Parabola

# Parabola | Mathematics (Maths) Class 11 - Commerce PDF Download

PARABOLA

A. Conic section

A conic section, or conic is the locus of a point which moves in a plane so that its distance from a fixed point is in a constant ratio to its perpendicular distance from a fixed straight line.

(a) The fixed point is called the FOCUS.

(b) The fixed straight line is called the DIRECTRIX.

(c) The constant ratio is called the ECCENTRICITY denoted by `e'.

(d) The line passing through the focus & perpendicular to the directrix is called the AXIS.

(e) A point of intersection of a conic with its axis is called a VERTEX.

B.   General equation of a conic : focal directrix property

The general equation of a conic with focus (p, q) & directrix lx + my + n = 0 is

(l2 + m2) [(x - p)2 + (y - q)2] = e2 (lx + my + n)2 ax2 + 2hxy + by2 + 2gx + 2fy + c = 0

C. Distinguishing between the Conic

The nature of the conic section depends upon the position of the focus S w.r.t. the directrix & also upon the value of the eccentricity e. Two different cases arise.

Case (i) When the focus lies on the directrix

In this case D  abc + 2 fgh - af2 - bg2 - ch2 = 0 & the general equation of a conic represents a pair of straight lines and if :

e > 1 the lines will be real & distinct intersecting at S.

e = 1 the lines will be coincident.

e < 1 the lines will be imaginary.

Case (ii) When the focus does not lie on the directrix

The conic represents :

 a parabola hyperbola an ellipse a hyperbola a rectangular e = 1 ; D   0 0 < e < 1 ; D  0 D  0 ; e > 1 ; e > 1;D 0 h2 = ab h2 < ab h2 > ab h2 > ab ; a + b = 0

D. Parabola

A parabola is the locus of a point which moves in a plane, such that its distance from a fixed point (focus) is equal to its perpendicular distance from a fixed straight line (directrix).

Standard equation of a parabola is y2 = 4 ax. For this parabola :

(i) Vertex is (0, 0)

(ii) Focus is (a, 0)

(iii) Axis is y = 0

(iv) Directrix is x + a = 0

(a) Focal distance : The distance of a point on the parabola from the focus is called the focal distance of the point.

(b) Focal chord : A chord of the parabola, which passes through the focus is called a focal chord.

(c) Double ordinate : A chord of the parabola perpendicular to the axis of the symmetry is called a double ordinate.

(d) Latus rectum : A double ordinate passing through the focus or a focal chord perpendicular to the axis of parabola is called the latus rectum. For y2 = 4ax.

• l Length of the latus rectum = 4a.
• l Length of the semi latus rectum = 2a.
• l Ends of the latus rectum are L (a, 2a) & L (a, -2a)

Note that :

(i) Perpendicular distance from focus on directrix = half the latus rectum.

(ii) Vertex is middle point of the focus & the point of intersection of directrix & axis.

(iii) Two parabolas are said to be equal if they have the same latus rectum.

E. type of parabola

Four standard forms of the parabola are y2 = 4ax ; y2 = - 4ax ; x2 = 4ay ; x2 = -4ay

 Parabola Vertex Focus Axis Directrix Length of Latus rectum Ends of Latus rectum Parametric eqution Focal length y2 = 4ax (0, 0) (a, 0) V = 0 x = -a 4a (a, ±2a) (at2, 2at) x + a V2 = -4ax (0, 0) (-a, 0) y = 0 x = a 4a (-a, ±2a) (-at2, 2at) x - a x2 = +4ay (0, 0) (0, a) x = 0 y = -a 4a (± 2a, a) (2at, at2) y + a x2 = -4ay (0, 0) (0, -a) x = 0 y = a 4a (± 2a, -a) (2at, - at2) y - a (y - k)2 = 4a(x - h) (h, k) (h + a, k) V = k k + a - h = 0 4a (h + a, k ± 2a) (h + at2, k + 2at) x - h + a (x - p)2 = 4b(y - q) (p, q) (p, b + q) x = p y + b - q = 0 4b (p ± 2a, q + a) (p + 2at, q + at2) y - q + b

F. Parametric representation

The simplest & the best form of representing the co-ordinates of a point on the parabola is (at2, 2at). The equation x = at2 & y = 2at together represents the parabola y2 = 4ax, t being the parameter.

