Solved Examples on Parabola

Parabola is an important head under coordinate geometry. It often fetches good number of questions in various competitive examinations like the JEE. 

Example 1:

Three normals are drawn from the point (c, 0) to the curve y² = x. Show that c must be greater than ½. One normal is always the x-axis. For what value of ‘c’ are the other two normals perpendicular to each other? (IIT JEE 1991)

Solution:

We know that the normal of a parabola is y = mx – 2am – am³.

Hence, the equation of normal for y² = x is y = mx – m/2 – m³/4 

Since the normal passes through (c, 0), we have

mc – m/2 – m³/4 = 0

This gives m(c – 1/2 – m²/4) = 0

This yields either m = 0 or m² = 4(c-1/2)

For m = 0, the equation of normal is y = 0.

Also, m² ≥ 0, so c-1/2 ≥ 0 or c ≥ ½.

At c = ½, we have m = 0.

For the other normals to be perpendicular to each other, we must have m₁m₂ = -1.

m²/4(1/2 – c) = 0 has m₁m₂ = -1.

This means (1/2 - c)/1/4 = -1.

This gives c = ¾.

Solution:

The equation of the normal to the parabola y² = 4ax.is y = mx – 2am – am³.

Hence, the equation of the normal to the parabola y² = 4x is y = mx – 2m – m³.

Now, if it passes through (h, k), then we have k =  mh – 2m – m³

This can be written as m³ + m(2-h) + k = 0 …. (1)

Here, m₁ + m₂ + m₃ = 0

m₁m₂ + m₂m₃ + m₃m₁ = 2 - h

Also, m₁m₂m₃ = -k, where m₁m₂ = α

Now, this gives m₃ = -k/α and this must satisfy equation (1).

Hence, we have (-k/α)³ + (-k/α)(2-h) + k = 0

Solving this, we get k² = α²h – 2α² + α³

And so we have y² = α²x – 2α² + α³

On comparing this equation with y² = 4x we get

α² = 4 and -2α² + α³ = 0

This gives α = 2.

Solution:

Let (x, y) be any point on the parabola. Then by definition, the distance between (x, y) and the focus (3, – 4) must be equal to the length of perpendicular from (x, y) on directrix. So first we will find the equation of the directrix.

The line parallel to 6x – 7y + 9 = 0 is  6x – 7y + 6 = 0 …… (1)

Since directrix passes through (3/2,2), this point wil satisfy equation (1) and hence 6(3/2) – 7(2) + k = 0

⇒ k = – 9 + 14 = 5.

 Equation of directrix is 6x – 7y + 5 = 0

 Now by definition of parabola,

√({(x – 3)² + (y + 4)²}) = (6x – 7y + 5)/√(6² + 7²)

⇒ 85{(x – 3)² + (y + 4)²} = (6x – 7y + 5)²

⇒ 49x² + 36y² + 84xy – 570x + 750y + 2100 = 0

Solution:

First we find the equation of directrix. Let the directrix form the isosceles triangle OAB with OA = OB = a.

Then according to the given condition, ar (? OAB) = 4

⇒ 1/2 a²= 4

⇒ a = ± 2√2 {? the triangle in 3rd quadrant so =-2√2}

Therefore the co-ordinate, of A and B are (–2√2, 0) and (0, –2√2) respectively.

 So, equation of directrix

 (y – 0) = ((0+2√2)/(-2√2-0)) (x + 2√2) ⇒ x + y + 2√2 = 0

 Now the focus S is on line y = x and 2 units away from the origin i.e. OS = 2 ⇒ point (√2, √2) by definition of parabola, we have

 √((x-√2)² + (y-√2)²) = |(x+y+2√2)/√(12+12)|

⇒ x² + y²– 8√2x – 8√2y – 2xy = 0.

Solution:

The given parabola is y² - 2y – 4x + 5 = 0.

This can be rewritten as (y-1)² = 4(x-1)

Its parametric coordinates are x-1 = t² and y -1 = 2t and hence we have P(1 + t², 1 + 2t)

Hence, the equation of tangent at P is

t(y-1) = x – 1 + t², which meets the directrix x = 0 at Q.

Hence, y = 1 + t – 1/t or Q(0, 1 + t – 1/t)

Let R(h, k) be the point which divides QP externally in the ratio 1/2:1. Q is the mid-point of RP

So, we have 0 = (h + t² + 1)/2 which gives t² = – (h + 1) …. (1)

and 1 + t – 1/t = (k + 2t + 1)/2 which gives t = 2/(1- k) ….. (2)

Hence, from equations (1) and (2) we get, 4/(1-k)² + (h + 1) = 0\or (k-1)²(h +1) + 4 = 0

Hence, the locus of a point is

(x + 1)(y - 1)² = 0.

