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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.

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

Three normals are drawn from the point (c, 0) to the curve y^{2} = 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)

We know that the normal of a parabola is y = mx – 2am – am^{3}.

Hence, the equation of normal for y^{2} = x is y = mx – m/2 – m^{3}/4

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

mc – m/2 – m^{3}/4 = 0

This gives m(c – 1/2 – m^{2}/4) = 0

This yields either m = 0 or m^{2} = 4(c-1/2)

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

Also, m^{2} ≥ 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 m_{1}m_{2} = -1.

m^{2}/4(1/2 – c) = 0 has m_{1}m_{2} = -1.

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

This gives c = ¾.

Normals are drawn from the point P with slopes m_{1}, m_{2}, m_{3} to the parabola y^{2} = 4x. If locus of P with m_{1}m_{2} = α is a part of parabola itself, then find α. (IIT JEE 2003)

The equation of the normal to the parabola y^{2} = 4ax.is y = mx – 2am – am^{3}.

Hence, the equation of the normal to the parabola y^{2} = 4x is y = mx – 2m – m^{3}.

Now, if it passes through (h, k), then we have k = mh – 2m – m^{3}

This can be written as m^{3 }+^{ }m(2-h) + k = 0 …. (1)

Here, m_{1} + m_{2} + m_{3} = 0

m_{1}m_{2} + m_{2}m_{3 }+_{ }m_{3}m_{1} = 2 - h

Also, m_{1}m_{2}m_{3} = -k, where m_{1}m_{2 }= α

Now, this gives m_{3 }= -k/α and this must satisfy equation (1).

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

Solving this, we get k^{2} = α^{2}h – 2α^{2} + α^{3}

And so we have y^{2} = α^{2}x – 2α^{2} + α^{3}

On comparing this equation with y^{2} = 4x we get

α^{2 }= 4 and -2α^{2 }+^{ }α^{3} = 0

This gives^{ }α^{ }= 2.

Find the equation of the parabola whose focus is (3, –4) and directrix is the line parallel to 6x – 7y + 9 = 0 and directrix passes through point (3/2,2).

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)/√(6^{2 }+ 7^{2})

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

⇒ 49x^{2} + 36y^{2} + 84xy – 570x + 750y + 2100 = 0

Find the equation of the parabola whose directrix makes an isosceles right angled triangle of area 4 square units with the axis in the 3rd quadrant and focus is on the line y = x, 2 units away from the origin.

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 a^{2}= 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)|

⇒ x^{2} + y^{2}– 8√2x – 8√2y – 2xy = 0.

At any point P on the parabola y^{2 }- 2y – 4x + 5 = 0, a tangent is drawn which meets the directrix at Q. Find the locus of point R, which divides QP externally in the ratio 1/2:1.(IIT JEE 2004)

The given parabola is y^{2 }- 2y – 4x + 5 = 0.

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

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

Hence, the equation of tangent at P is

t(y-1) = x – 1 + t^{2}, 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 + t^{2 }+ 1)/2 which gives t^{2} = – (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.

Find the equation of the common tangents to the parabola y^{2} = 32x and x^{2} = 108y.

The equation of the tangent to the parabola y^{2} = 4ax, is

y = mx + a/m …… (1)

The equation of tangent to the parabola y^{2} = 32x …… (2) is

y = mx + 8/m …… (3)

If this line given by (3) is also a tangent to the parabola x^{2} = 108y, then (3) meets x^{2} = 108y …… (4) in two coincident points

⇒ Substituting the value of y from (3) in (4) we get x^{2} = 108 [mx + (8/m)]

⇒ mx^{2} – 108m^{2}x – 864 = 0

The roots of this quadratic are equal provided b^{2} = 4ac.

i.e. (– 108 m^{2})^{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

Find the angle of intersection of the parabola y^{2} = 8x and x^{2} = 27y.