Ex.1 Find the vertex, axis, directrix, focus, latus rectum and the tangent at vertex for the parabola 9y2 – 16x – 12y – 57 = 0.

Sol. The given equation can be rewritten as   which is of the form Y= 4AX.

Hence the vertex is (-61/16,2/3) The axis the y -2/3 = 0 ⇒ y =2/3

The directrix is X + A = 0

The focus is X = A and Y = 0

Length of the latus rectum = 4A = 16/9 .The tangent at the vertex is X = 0  ⇒ x = -61/16

Ex.2 The length of latus rectum of a parabola, whose focus is (2, 3) and directrix is the line x – 4y + 3 = 0 is

Sol. The length of latus rectum = 2 × perp. from focus to the directrix

Ex.3 Find the equation of the parabola whose focus is (–6, –6) and vertex (–2, 2).

Sol. Let S(–6, –6) be the focus and A(–2, 2) is 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 directrix of the parabola.

–6 + x1 = –4 and –6 + y1 = 4 or (x1, y1) = (2, 10)

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

Also gradient of SK

x + 2y – 22 = 0 is the directrix

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

or 5(x2 + y2 + 12x + 12y + 72) = (x + 2y – 22)

or 4x2 + y2 – 4xy + 104x + 148y – 124 = 0

or (2x – y)2 + 104x + 148y – 124 = 0.

Ex.4 The extreme points of the latus rectum of a parabola are (7, 5) and (7, 3). Find the equation of the parabola

Sol. Focus of the parabola is the mid-point of the latus rectum.

⇒ S is (7, 4). Also axis of the parabola is perpendicular to the latus rectum and passes through the focus. Its equation is y – 4 =

Length of the latus rectum = (5 – 3) = 2

Hence the vertex of the parabola is at a distance 2/4 = 0.5 from the focus. We have two parabolas, one concave rightward and the other concave leftward

The vertex of the first parabola is (6.5, 4) and its equation is (y – 4)2 = 2(x – 6.5) and it meets the x-axis at (14.5, 0). The equation of the second parabola is (y – 4)2 = –2(x – 7.5). It meets the x-axis at (–0.5, 0)

G. POSITION OF A POINT RELATIVE TO A PARABOLA

The point (x1, y1) lies outside, on or inside the parabola y2 = 4ax according as the expression y2 - 4axi is positive, zero or negative

Ex.5 Find the value of a for which the point (α –1, α ) lies inside the parabola y2 = 4x.

Sol. Q Point (α – 1, α) lies inside the parabola y2 = 4x

y12 - 4 x< 0 ⇒ α2 -4(α- 1) < 0

α2 - 4α + 4 < 0 ⇒ (α - 2)2 < 0  ⇒ α∈ Ø

H. CHORD JOINING TWO POINTS

The equation of a chord of the parabola y2 = 4ax joining its two points P(t1) and Q(t2) is y(t1 + t2) = 2x + 2at1t2

Note : (i) If PQ is focal chord then t1t2 = –1.

(ii) Extremities of focal chord can be taken as (at2, 2at) &

Ex.6 Through the vertex O of a parabola y2 = 4x chords OP and OQ are drawn at right angles to one another. Show that for all position of P, PQ cuts the axis of the parabola at a fixed point.

Sol. The given parabola is y2 = 4x ....(1)

variable line PQ passes through a fixed point which is point of intersection of L1 = 0 & L2 = 0 i.e. (4, 0)

I. LINE & A PARABOLA

(a) The line y = mx + c meets the parabola y2 = 4ax in two points real, coincident or imaginary according as a > = < cm  ⇒  condition of tangency is, c = a/m.

Note : Line y = mx + c will be tangent to parabola x2 = 4ay if c = – am2

(b) Length of the chord intercepted by the parabola y2 = 4ax on the line y = mx + c is

Note : length of the focal chord making an angle a with the x-axis is 4a cosec2α

Ex.7 If the line y = 3x + λ intersect the parabola y2 = 4x at two distinct points then set of values of λ is

Sol. Putting value of y from the line in the parabola –

(3x + λ)2 = 4x ⇒ 9x2 + (6λ – 4)x + λ2 = 0

line cuts the parabola at two distinct points

D > 0   ⇒ 4(3λ – 2)2 – 4.9λ2 > 0 ⇒  9λ2 - 12λ + 4 - 9λ2 > 0  ⇒  λ < 1/3 Hence , λ E (-∝, 1/3)

