Practicing the Intermediate 2nd Year Maths 2B Textbook Solutions Inter 2nd Year Maths 2B Parabola Solutions Exercise 3(a) will help students to clear their doubts quickly.

## Intermediate 2nd Year Maths 2B Parabola Solutions Exercise 3(a)

I.

Question 1.

Find the vertex and focus of 4y² + 12x – 20y + 67 = 0.

Solution:

Given equation 4y² + 12x- 20y + 67 = 0

4y² – 20y = -12x – 67

y² – 5y = – 3x – \(\frac{67}{4}\)

Adding \(\frac{25}{4}\) on both sides

Question 2.

Find the vertex and focus of x2 -16x- 6y + 6 = 0.

Solution:

Given equation is

x² – 6x – 6y + 6 = 0

x² – 6x = 6y – 6

Adding 9 on both sides

x² – 6x + 9 = 6y + 3

Question 3.

Find the equations of axis and directrix of the parabola y² + 6y – 2x + 5 = 0.

Solution:

y² + 6y = 2x – 5

Adding ‘9’ on both sides we get,

y² + 6y + 9 = 2x – 5 + 9

[y – (-3)]² = 2x + 4

[y – (-3)]² = 2[x – (-2)]

Comparing with (y – k)² = 4a (x- h) we get,

(h, k) – (-2, -3), a = \(\frac{1}{2}\)

Equation of the axis y – k = 0 i.e. y + 3 = 0

Equation of the directrix x – h + a = 0

i.e., x – (-2) + \(\frac{1}{2}\) = 0

2x + 5 = 0.

Question 4.

Find the equation of axis and directrix of the parabola 4x² + 12x – 20y + 67 = 0.

Solution:

4x² + 12x = 20y – 67

x² + 3x = 5y – \(\frac{67}{4}\)

Adding \(\frac{9}{4}\) on both sides we get

Comparing with (x – h)² = 4a(y – k)

(h, k) = (\(\frac{-3}{2},\frac{29}{10}\)) ; a = \(\frac{5}{4}\)

Equation of the axis x – h = 0, i.e., x + \(\frac{3}{2}\) = 0

2x + 3 = 0

Equation of the directrix, y – k + a = 0

y – \(\frac{29}{10}\) + \(\frac{5}{4}\) = 0

⇒ 20y – 33 = 0.

Question 5.

Find the equation of the parabola whose focus is S (1, -7) and vertex is A(1, -2).

Solution:

Let S = (1, -7), A(1, -2)

h = 1, k = -2, a = -2 + 7 = 5

Axis of the parabola is parallel to y-axis Equation of the parabola is

(x – h)² = -4a (y – k)

(x – 1)² = – 20(y + 2)

x² – 2x + 1 = – 20y – 40

⇒ x² – 2x + 20y + 41 = 0.

Question 6.

Find the equation of the parabola whose focus is S(3, 5) and vertex is A(1, 3).

Solution:

Equation of the axis y – 3 = \(\frac{3-5}{1-3}\) (x – 1)

= x – 1

x – y + 2 = 0

The directrix is perpendicular to the axis. Equation of the directrix is x + y + k = 0

Co-ordinates of Z be (x, y)

A is the midpoint of SZ

Co-ordinates of A are \(\frac{3+x}{2},\frac{5+y}{2}\)) = (1, 3)

Co-ordinates of Z are (-1, 1)

The directrix passes through Z (-1, 1)

-1 + 1 + k = 0 ⇒ k = 0

Equation of the directrix is x – y = 0

Equation of the parabola is

⇒ 2(x² – 6x + 9 + y2 – 10y + 25) = (x + y)²

⇒ 2x² + 2y² – 12x – 20y + 68 = x² + 2xy + y²

i.e., x² – 2xy + y² – 12x – 20y + 68 = 0.

Question 7.

Find the equation of the parabola whose latus rectum is the line segment of joining the points (-3, 2) and (-3, 1).

Solution:

L (-3, 2) and L(-3, 1) are the ends of the latus rectum.