The given parabolas are y^{2} = 8x …… (1)

and x^{2} = 27y …… (2)

Solving (1) and (2) we get

(x^{2}/27)^{2} = 8x

⇒ x^{4} = 5832 x

⇒ x^{4} – 5832 x = 0

⇒ x(x^{3} – 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)

y^{2} = 8x ⇒ dy/dx = 4/y ⇒ dy/dx|(0,0) = ∞

x^{2 }= 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)

y^{2 }= 8x ⇒ dy/dx|(18,12) = 4/12 = 1/3 = m_{1} (say)

x^{2 }= 27y ⇒ dy/dx = 2x/27 ⇒ dy/dx|(18,12) = (2×18)/27 = 4/3 = m_{2} (say)

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

Then tan θ_{acute} = |(m_{2 }- m_{1})/(1 + m_{1} m_{2})| = 3/13

θ_{acute} = tan^{-1} (3/13).

Prove that (x + a)^{2} = (y^{2 }- 4ax), is the locus of the point of intersection of the tangents to the parabola y^{2 }= 4ax, which includes an angle π/4.

Let two tangent to the parabola y^{2} = 4ax …… (1)

be yt_{1} = x + at_{1}^{2}…… (2)

and yt_{2} = x + at_{2}^{2 } …… (3)

Let the point of intersection of the tangent be (x_{1}, y_{1}) then solving equation (1) and (2) we get, x_{1} = at_{1}t_{2}

y_{1} = a(t_{1} + t_{2})

Also the slope of these tangents are 1/t_{1} and 1/t_{2}

∴ If α be the angle between these two tangents then

tan α = + (m_{1 }- m_{2})/(1 + m_{1}m_{2}) = (±(1/t_{1} -1/t_{2}))/(1 + 1/t_{1} × 1/t_{2}).

⇒ tan α = ± ((t_{2 }- t_{1})/(1 + t_{2}t_{1}))

We are given α = π/4

∴ tan π/4 = 1 = ± ((t_{2 }- t_{1})/(1 + t_{2}t_{1}))

⇒ (1 + t_{1}t_{2})^{2} = (t_{2} – t_{1})

⇒ {1 + (x_{1}/a)^{2}} = (t_{1} – t_{2})^{2} – 4t_{1}t_{2} = (y_{1}/a)^{2 }- 4x_{1}/a

⇒ (x_{1} + a)^{2} = y_{1}^{2}– 4ax_{1}

⇒ Required locus of (x_{1}, y_{1}) is (x + a)^{2} = (y^{2 }- 4ax).

If two tangents to a parabola intercept a constant length on any fixed tangent, find the locus of their point of intersection.

Let yt = x + at^{2} …… (1)

be a fixed tangent to the parabola y^{2 }= 4ax …… (2)

Let the other two tangent to (2) be yt_{1} = x + at_{1}^{2}…… (3)

And yt_{2} = x + at_{2}^{2}…… (4)

The point of intersection of (1) with (3) and (4) are P {att1, a(t_{1} + t_{2})} and Q{at t_{2}, a (t + t_{2})}

Given PQ is constant or PQ2= constant

⇒ a^{2}t_{2} (t_{1} – t_{2})^{2 }+ a^{2}(t_{1} – t_{2})^{2} = constant

⇒ a^{2}(t_{2} + 1) (t_{1} – t_{2})^{2}= constant

⇒ (t_{1} – t_{2})^{2} = constant since t is constant as (1) is a fixed tangent.

⇒ (t_{1} – t_{2})^{2} = c (say) …… (5)

Let (x_{1}, y_{1}) be the point of intersection of (3) and (4),

then x_{1} = at_{1}t_{2} and y_{1} = a(t_{1} + t_{2}) …… (6)

We know that

(t_{2} – t_{1})^{2} = (t_{1} + t_{2})^{2}

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

c = (y_{1}/a)^{2}- 4x_{1}/a or y_{1}^{2 }= 4x_{1} a – a^{2}

c = 4a (x_{1 }+ 1/4 ac)

⇒ the locus of (x_{1}, y_{1}) is y^{2 }= 4a (x + ¼ ac)

⇒ The required locus is a parabola whose latus rectum is 4a i.e. equal to latus rectum of y^{2} = 4ax.

