4ax at a point P(x1,y1) Let P(x1,y1) be a point on the parabola y² = 4ax and Q(x2,y2) be its neighboring point on the parabola.

y1² = 4ax1

and, y2² = 4ax2

Subtracting them,

y2² - y1² = 4ax2 - 4ax

or, = Slope of PQ

Now,

Equation of chord PQ is or, The chord PQ becomes tangent to the parabola at point P if

Q → P i.e. y2 → y1 and x2 → x1

or, or,        yy1 - y1² = 2ax - 2ax1

or,        yy1 - 4ax1 = 2ax - 2ax1

or,        yy1 = 2ax + 2ax1

or,        yy1 = 2a(x + x1)

Now,

Slope of tangent = 2a/y1

Slope of normal = - y1/2a

Equation of normal to the parabola at a point P(x1,y1) is

y - y1 = - y1/2a (x - x1) ___

Let m = - y1/2a            y1 = - 2am

From y1² = 4ax1

or,        (- 2am)² = 4ax1

or,          x1 = am²

From

y + 2am = m(x - am)²

or,       y = mx - 2am - am³      is the required normal in m form.

<Here is a link to my resource of Parabola made in Geogebra, which you might find it helpful.>

<If you find any difficulty or mistake, do reach out to me through the comment section or social media.>

<notes source: from Rakesh Kumar Jha (RK) sir and book>