J. TANGENT TO THE PARABOLA y2 = 4ax

(a) Point form : Equation of tangent to the given parabola at its point (x1, y1) is yy1 = 2a (x + x1)

(b) Slope form : Equation of tangent to the given parabola whose slope is ‘m’, is y = mx + a/m, (m≠0) & Point of contact is

(c) Parametric form : Equation of tangent to the given parabola at its point P(t), is ty = x + at2

Note : Point of intersection of the tang ents at the point t1 & t2 is [at1 t2, a(t+ t2)], (i.e. G.M. and A.M. of abscissa and ordinates of the points)

Ex.8 A tangent to the parabola y2 = 8x makes an angle of 45º with the straight line y = 3x + 5. Find its equation and its point of contact.

Sol. Let the slope of the tangent be m

As we know that equation of tangent of slope m to the parabola y2 = 4ax is y = mx + a/m. and point of contact is

for m = –2, equation of tangent is y = –2x – 1 and point of contact is (1/2,-2)

for m = 1/2,  equation of tangent is y =  1/2x+4 and point of contact is (8, 8)

Ex.9 Find the equation to the tangents to the parabola y2 = 9x which go through the point (4, 10).

Sol. Equation of tangent to parabola y2 = 9x is y = mx + 9/4m

Since it passes through (4, 10)

∴ 10 = 4m + 9/4m

16m2 – 40 m + 9 = 0

= m -- 1/4,9/4

∴equation of tangent’s are

Ex.10 Find the locus of the point P from which tangents are drawn to the parabola y2 = 4ax having slopes m1 and m2 such that

where θ1 and θ2 are the inclinations of the tangents from positive x-axis.

Sol. Equation of tangent to y2 = 4ax is y = mx + a/m

Let it passes through P(h, k).

∴ m2h – mk + a = 0

locus of P(h, k) is y2 – 2ax = λx2

K. DIRECTOR CIRCLE

Locus of the point of intersection of the perpendicular tangents to the parabola y2 = 4ax is called the DIRECTOR CIRCLE. It’s equation is x + a = 0 which is parabola’s own directrix.

Ex.11 The angle between the tangents drawn from a point (–a, 2a) to y= 4ax is

Sol. The given point (–a, 2a) lies on the directrix x = –a of the parabola y2 = 4ax. Thus, the tangents are at right angle.

Ex.12 The circle drawn with variable chord x + ay – 5 = 0 (a being a parameter) of the parabola y2 = 20x as diameter will always touch the line

Sol. Clearly x + ay – 5 = 0 will always pass through the focus of y2 = 20x i.e. (5, 0). Thus the drawn circle will always touch the directrix of the parabola i.e.. the line x + 5 = 0.

L. NORMAL TO THE PARABOLA   y2 = 4ax

(a) Point form : Equation of normal to the given parabola at its point (x1, y1) is

(b) Slope form : Equation of normal to the given parabola whose slope is ‘m’, is

y = m x – 2am – am3 & foot of the normal is (am2, – 2am)

(c) Parametric form : Equation of normal to the given parabola at its point P(t), is y + tx = 2at + at3

Note :

(i) Point of intersection of normals at t1 & t2 is,

(ii) I f the normal to the parabola y2 = 4 ax at the point t1, meets the para bola again at the point t2

(iii)If the normals to the parabola y2 = 4ax at the points t1 & t2 intersect again on the parabola at the point ‘t3’ then t1t2 = 2; t= – (t1 + t2) and the line joining t1 & t2 passes through a fixed point (–2a, 0).

(iv) If normal drawn to a parabola passes through a point P(h, k) then k = mh – 2 am – am3 i.e. am3 + m (2a – h) + k = 0.

This gives m1 + m2 + m3 = 0 ;    m1m2 + m2m3 + m3m1

where m1, m2, & m3 are the slopes of the three concurrent normals :

• Algebraic sum of slopes of the three concurrent normals is zero
• Algebraic sum of ordinates of the three co-normal points on the parabola is zero.
• Centroid of the D formed by three co-normal points lies on the axis of parabola (x-axis)

Ex.13 Prove that the normal chord to a parabola y2 = 4ax at the point whose ordinate is equal to abscissa subtends a right angle at the focus.

