1. Find the equation of the parabola whose axis is parallel to X - axis and which passes through (1, 2), (4, -1) and (2, 3).
Sol: Given points: (1, 2), (4, -1), (2, 3)
Parabola: x = l y2 + my + n (Axis 
X- axis)
At (1, 2): 1 = 4 l + 2m + n ........... (1)
At (4, -1): 4 = l - m + n ........... (2)
At (2, 3): 2 = 9 l + 3m + n ........... (3)
⇒ 2x = y2 - 3y + 4
⇒ y2 - 2x - 3y + 4 = 0
2. Find the equation of the parabola whose axis is parallel to Y- axis and which passes through the points (4, 5), (-2, 11) and (-4, 21)
Sol: Given Points: (4, 5), (-2, 11), (-4, 21)
Parabola: y = lx2 + mx + n (Axis Y - axis)
At (4, 5): 5 = 16 l + 4m + n ........... (1)
At (-2, 11): 11 = 4 l - 2m + n ........... (2)
At (-4, 21): 21 = 16 l - 4m + n ........... (3)
Solving (1) and (2)
⇒ 2y = x2 - 4x + 10
⇒ x2 - 4x - 2y + 10 = 0
3. Find the equation of the parabola whose focus is (4, 5) and vertex at (3, 6). Also find the length of the Latus rectum.
Given Focus : S(4, 5)
Vertex: A(3, 6)
A point on directrix: Z (2, 7)
slope of directrix: 1
Directrix: y - 7 = 1 (x - 2)
⇒ x - y + 5 = 0
Let P(x, y) be any point on the Parabola
From the definition of Parabola: PS = PM
4. Find the vertex, focus, directrix, axis, tangent at the vertex and length of latus rectum of the parabola 4x2 + 12x - 20y + 67 = 0.
Sol: Given Parabola: 4x2 + 12x - 20y + 67 = 0
5. Find the vertex, focus, directrix, axis, tangent at the vertex and length of latus rectum of the parabola y2 - x + 4y + 5 = 0
Sol: Given Parabola: y2 - x + 4y + 5 = 0
⇒ y2 + 4y = x - 5
⇒ y2 + 2 (y) (2) + 4 = x - 5 + 4
⇒ (y + 2)2 = x - 1
⇒ [y-(-2)]2 = x - 1
Comparing with (y - β)2 = 4a (x - α)
β = -2; 4a = 1 ; α = 1
⇒ a = 1/4
Vertex: A (α, β) = A (1, -2)
Focus: S(a + α, β) = S ( 1/4+ 1, -2)
= S (5/4, - 2)
Directrix: x + a - α = 0
⇒ x + 1/4 - 1 = 0
⇒ x - 3/4 = 0
⇒ 4x - 3 = 0
Axis: y - β = 0
⇒ y + 2 = 0
Tangent at the vertex: x - α = 0
⇒ x - 1 = 0
Length of latus rectum = 4a
= 1 unit.
6. Show that the condition that the line y = mx + c may be a tangent to the parabola
y2 = 4ax is c = a/m
Sol: Given Parabola: y2 = 4ax
Tangent at (x1, y1): yy1 = 2a (x + x1)
⇒ yy1 - 2ax - 2ax1 = 0 ........... (1)
Given tangent: y = mx + c
y - mx - c = 0 ........... (2)
7. Show that the equation of the common tangent to the circle x2 + y2 = 2a2 and the parabola
y2 = 8ax is y = (x + 2a)
Sol: Parabola; y2 = 8ax
Centre: (0 , 0)
⇒ m4 + m2 - 2 = 0
⇒ m4 + 2m2 - m2 - 2 = 0
⇒ m2 ( m2 + 2) - 1 (m2 + 2) = 0
⇒ (m2 - 1) (m2 + 2) = 0
⇒ m2 - 1 = 0; m2 + 2 = 0
⇒ m2 = 1 ; m2 = -2
8. From an external point P tangents are drawn to the parabola y2 = 4ax and these tangents make angels θ1, θ2 with it axis, such that cot θ1 + cot θ2 is a constant 'a'. Then show that 'P' lies on a horizontal line.