Find the locus of the poles of normals to parabola y^{2 }= 4ax.

Any normal to the parabola y^{2 }= 4ax is …… (1)

y = mx – 2am – am^{3} …… (2)

Let (x_{1}, y_{1}) be the pole of (2) with respect to (1), then (2) is the polar of (x_{1}, y_{1}) w.r.t (1) i.e.

yy_{1} = 2a (x + x_{1}) comparing (2) & (3), we get 2a/m = y_{1}/1 = (2ax_{1})/(-2am – am^{3})

Hence we get x_{1} = –2a – am^{2}…… (4)

and y_{1} = 2a/m …… (5)

Eliminating m between (4) & (5) we get y_{1}^{2}(x_{1} + 2a) + 4a^{3} = 0

∴ The required locus of (x_{1}, y_{1}) is (x + 2a)y^{2 }+ 4a^{3} = 0

If the line x -1 = 0 is the directrix of the parabola y^{2} – kx + 8 = 0, then one of the values of k is (IIT JEE 2000)

The given parabola is y^{2} – kx + 8 = 0.

This can be written as y^{2} = kx - 8

This gives y^{2} = k (x - 8/k)

Shifting the origin we get Y^{2} = 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.

A tangent to the parabola y^{2} + 12x = 0 cuts the parabola y^{2} = 4ax at P and Q. Find the locus of middle points of PQ.

Any tangent to the parabola y^{2} = – 4bx is y = mx – b/m

y = mx – 3/m

Let (x_{1}, y_{1}) be the mid point of PQ, where P and Q are point of intersection o line (1) and y^{2} = 4ax

Equation of chord PQ is

S_{1} = T

y1^{2} – 4ax_{1} = y_{1}y – 2a(x + x_{1})

y_{1}y – 2ax – y_{1}^{2} + 2 ax_{1 }= 0 …… (2)

Equation (1) can be written as

my – m^{2}x + 3 = 0 …… (3)

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

⇒ m/y_{1} = m^{2}/2a = 3/(2ax_{1 }- y_{1}^{2}) …… (4)

⇒ m = 2a/y_{1} (from 4)

Again, from (4), we get

m/y_{1} = 3/(2ax_{1 }- y_{1}^{2} )

⇒ 2a/y_{1} = 3/(2ax_{1 }- y_{1}^{2} )

⇒ Locus of (x_{1}, y_{1}) is 4a^{2}x = y^{2} (3 + 2a)

The point of intersection of the tangents at the ends of the latus rectum of the parabola y^{2} = 4x is ..?(IIT JEE 1994)

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).

The normal at any point P of the y^{2} = 4ax meets the axis in G and to the tangent at the vertex at H. If A be the vertex and the rectangle AGQH be completed, prove that the locus of Q is x^{3 }= 2ax^{2} + ay^{2}.

Any normal to the parabola y^{2 }= 4ax … (1)

is y = mx – 2am – am^{3} … (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 + am^{2}, 0) and (0, –2am – am^{3}) respectively. Also the vertex A is (0, 0). Let Q be (x_{1}, y_{1})

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 x_{1},1/2 y_{1} ) and that of GH is [1/2 (2a + am^{2}+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 x_{1} = 1/2 (2a + am2), 1/2 y1 = -1/2 (2am + am3)

or x_{1} = 2a + am2, y_{1} = – (2am + am3)

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

Now y_{1}^{2} = (m2a2) (2 + m2)^{2} = a (m2a) [2 + am2/a]^{2} = a(x_{1} – 2a) (x_{1}/a)^{2}

⇒ ay_{1}^{2} = (x_{1} – 2a) x_{1}^{2}

So the locus of Q is ⇒ ay^{2} + 2ax^{2} = x^{3}

You may wish to refer working rule to calculate area.

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