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- by Prabal Barahi 9 months ago

- 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(x
_{1},y_{1})

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

[Image missing]

**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(x_{1},y_{1}) be any point. Draw PM perpendicular to the axis of parabola.

∴ OM = x_{1}, PM = y_{1} ___ⓐ

Let PM meet the parabola at point Q. Then,

QM² = 4ax_{1}___ⓑ

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

PM² > QM²

or, PM² = QM²

or, PM² < QM²

i.e.

y_{1}² > 4ax_{1}_{}

or, y_{1}² = 4ax_{1}_{}

or, y_{1}² < 4ax_{1}

** 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 ___ⓒ

which is quadratic in x-axis.

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(x _{1},y_{1})__

Let P(x_{1},y_{1}) be a point on the parabola y² = 4ax and Q(x_{2},y_{2}) be its neighboring point on the parabola.

∴ y_{1}² = 4ax_{1}

and, y_{2}² = 4ax_{2}_{}

Subtracting them,

y_{2}² - y_{1}² = 4ax_{2 }- 4ax_{1 }

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. y_{2} → y_{1 }and x_{2 }→ x_{1}

or, _{}

or, yy_{1} - y_{1}² = 2ax - 2ax_{1}

or, yy_{1} - 4ax_{1} = 2ax - 2ax_{1}

or, yy_{1} = 2ax + 2ax_{1}

or, yy_{1} = 2a(x + x_{1})

Now,

Slope of tangent = 2a/y_{1}

Slope of normal = - y_{1}/2a

Equation of normal to the parabola at a point P(x_{1},y_{1}) is

y - y_{1} = - y_{1}/2a (x - x_{1}) ___ⓐ

Let m = - y_{1}/2a ⇒ y_{1} = - 2am

From y_{1}² = 4ax_{1}

or, (- 2am)² = 4ax_{1}

or, x_{1} = 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.>

Geogebra - Conic Section Parabola

<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>

<images made from Geogebra>

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