Solution: 

The equation of the tangent to the parabola y² = 4ax, is

 y = mx + a/m …… (1)

 The equation of tangent to the parabola y² = 32x …… (2) is

 y = mx + 8/m …… (3)

 If this line given by (3) is also a tangent to the parabola x² = 108y, then (3) meets x² = 108y …… (4)  in two coincident points

⇒ Substituting the value of y from (3) in (4) we get  x² = 108 [mx + (8/m)]

⇒ mx² – 108m²x – 864 = 0

The roots of this quadratic are equal provided b² = 4ac.

 i.e. (– 108 m²)² = 4m (–864)

⇒ m = (-2)/3 (m ≠ 0, from geometry of curves)

Substituting their value of m in (3), the required equation is

 y = (-2)/3 x + 8/(–2/3)

 y = (-2)/3 x – 12

⇒ 2x + 3y + 36 = 0 

Solution:

The given parabolas are y² = 8x …… (1)

and x² = 27y …… (2)

Solving (1) and (2) we get

(x²/27)² = 8x

⇒ x⁴ = 5832 x

⇒ x⁴ – 5832 x = 0

⇒ x(x³ – 5832) = 0

⇒ x = 0, x = 18

Substituting these values of x in (2) we get y = 0, 12

The point of intersection are (0, 0), (18, 12).

Now, we consider these points one by one

At the point (0, 0)

y² = 8x ⇒ dy/dx = 4/y ⇒ dy/dx|(0,0) = ∞

x² = 27y ⇒ dy/dx = 2x/27 ⇒ dy/dx|(0,0) = 0

⇒ The two curves intersect at the point (0, 0) at right angle.

At the point (18, 12)

y² = 8x ⇒ dy/dx|(18,12) = 4/12 = 1/3 = m₁ (say)

x² = 27y ⇒ dy/dx = 2x/27 ⇒ dy/dx|(18,12) = (2×18)/27 = 4/3 = m₂ (say)

Let θ be the angle at which the two curves intersect at the point (18, 12)

Then tan θ_acute = |(m₂ - m₁)/(1 + m₁ m₂)| = 3/13

θ_acute = tan^-1 (3/13).

Solution:

Let two tangent to the parabola y² = 4ax …… (1)

be yt₁ = x + at₁²…… (2)

and yt₂ = x + at₂²  …… (3)

Let the point of intersection of the tangent be (x₁, y₁) then solving equation (1) and (2) we get, x₁ = at₁t₂

y₁ = a(t₁ + t₂)

Also the slope of these tangents are 1/t₁ and 1/t₂

∴ If α be the angle between these two tangents then

tan α = + (m₁ - m₂)/(1 + m₁m₂) = (±(1/t₁ -1/t₂))/(1 + 1/t₁ × 1/t₂).

⇒ tan α = ± ((t₂ - t₁)/(1 + t₂t₁))

We are given α = π/4

∴ tan π/4 = 1 = ± ((t₂ - t₁)/(1 + t₂t₁))

⇒ (1 + t₁t₂)² = (t₂ – t₁)

⇒ {1 + (x₁/a)²} = (t₁ – t₂)² – 4t₁t₂ = (y₁/a)² - 4x₁/a

⇒ (x₁ + a)² = y₁²– 4ax₁

⇒ Required locus of (x₁, y₁) is (x + a)² = (y² - 4ax).

Solution:

 Let yt = x + at² …… (1)

 be a fixed tangent to the parabola y² = 4ax …… (2)

 Let the other two tangent to (2) be yt₁ = x + at₁²…… (3)

 And yt₂ = x + at₂²…… (4)

 The point of intersection of (1) with (3) and (4) are P {att1, a(t₁ + t₂)} and Q{at t₂, a (t + t₂)}

 Given PQ is constant or PQ2= constant

⇒ a²t₂ (t₁ – t₂)² + a²(t₁ – t₂)² = constant

⇒ a²(t₂ + 1) (t₁ – t₂)²= constant

⇒ (t₁ – t₂)² = constant since t is constant as (1) is a fixed tangent.