Sol. Let the normal at P (at12 , 2at) meet the curve at Q (a t 22 , 2 a t2 )

PQ is a normal chord and t2 = -t1 -2/t1 ....(i)

By given condition 2at1 = at12

t1 = 2 from equation (i), t2 = –3

then P(4a, 4a) and Q(9a, –6a) but focus S(a, 0)

Ex.14 If two normals drawn from any point to the parabola y2 = 4ax make angle a and b with the axis such that tan α . tan β = 2, then find the locus of this point,

Sol. Let the point is (h, k). The equation of any normal to the parabola y2 = 4ax is y = mx – 2am – am3 passes through (h, k)  ⇒ k = mh – 2am – am3

⇒   am3 + m(2a – h) + k = 0 ...(i)

m1, m2, m3 are roots of the equation, then m1. m2. m

⇒  k2 = 4ah. Thus locus is y2 = 4ax.

Ex.15 Three normals are drawn from the point (14, 7) to the curve y2 – 16x – 8y = 0. Find the coordinates of the feet of the normals.

Sol. The given parabola is y– 16x – 8y = 0 .....(i)

Let the co-ordinates of the feet of the normal from (14, 7) be P(α, β). Now the equation of the tangent at P(α, β) to parabola (i) is

yb – 8(x + a) – 4(y + b) = 0 or (b – 4)y = 8x + 8a + 4b ...(ii)

Its slope = 8/β-4 ....(2)

Equation of the normal to parabola (i) at

It passes through (14, 7)

Also (a, b) lies on parabola (i) i.e. b2 - 16a - 8b = 0

Putting the value of a from (iii) in (iv), we get

⇒ β2(β - 4) - 96β - 8β(β - 4) = 0

⇒ β(β2 - 4β - 96 - 8β + 32) = 0

⇒ β(β2 - 12β - 64) = 0

⇒ β(β - 16)(β + 4) = 0 ⇒ β = 0, 16, - 4

from (iii), a = 0 when β = 0; α = 8, when β = 16 ; α = 3 when β = –4

Hence the feet of the normals are (0, 0) (8, 16) and (3, –4)

M. LENGTH OF SUBTANGENT & SUBNORMAL

PT and PG are the tangent and normal respectively at the point P to the parabola y2 = 4ax. Then

TN =  length of subtangent = twice the abscisse of the point P   (Subtangent is always bisected by the vertex)

NG = length of subnormal which is constant for all points on the parabola & equal to its semi latusrectum (2a).

N. PAIR OF TANGENTS

The equation of the pair of tangents which can be drawn from any point P(x1, y1) out side the parabola to the parabola y2 = 4ax is given by : SS1 = T2 where :

S ≡ y2 – 4ax ; S1 ≡ y12 – 4ax1 ; T≡ yy1 – 2a(x + x1).

O. CHORD OF CONTACT

Equation of the chord of contact of tangents drawn from a point P(x1, y1) is yy1 = 2a(x + x1) Remember that the area of the triangle formed by the tangents from the point (x1, y1) & the chord of contact is    Also note that the chord of contact exists only if the point P is not inside.

Ex.16 If the line x – y – 1 = 0 intersect the parabola y2 = 8x at P & Q, then find the point of intersection of tangents at P & Q

Sol. Let (h, k) be point of intersection of tangents then chord of contact is

yk = 4(x + h) ⇒ 4x – yk + 4h = 0 ............(i)

But given line is ⇒ x – y – 1 = 0 ............(ii)

Comparing (i) and (ii),

h = – 1, k = 4

Therefore, point ≡ (–1, 4)

Ex.17 Find the locus of point whose chord of contact w.r.t. to the parabola y2 = 4bx is the tangent of the parabola y2 = 4ax

Sol. Equation of tangent to y2 = 4ax is y = mx + a/m ....(1)

Let it is chord of contact for parabola y2 = 4bx w.r.t. the point P(h, k)

Equation of chord of contact is yk = 2b(x + h)

From (i) & (ii),

P. CHORD WITH A GIVEN MIDDLE POINT

Equation of the chord of the parabola y2 = 4ax whose middle point is (x1, y1) is y – y1 = 2a/y1 (x-x1)

The reduced to T = S1 where T ≡ yy1 – 2a (x + x1)  &  S1 ≡ y12 – 4ax1.

Ex.18 Find the locus of middle of the chord of the parabola y2 = 4ax which pass through a given (p, q).