Sol: Parabola: y2 = 4ax
Tangent:
⇒ m2x − my + a = 0
This tangent passes through P (x1, y1)
⇒ m2x1 - my1 + a = 0
9. If lx + my + n = 0 is a normal to the parabola y2 = 4ax, then show that the
condition is al 3 + 2alm2 + nm2 = 0
Sol: Parabola: y2 = 4ax
Normal at 't' : y + xt = 2at + at3 ........ (1)
Given normal: lx + my + n = 0
⇒ lx + my = -n ........ (2)
Comparing (1) and (2)
⇒ 2alm2 + al 3 = - nm2
⇒ al 3 + 2alm2 + nm2 = 0
10. The normal at 't1' on the parabola y2 = 4ax meets the curve again at 't2' then show that 
Sol: Given parabola: y2 = 4ax
Normal: y - 2at1 = - t1 (x - at12)
⇒ y - 2at1 = -xt1 + at13
⇒ y + xt1 = 2at1 + at13
This equation again passing through (at22, 2at2)
⇒ 2at2 + at22t1 = 2at1 + at13
⇒ 2at2 - 2at1 = at13 - at22 t1
⇒ 2a (t2 - t1) = at1 (t12 - t22)
⇒ -2(t1 - t2) = t1(t1 - t2) (t1 + t2)
11. Show that the Locus of poles of chords of the parabola y2 = 4ax which subtend a right angle at the vertex is x + 4a = 0
Sol: Given parabola: y2 = 4ax
Polar of (x1, y1), yy1 = 2 a (x + x1)
⇒ yy1 - 2ax = 2ax1
Pair of straight line: y2x1 - 2xyy1 + 4ax2 = 0
Coefficient of x2 : 4a
Co-efficient of y2 : x1
Angle: 90°
⇒ Coefficient of x2 + coefficient of y2 = 0
⇒ 4a + x1 = 0
Locus : x + 4a = 0
12. The Polar of P w.r.t the parabola y2 = 4ax touches the circle x2 + y2 = 4a2. Find the locus of P.
Sol: Given Parabola: y2 = 4ax
Polar of (x1, y1): yy1 = 2a (x + x1)
⇒ yy1 - 2ax - 2ax1 = 0
Given Circle: x2 + y2 = 4a2
Centre: (0, 0)
Radius: 2a
⇒ x12 = y12 + 4a2
⇒ x12 - y12 = 4a2
Locus of P (x1, y1): x2 - y2 = 4a2
13. Show that the locus of the chord of the parabola y2 = 4ax which subtend a constant angle α , at the vertex is (x + 4a)2 = 4 cot2 α (y2 - 4ax)
Sol: Let Pole: (x1, y1)
Parabola: y2 = 4ax
Polar of (x1, y1): yy1 = 2a (x + x1)
⇒ yy1 - 2ax = 2ax1
Homogeneous equation: y2 - 4ax (1) = 0
⇒ y2x1 - 2xyy1 + 4ax2 = 0
⇒ 4ax2 - 2y1xy + x1y2 = 0
⇒ a = 4a ; 2h = -2y ; b = x1
h = -y1
⇒ tan2 α (4a + x1)2 = 4 (y12 - 4ax1)
Locus of (x1, y1): tan2 α (4a + x)2 = 4(y2 - 4ax)
⇒ (x + 4a)2 = 4 cot2 α (y2 - 4ax)
14. The Coordinates of the ends of a focal chord of the parabola y2 = 4ax are (x1,y1) and (x2, y2). Prove that x1x2 = a2, y1y2 = - 4a2
Sol: Equation of the chord joining the points (at12, 2at1) and (at22, 2at2) is
⇒ y(t2 + t1) - 2at1(t2 + t1) = 2x - 2at12
⇒ y(t2 + t1) - 2at1t2 - 2at12 = 2x - 2at12
⇒ 2x - y(t2 + t1) + 2at1t2 = 0
This Chord passes through S(a, 0)
⇒ 2a + 2at1t2 = 0
⇒ 1 + t1 t2 = 0
Given: P(x1, y1) and Q(x2, y2) are the ends of the focal chord
⇒ x1x2 = a2 t12 t22
⇒ x1x2 = a2 (t1t2)2
⇒ x1 x2 = a2 (-1)2
⇒ y1y2 = 4a2 t1 t2
⇒ y1 y2 = 4a2 (-1)
15. Tangents are drawn to the parabola y2 = 4ax at points whose abscissae are in the ratio k : 1. Prove that they intersect on the curve 
Sol: Given Parabola: y2 = 4ax
Let Points: (at12, 2at1 ), (at22, 2at2)
By data: at12 : at22 = k : 1
t1 = t2 
Point of intersection of tangents at t1, t2 : [at1t2, a(t1 + t2)]
Let Point of intersection of the tangents: (x1, y1)
x1 = at1t2 ; y1 = a (t1 + t2)
⇒ x1 = at22 y1 = a (t2 + t2)
