Parabola

Conormal Points of a Parabola

Properties of Tangents to a Parabola

Position of a Point with respect to Parabola

Focal Chord and Focal Distance

Focal chord:  Any chord passes through the focus of the parabola is a fixed chord of the parabola.

Focal Distance: The focal distance of any point p(x, y) on the parabola y² = 4ax is the distance between point ‘p’ and focus.

PM = a + x

PS = Focal distance = x + a

General Equations of Parabola

Equation of parabola by definition.

SP = PM

The general equation of 2^nd degree i.e. ax² + 2hxy + by² + 2gx + 2fy + c = 0 if

Position of a point with respect to parabola

For parabola

Intersection of a straight line with the parabola y² = 4ax

Straight line y = mx + c

m slope of straight line

(mx + c)² – 4ax = 0

m²x² + 2x(mc – 2a) + c² = 0

Ax² Bx + c = 0

B² – 4AC = discriminant D

D = 0

D > 0

mc – a > 0: Straight line intersect the curve

D < 0 (mc – a) < 0: Straight line not touching the curve

Tangent to a Parabola

Tangent at point (x₁, y₁)

y² = 4ax (parabola)

equation of Tangent

⇒ Point

Tangent in slope (m) form:

y² = 4ax

Let equation of Tangent y = mx + c

From the previous illustration

y = mx + c touches curve at a point

so ,

equation of Tangent :- y = mx +

so, point of Tangency is

Tangent in parameter form (at², 2at)

ty = x + at² where ‘t’ is

parameter

Pair of Tangents from (x₁, y₁) external points

Let y² = 4ax, (parabola)

P(x₁, y₁) external point then equation of Tangents is given by

SS₁ = T²

Chord of contact:

Equation of chord of contact of Tangents from a point p(x₁, y₁) to the parabola y² = 4ax is given by T = 0

i.e., yy₁ – 2a(x + x₁) = 0

Equation of QS T = 0

Normal to the parabola:

Normal to the point p(x₁, y₁) since normal is perpendicular to Tangent so slope of normal be will

slope of normal at ‘p’ (x₁ y₁) is

equation of normal

Normal in term of ‘m’:

Equation of normal at point (am², – 2am)

Normal at point (at², 2at)

T parameter

y = tx + 2at + at³

Important Properties of Focal Chord

1. If chord joining
   \(\begin{array}{l}P=(at_{1}^{2},2a{{t}_{1}})\end{array} \)
   and
   \(\begin{array}{l}Q=(at_{2}^{2},2a{{t}_{2}})\end{array} \)
   is focal chord of parabola
   \(\begin{array}{l}{{y}^{2}}=4ax\end{array} \)
   then
   \(\begin{array}{l}{{t}_{1}}{{t}_{2}}=-1\end{array} \)
   .
2. If one extremity of a focal chord is
   \(\begin{array}{l}(at_{1}^{2},2a{{t}_{1}})\end{array} \)
   then the other extremity
   \(\begin{array}{l}(at_{1}^{2},2a{{t}_{2}})\end{array} \)
   becomes
   \(\begin{array}{l}\left( \frac{a}{t_{1}^{2}},-\frac{2a}{{{t}_{1}}} \right)\end{array} \)
   .
3. If point
   \(\begin{array}{l}P(a{{t}^{2}},2at)\end{array} \)
   lies on parabola
   \(\begin{array}{l}{{y}^{2}}=4ax\end{array} \)
   , then the length of focal chord PQ is
   \(\begin{array}{l}a{{(t+1/t)}^{2}}\end{array} \)
   .
4. The length of the focal chord which makes an angle θ with positive x-axis is
   \(\begin{array}{l}4a\cos e{{c}^{2}}\theta\end{array} \)
   .
5. Semi latus rectum is harmonic mean of SP and SQ, where P and Q are extremities of latus rectum. i.e.,
   \(\begin{array}{l}2a=\frac{2SP\times SQ}{SP+SQ}\,or\frac{1}{SP}+\frac{1}{SQ}=\frac{1}{a}\end{array} \)
6. Circle described on focal length as diameter touches tangent at vertex. Circle described on focal chord as diameter touches directrix.

Important Properties of focal chord, Tangent and normal of Parabola

• The tangent at any point P on a parabola bisects the angle between the focal chord through P and the perpendicular from P on the directix.

• The portion of a tangent to a parabola cut off between the directrix and the curve subtends a right angle at the focus.

• Tangents at the extremities of any focal chord intersect at right angles on the directrix.

• Any Tangent to a parabola and perpendicular on it from the focus meet on the Tangent at its vertex.

Intersect at y-axis, at u = 0

Four Common Forms of a Parabola:

Equations of Normal to a Parabola

Tangent to a Parabola

Practice Problems on Parabola

Illustration 1: Find the vertex, axis, directrix, tangent at the vertex and the length of the latus rectum of the parabola

Solution: The given equation can be re-written as

which is of the form

Hence the vertex is

The axis is

The directix is

The tangent at the vertex is

Length of the latus rectum = 4a = 2.