Sol. Let P(h, k) be the mid point of chord of the parabola y2 = 4ax.

so equation of chord is yk – 2a(x + h) = k2 – 4ah. Since it passes through (p, q)

qk – 2a(p + h) = k2 – 4ah

Required locus is y2 – 2ax – qy + 2ap = 0.

Ex.19 Find the locus of the middle point of a chord of a parabola y2 = 4ax which subtends a right angle at the vertex.

Sol.  The equation of the chord of the parabola whose middle point is (α, β) is

points of intersection P and Q of the chord with the parabola y2 = 4ax is obtained by making the equation homogeneous by means of (i). Thus the equation of lines OP and OQ is

If the lines OP and OQ are at right angles, then the coefficient of x2 + the coefficient of y2 = 0. Therefore, β2 - 2aα + 8a= 0  ⇒ b2 = 2a(α - 4a). Hence the locus of (a, β) is y2 = 2a(x - 4a)

Q. AN IMPORTANT CONCEPT

If a family of straight lines can be represented by an equation λ2P + λQ + R = 0 where l is a parameter and P, Q, R are linear functions of x and y then the family of lines will be tangent to the curve Q2 = 4PR.

Ex.20 If the equation m2(x + 1) + m(y – 2) + 1 = 0 represents a family of lines, where ‘m’ is parameter then find the equation of the curve to which these lines will always be tangents.

Sol. m2(x + 1) + m(y – 2) + 1 = 0. The equation of the curve to which above lines will always be tangents can be obtained by equating its discriminant to zero.

(y – 2)2 – 4(x + 1) = 0

⇒ y2 – 4y + 4 – 4x – 4 = 0

⇒ y2 = 4(x + y)

R. DIAMETER

The locus of the middle points of a system of parallel chords of a Parabola is called DIAMETER. Equation to the diameter of a parabola is y = 2a/m, where m = slope of parallel chords.

Ex.21 The common tangent of the parabola y2 = 8ax and the circle x2 + y2 = 2a2 is

Sol. Any tangent to parabola is y = mx + 2a/m

B2 - 4AC, gives m = +- 1,

Tangent y = +-x,+-a

Ex.23 If P(–3, 2) is one end of the focal chord PQ of the parabola y2 + 4x + 4y = 0, then the slope of the normal at Q is

Sol. The equation of the tangent at (–3, 2) to the parabola y2 + 4x + 4y = 0 is

2y + 2(x – 3) + 2(y + 2) = 0 or 2x + 4y – 2 = 0  =  x + 2y – 1 = 0

Since the tangent at one end of the focal chord is parallel to the normal at the other end, the slope of the normal at the other end of the focal chord is =-1/2

Ex.25 If r1, r2 be the length of the perpendicular chords of the parabola y2 = 4ax drawn through the vertex, then show that

Sol. Since chord are perpendicular, therefore if one makes an angle θ then the other will make an angle (90º – θ) with x-axis

Let AP = r1 and AQ = r2

If ∠PAX = θ  then  ∠QAX = 90º – θ

Co-ordinates of P and Q are (r1 cosq, r1 sinq) and (r2 sinθ, – r2 cosθ) respectively.

Since P and Q lies on y2 = 4ax

S. IMPORTANT HIGHLIGHTS OF PARABOLA :

(1) If the tangent & normal at any point ‘P’ of the para bola intersect the axis at T & G then ST = SG = SP where ‘S’ is the focus. In other words the tangent and the normal at a point P on the parabola are the bisectors of the angle between the focal radius SP & the perpendicular from P on the directrix. From this we conclude that all rays emanating from S will become parallel to the axis of the parabola after reflection.

(2) The portion of a tangent to a parabola cut off between the directrix & the curve subtends a right angle a the focus.

(3) The tangents at the extremities of a focal chord intersect at right angles on the directrix, and a circle on any focal chord as diameter touches the directrix. Also a circle on any focal radii of a point P (at2, 2at) as diameter touches the tangent at the vertex and intercepts a chord of length    on a normal at the point P.

(4) Any tangent to a parabola & the perpendicular on it from the focus meet on the tangent at the vertex.

(5) If the tangents at P and Q meet in T, then

(i) TP and TQ subtend equal angles at the focus S

(ii) ST2 = SP.SQ, and

(iii)the triangles SPT and STQ are similar

Proof. Let P be the point  (at2, 2at) and Q be the point (at22  2at2 ) . Co-ordinates of T which is the point of intersection of tangents at P and Q is  {at1t2, a(t1 + t2)}

(i) The equation of SP is y

The perpendicular distance TU, from T on the straight line

Similarly TU has the same numerical value. The angles PST and QST are therefore equal.