Locus of (x1, y1) : 
16. Show that the length of the chord of the contact of the (x1, y1) w.r.t. the Parabola y2 = 4ax is 
and also show that the area of the triangle formed by the tangents from (x1, y1) and the chord of contact 
Sol: Given Parabola: y2 = 4ax
Given Point: P (x1, y1)
Let Contact Points: Q (at12, 2at1)
and R (at22, 2at2)
Point of Intersection of
tangents at Q and R : (at1t2, a (t1 + t2))
By data: x1 = a t1t2 ; y1 = a (t1 + t2)
Length of the Chord of Contact:
17. Prove that the locus of the point of interaction of two perpendicular normals to the parabola y2 = 4ax is the parabola y2 = a (x - 3a)
Sol: Parabola; y2 = 4ax
Given point: P (x1, y1)
Normal at 't': y + xt = 2at + at3
If it passes through P (x1, y1)
⇒ y1 + x1t = 2at + at3
⇒ at3 + 2at - x1t - y1 = 0
⇒ at3 + (2a - x1)t - y1 = 0
Let roots; t1, t2, t3
t1t2t3 = 
.......... (1)
Slope of the normal at 't1': - t1
Slope of the normal at 't2': - t2
Given: PA 
PB
(Slope of PA) (Slope of PB) = -1
(-t1) (-t2) = -1
t1t2 = -1 ............(2)
From (1) and (2) : (-1) t3 = 
t3 = 
Since the normal at t3 also passes through P(x1, y1)
⇒ y1 + x1t3 = 2at3 + at33
⇒ a2 - ax1 = - 2a2 - y12
⇒ y12 - ax1 + 3a2 = 0
⇒ y12 - a (x1 - 3a) = 0
⇒ y12 - a (x1 - 3a)
Locus of P (x1, y1): y2 = a (x - 3a)
18. Find the equations of tangent and normal to the parabola y2 = 8x at (2, 4).
Sol: Given Parabola: y2 = 8x
4a = 8
19. Write down the equation of the tangent to the parabola y2 = 16x inclined at 60° to X -axis.
Sol : Given parabola: y2 = 16x
⇒ 4a = 16
Slope of the tangent : m = tan60°
20. Show that the line 2x - y + 2 = 0 is tangent to the parabola y2 = 16x. Find the point of contact also.
Sol: Given Parabola: y2 = 16x
⇒ 4a = 16
Given line: 2x - y + 2 = 0
⇒ y = 2x + 2
⇒ m = 2; c = 2
Condition 
is satisfied with the 2x - y + 2 = 0 and the parabola
y2 = 16x.
The given line is a tangent to the given parabola.
21. Find the equation of the tangent of the parabola y2 = 8x which is parallel to the line x - y + 3 = 0
Sol: Given parabola: y2 = 8x
⇒ 4a = 8
⇒ a = 2 Given line: x - y + 3 = 0
⇒ y = x + 3
⇒ m = 1
Equation to the tangent parallel to x - y + 3 = 0
22. Find the equation of the normal to the parabola y2 = 16x which is perpendicular to the line 2x - y + 5 = 0.
Sol. Given parabola: y2 = 16x
⇒ 4a = 16
⇒ a = 4
Given line: 2x - y + 5 = 0
⇒ y = 2x + 5
⇒ m = 2
Equation to the normal
Perpendicular to 2x - y + 5 = 0
⇒ x + 2y + 16 = 0
23. Find the equation of the normal to the parabola y2 = 4x and whose slope is 2
Sol: Given Parabola: y2 = 4x
⇒ 4a = 4
⇒ a = 1
Given slope to the normal: m = 2
Required normal: y = mx - 2 am - am3
⇒ y = 2x - 2 (1) (2) - (1) (2)3
⇒ y = 2x - 12
⇒ 2x - y - 12 = 0
24. Find the coordinates of the points on the parabola y2 = 2x, whose focal distance is 
Sol: Given Parabola: y2 = 2x
⇒ 4a = 2
⇒ a = 
let P (x1, y1) be any point on the parabola
y12 = 2x1 .............. (1)
Given focal distance:
⇒ x1 + a =
⇒ x1 + =
⇒ x1 = 2
From (1): y12 = 4
⇒ y1 = 
2
Required Points: (2, 2) and (2, -2)