Illustration 2: Find the equation of the parabola whose focus is (3, -4) and directix x – y + 5 = 0.

Solution: Let P(x, y) be any point on the parabola. Then

Illustration 3: Find the equation of the parabola having focus (-6, -6) and vertex (-2, 2).

Solution: Let S(6, -6) be the focus and A(-2, 2) the vertex of the parabola. On SA take a point K (x₁ , y₁) such that SA = AK. Draw KM perpendicular on SK. Then KM is the directix of the parabola.

Since A bisects SK,

Hence the equation of the directrix KM is y – 10 = m(x+2) ……(1)

Also gradient of

So that equation (1) becomes

Next, let PM be a perpendicular on the directrix KM from any point P(x, y) on the parabola.

From SP = PM, the equation of the parabola is

Illustration 4: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for

Solution: The given equation is

Here, the coefficient of x is positive. Hence, the parabola opens towards the right.

On comparing this equation with

Coordinates of the focus are given by (a, 0) i.e., (3, 0).

Since the given equation involves y², the axis of the parabola is the y-axis.

Equation of directix is

Length of latus rectum = 4a = 4 x 3 = 12.

Illustration 5: Find the coordinates of the focus, axis of the parabola, the equation of directrix and the length of the latus rectum for

Solution: The given equation is

Here, the coefficient of y is negative. Hence, the parabola opens downwards.

On comparing this equation with

Coordinates of the focus = (0, -a) = (0, -4).

Since the given equation involves

Equation of directrix, y = a i.e. = 4.

Length of latus rectum = 4a = 16.

Illustration 6: If the parabola y² = 4x and x² = 32y intersect at (16, 8) at an angle θ, then find the value of θ.

Solution: The slope of the tangent to y² = 4x at (16, 8) is given by

The slope of the tangent to x² = 32y at (16, 8) is given by

∴

Illustration 7: Find the equation of common tangent of y² = 4ax and x² – 4ay.

Solution: Equation of tangent to y² = 4ax having slope m is

It will touch x² – 4ay, if

Thus,

Thus, common tangent is y + x + a = 0.

Illustration 8: Find the equation of normal to the parabola y² = 4x passing through the point (15, 12).

Solution: Equation of the normal having slope m is

If it passes through the point (15, 12) then

Hence, equations of normal are:

Illustration 9: Find the point on y² = 8x where line x + y = 6 is a normal.

Solution: Slope m of the normal x + y = 6 is -1 and a = 2

Normal to parabola at point (am², -2am) is

Illustration 10: Tangents are drawn to y² = 4ax at point where the line lx + my + n = 0 meets this parabola. Find the intersection of these tangents.

Solution: Let the tangents intersects at P (h, k). Then lx + my + n = 0 will be the chord of contact. That means lx + my + n = 0 and yk – 2ax – 2ah = 0 which is chord of contact, will represent the same line.

Comparing the ratios of coefficients, we get

Illustration 11: If the chord of contact of tangents from a point P to the parabola If the chord of contact of tangents from a point P to the parabola y² = 4ax touches the parabola x²=4by, then find the locus of P.

Solution: Chord of contact of parabola y² = 4ax w.r.t. point P(x₁ , y₁)

yy₁ = 2a(x + x₁) ……(1)

This line touches the parabola x² = 4by.

Solving line (1) with parabola, we have

or

According to the question, this equation must have equal roots.

Illustration 12:

Illustration 13:

Illustration 14:

Real-life applications of a Parabola

1] A liquid is rotated and the forces of gravity in the liquid are responsible for the formation of a parabola. For example when any juice is stirred round along its axis. The level of the juice increases around the edges while it slightly falls in the centre of the glass.

2] The concept of the parabola is used in satellite dishes which helps to reflect the signals and then go to the receiver. Due to parabola’s reflective properties, the signals that go to the satellite will mirror off and back to the receiver soon after reflecting off the focus.

3] The Golden Gate Bridge cables that act as a suspension are parabolas.

4] Water in The Bellagio’s fountains in Las Vegas are exhibited in the shape of parabolas.

5] The reflective properties of parabolas are used in some heaters. The source of heat is at the focus and the heat
is focused in the form of parallel rays.

6] Parabolas are applied in the field of architecture and engineering projects. Parabolic shapes include:
a] The Parabola, a structure in London built in 1962 that consists of a roof made of copper with parabolic and hyperbolic lines.
b] There exists parabolas on each side of its side spans on The Golden Gate Bridge in San Francisco, California.

7] The use of parabolas are extensive when the light needs to be focused. A parabola-shaped reflector helps in focusing light into a beam that can be seen from long distances. It helps in reducing more usage of light and thus improves the surface of the parabola.

8] Solar power industry is aided by the parabolic reflectors to concentrate light.

9] The most frequent example of a parabola is that of the stretched arc of a rocket launch.

10] The parabolic trajectories are in use from decades.

Parabola Solved Questions