(ii) We have SP = a (1 + t12 ) and SQ = a (1 + t22

Also ST= (at1t2 – a)+ a2 (t1 + t2)2

Hence ST2 = SP . SQ.

(iii)Since ST/SP = SQ/ST and the angles TSP and TSQ are equal, the triangles SPT and STQ are similar, so that ∠SQT = ∠STP and∠STQ = ∠SPT.

(6) Tangents and normals at the extremities of the latus rectum of a parabola y2 = 4ax constitute a square, their point of intersection being (–a, 0) & (3a, 0)

(7) Semi latus rectum of the parabola y2 = 4ax, is the harmonic mean between segments of any focal chord of the parabola is

(8) The circle circumscribing the D formed by any 3 tangents to a parabola passes through the focus.

(9) The orthocentre of any triangle formed by three tangents to a parabola lies on the directrix.

Proof. Let the equations to the three tangents be t1y = x + at12  .....(i)

t2y = x + at22 ....(ii)   and

t3y = x + at32 ......(iii)

The point of intersection of (ii) and (iii) is found, by solving them, to be (at2t3, a(t2 + t3)) The equation to the straight line through this point perpendicular to (i) is

y – a(t2+ t3) = –t1(x – at2t3) i.e. y + t1x = a(t2 + t3 + t1t2t3) ...(iv)

Similarly, the equation to the straight line through the line intersection of (iii) and (i) perpendicular to (ii) is y + t2x = a(t3 + t1 + t1t2t3) ...(v)

and the equation to the straight line through the intersection of (i) and (ii) perpendicular to (iii) is y + t1x = a(t1 + t2 + t1t2t3) (vi)

The point which is common to the straight lines (iv), (v) and (vi)

i.e. the orthocentre of the triangle, is easily seen to be the point whose coordinates are x = –a, y = a(t1 + t2 + t3 + t1t2t3) and this point lies on the directrix.

(10) The area of the D formed by 3 points on a parabola is twice the area of the D formed by the tangents at these points

Proof. Let the three points on the parabola be (at12, 2at1) , (at22  2at 2 ) and (at32 2at3 )

The area of the triangle formed by these points

The points of intersection of the tangents at these points are

(at2t3, a(t2 + t3)), (at3t1, a(t3 + t1)) and (at1t2, a(t1 + t2))

The area of the triangle formed by these three points

(11) A circle circumscribing the D formed by the 3 co-normal points. normals at which meet at (h, k) passes through the vertex of the parabola and its equation is 2(x+ y2) – 2(h + 2a) x – ky = 0

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## FAQs on Parabola - Mathematics (Maths) Class 11 - Commerce

 1. What is a parabola?
Ans. A parabola is a type of curve that is commonly found in mathematics, physics, and engineering. It is a two-dimensional object that is symmetrical and has a U-shape. The parabola is defined by its focus point and its directrix line.
 2. What is the equation of a parabola?
Ans. The equation of a parabola depends on its orientation. If the parabola opens upward or downward, its equation will be in the form of y = a(x-h)^2 + k. If the parabola opens left or right, its equation will be in the form of x = a(y-k)^2 + h. The values of a, h, and k will determine the size, location, and shape of the parabola.
 3. What are some real-life examples of parabolas?
Ans. There are many real-life examples of parabolas. Some common examples include the shape of satellite dishes, the trajectory of a ball thrown in the air, the shape of a suspension bridge, and the shape of a reflector in a car headlight. Parabolas can also be seen in the design of roller coasters and in the shape of some musical instruments.
 4. What is the focus of a parabola?
Ans. The focus of a parabola is a point on the interior of the curve that is equidistant from the directrix line and the vertex of the parabola. This point is important because it helps to define the shape of the parabola and its properties, such as its axis of symmetry and its focal length.
 5. How is the vertex of a parabola calculated?
Ans. The vertex of a parabola is located at the point (h,k), where h and k are the x- and y-coordinates of the point where the axis of symmetry intersects the parabola. The formula for finding the vertex is given by (h,k) = (-b/2a, c - b^2/4a), where a, b, and c are the coefficients of the quadratic equation that defines the parabola.

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