25. Find the equation of the parabola whose vertex is (3, -2) and focus is (3, 1).
Sol: Given; Vertex; A (3, -2)
Focus: S (3, 1)
AS = a = 
= 3
X- coordinates are equal in A and S
⇒ Axis is parallel to Y-axis
Required parabola: (x - α)2 = 4a (y - β)
⇒ (x - 3)2 = 4(3) (y + 2)
= x2 + 9 - 6x = 12y + 24
⇒ x2 - 6x - 12y -1 5 = 0
26. ( 
, 2 ) is one extremity of a focal chord of the parabola y2 = 8x. Find the Coordinates of the other extremity
Sol: Given parabola: y2 = 8x
⇒ 4a = 8
⇒ a = 2
⇒ Focus = S( 2 , 0 )
Given extremity of a focal chord: P ( , 2 )
⇒ - 4y12 + 64 = 24y1
⇒ y12 + 6y1 - 16 = 0
⇒ y12 + 8y1 - 2y1 - 16 = 0
⇒ y1(y1 + 8) - 2 (y1 + 8) = 0
⇒ (y1 - 2) (y1 + 8) = 0
⇒ y1 = 2 ; y1 = -8
Other extremity: ( 8, -8 )
27. A comet moves in parabolic orbit with sun as focus. When the comet is 2 × 107 km. From the Sun, the line from Sun to it makes an angle 
with the axes of the orbit. Find how near the comet comes to the sun.
Sol: Let the equation of the parabolic orbit of the comet y2 = 4ax
Let the position of the comet: P
SP is perpendicular to the axis of the parabola Also, SP is the semi latus rectum
⇒ 2a = 2 × 107
⇒ a = 107 km
A is the nearest point on the parabola form focus.
⇒ AS = a = 107 km
The nearest point on the parabola is 107 km from the Sun.
28. Prove that the point on the parabola y2 = 4ax (a > 0) nearest to the focus is the vertex.
Sol: Let P (at2, 2at ) be any point on the parabola y2 = 4ax, which is nearest to the focus
⇒ S (a , 0)
⇒ SP2 = a2 (t2 - 1)2 + 4a2t2 = ƒ (t) (say)
⇒ ƒ' (t) = a2 2(t2 - 1) ( 2t ) + 4a2 ( 2t ) = 4a2t [t2 - 1 + 2]
= 4a2t (t2 + 1)
For Maximum (or) minimum: ƒ'(t) = 0
⇒ 4a2t (t2 + 1) = 0
⇒ t = 0
Also ƒ'' (t) = 4a2 (3t2 + 1)
⇒ ƒ'' (t) = 4a2 > 0 [ t = 0]
⇒ ƒ (t) has minimum value at t = 0
P = (0, 0 )
The point on the parabola y2 = 4ax, which is the nearest to the focus is its vertex.
29. Find the pole of the line 2x + 3y + 4 = 0 with respect to the parabola y2 = 8x.
Sol. Given parabola: y2 = 8x
polar of (x1, y1): yy1 = 4 (x + x1)
⇒ 4x - yy1 + 4x1 = 0 ......... (1)
Given Polar: 2x + 3y + 4 = 0 ......... (2)
Comparing (1) and (2)
30. Show that the lines 2x - y = 0 and 6x - 2y + 1 = 0 are conjugate lines w.r.t the parabola y2 = 2x.
Sol: Given parabola: y2 = 2x
⇒ 4a = 2
⇒ a = 
Given line: 2x - y = 0
l = 2 ; m = -1 ; n = 0
31. Find the value of k if the points ( 1 , 2 ) and ( k, -1 ) an conjugate w.r.t the parabola y2 = 8x.
Sol: Given parabola: y2 = 8x
Given conjugate points: (1, 2) and (k, -1)
By data condition: S12 = 0
⇒ y1y2 = 4 ( x1 + x2 )
⇒ 2(-1) =4 ( 1 + k )
⇒ -2 = 4 + 4k
32. Find the value of k, if the lines 2x + 3y + 4 = 0 and x + y + k = 0 are conjugate w.r.t the parabola y2 = 8x.
Sol: Given parabola: y2 = 8x
⇒ 4a = 8
⇒ a = 2
Given conjugate lines: 2x + 3y + 4 = 0
⇒ l1 = 2 ; m1 = 3 ; n1 = 4
and x + y + k = 0
⇒ l2 = 1 ; m2 = 1 ; n2 = k
By date condition: l1n2 + l2n1 = 2am1m2
⇒ 2(k) + 1(4) = 2 (2) (3) (1)
⇒ 2k + 4 = 12
⇒ k = 4.