⇒ (t₁ – t₂)² = c (say) …… (5)

 Let (x₁, y₁) be the point of intersection of (3) and (4),

 then x₁ = at₁t₂ and y₁ = a(t₁ + t₂) …… (6)

 We know that

 (t₂ – t₁)² = (t₁ + t₂)²

 From equation (5), (6) and 97) we get

 c = (y₁/a)²- 4x₁/a  or y₁² = 4x₁ a – a²

c = 4a (x₁ + 1/4 ac)

⇒ the locus of (x₁, y₁) is  y² = 4a (x + ¼ ac)

⇒ The required locus is a parabola whose latus rectum is 4a i.e. equal to latus rectum of y² = 4ax. 

Solution:

Any normal to the parabola y² = 4ax is …… (1)

y = mx – 2am – am³ …… (2)

Let (x₁, y₁) be the pole of (2) with respect to (1), then (2) is the polar of (x₁, y₁) w.r.t (1)  i.e.

yy₁ = 2a (x + x₁) comparing (2) & (3), we get  2a/m = y₁/1 = (2ax₁)/(-2am – am³)

Hence we get  x₁ = –2a – am²…… (4)

and y₁ = 2a/m …… (5)

Eliminating m between (4) & (5) we get y₁²(x₁ + 2a) + 4a³ = 0

∴ The required locus of (x₁, y₁) is  (x + 2a)y² + 4a³ = 0 

Solution:

The given parabola is y² – kx + 8 = 0.

This can be written as y² = kx - 8

This gives y² = k (x - 8/k)

Shifting the origin we get Y² = kX, where Y = y and X = x-8/k

The directrix of the standard parabola is X = -k/4

The directrix of the original parabola is x = 8/k – k/4

Also, x = 1 coincides with x = 8/k – k/4.

On solving these we get k = 4.

Solution:

Any tangent to the parabola y² = – 4bx is y = mx – b/m

y = mx – 3/m

Let (x₁, y₁) be the mid point of PQ, where P and Q are point of intersection o line (1) and y² = 4ax

Equation of chord PQ is

S₁ = T

y1² – 4ax₁ = y₁y – 2a(x + x₁)

y₁y – 2ax – y₁² + 2 ax₁ = 0 …… (2)

Equation (1) can be written as

my – m²x + 3 = 0 …… (3)

equation (2) and (3) represent the same line

⇒ m/y₁ = m²/2a = 3/(2ax₁ - y₁²) …… (4)

⇒ m = 2a/y₁ (from 4)

Again, from (4), we get

m/y₁ = 3/(2ax₁ - y₁² )

⇒ 2a/y₁ = 3/(2ax₁ - y₁² )

⇒ Locus of (x₁, y₁) is  4a²x = y² (3 + 2a) 

Solution:

The coordinates of the xetremities of the latus rectum of y2 = 4ax are (1, 2) and (1, -2).

The equations of tangents at these points are gievn by 

y.2 = 4(x+1)/2

This gives 2y = 2(x + 1)...... (1)

and y(-2) = 4(x+1)/2 

which gievs -2y = 2(x + 1) ….. (2)

The points of intersection of these tangents can be obtained by solving these two equatiosn simultaneously.

Therefore, -2(x +1) = 2(x + 1)

which gives 0 = 4(x +1)

this yields x = -1 and y = 0.

Hence, the required point is (-1, 0). 

Solution:

Any normal to the parabola y² = 4ax … (1)

is y = mx – 2am – am³ … (2)

This normal (2) meets the axis y = 0 of (1) in G and the tangent at the vertex i.e. x = 0 in H

∴ The coordinates of G and H are (2a + am², 0) and (0, –2am – am³) respectively. Also the vertex A is (0, 0). Let Q be (x₁, y₁)

Given that AGQH is a rectangle. AQ and GH are its diagonals and therefore there mid points are same. Now the mid point of AQ is (1/2 x₁,1/2 y₁ ) and that of GH is  [1/2 (2a + am²+0) 1/2 (0-2am-am3 ) ]

i.e. [1/2 (2a+am2 )-1/2 (2am+am3 ) ]

⇒ The mid points coincide so we have

1/2 x₁ = 1/2 (2a + am2), 1/2 y1 = -1/2 (2am + am3)

or x₁ = 2a + am2, y₁ = – (2am + am3)

The required locus of Q is obtained by eliminating m between these.

Now y₁² = (m2a2) (2 + m2)² = a (m2a) [2 + am2/a]² = a(x₁ – 2a) (x₁/a)²

⇒ ay₁² = (x₁ – 2a) x₁²

So the locus of Q is ⇒ ay² + 2ax² = x³

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