S is the midpoint of LL’

Co-ordinates of S are (-3, \(\frac{3}{2}\))

LL’ = \(\sqrt{(-3+3)^{2}+(2 – 1)^{2}}=\sqrt{0+1}\) = 1

4|a| = 1, ⇒ |a| = \(\frac{1}{4}\) ⇒ a = ± \(\frac{1}{4}\)

Case (i) a = –\(\frac{1}{4}\)

Co-ordinates of A are [-3 + \(\frac{1}{4}\), \(\frac{3}{2}\)]

Equation of the parabola is

(2y – 3)² = -(4x + 11)

Case (ii) a = –\(\frac{1}{4}\)

Co-ordinates of A are [-3 – \(\frac{1}{4}\), \(\frac{3}{2}\)]

Equation of the parabola is

i.e., (2y – 3)² = 4x + 13.

Question 8.

Find the position (interior or exterior or on) of the following points with respect to the parabola y² = 6x,

i) (6, -6)

Solution:

Equation of the parabola is

y² = 6x

i.e., S ≡ y² – 6x

S_{11} = (-6)² – 6.6 = 36 – 36 = 0

∴ (6, – 6) lies on the parabola.

ii) (0,1)

S_{11} = 1² – 6.0 = 1 > 0

∴ (0, 1) lies outside the parabola.

iii) (2, 3)

S_{11} = 32 – 6.2

= 9 – 12

= – 3 < 0

∴ (2, 3) lies inside the parabola.

Question 9.

Find the co-ordinates of the point on the parabola y² = 8x whose focal distance is 10.

Solution:

Equation of the parabola is y² = 8x

4a = 8 ⇒ a = 2

Co-ordinates of the focus S are (2, 0)

Suppose P(x, y) is the point on the parabola.

Given SP = 10 ⇒ SP² = 100

(x – 2)² + y2 = 100

But y² = 8x

⇒ (x – 2)² + 8x = 100

⇒ x² – 4x + 4 + 8x – 100 = 0

⇒ x² + 4x – 96 = 0 ⇒ (x + 12) (x – 8) = 0

x + 12 = 0 or x – 8 = 0

x = -12, or 8

Case (i) x = 8

y² = 8.x = 8.8 = 64

y = ±8

Co-ordinates of the required points are (8, 8) and (8, -8)

Case (ii) x = -12

y² = 8(-12) = -96 < 0

y is not real.

Question 10.

If (\(\frac{1}{2}\), 2) j is one extermity of a focal chord of the parabola y² = 8x. Find the co-ordinates of the other extremity.

Solution:

A = \(\frac{1}{2}\), 2); S = (2, 0)

B = (x_{1}, y_{1}) ⇒ (\(\frac{y^{2}_{2}}{8}\), y_{1})

ASB is a focal chord.

∴ Slopes of SA and BS are same.

24y_{1} = -4y²_{1} + 64

or 4y²_{1} + 24y_{1} – 64 = 0

⇒ y²_{1} + 6y_{1} – 16 = 0 ⇒ (y_{1} + 8) (y_{1} – 2) = 0

y_{1} = 2, -8

x_{1} = \(\frac{1}{2}\), 8; So (8, -8) other extremity.

Question 11.

Prove that the parabola y2 – 4ax, (a > O) Nearest to the focus is its vertex.

Solution:

Let P(at², 2at) be the point on the parabola y² = 4ax, which is nearest to the

focus S(a, 0) then

fSp²= (at² – a)² + (2at – 0)²

f(t) = a²2(t² – 1)(2t) + 4a²(2t)

= 4a²t(t² – 1 + 2) = 4a²t(t² + 1)

For minimum value of f(t) = 0 ⇒ t = 0

f”(L) = 4a²(3t² +1) s

f(0) = 4a² > 0

∴ At t-= 0, f(t) is minimum Then P = (0, 0)

∴ The point on the parabola y² = 4ax, which is nearest to the focus is its vertex A(0, 0).

Question 12.

A comet moves in a Parabolic orbit with the sun as focus when the comet is 2 × 10^{7} K.M from the sun, the line from the sun to it makes an fmgle \(\frac{\pi}{2}\) with the axis of the orbit. Find how near the comet comes to the sun.

Solution:

Suppose the equation of the parabolic orbit of the comet is y² = 4ax

P is the position of the comet.

Given ∠XSP = \(\frac{\pi}{2}\)

SP is perpendicular to the axis of the parabola.

SP is the semi-latus rectum

2a = 2 × 10^{7}

⇒ a = 10^{7} km

A is the nearest point on the parabola from focus.

AS = a = 10^{7} km

∴ The nearest point on the parabola is 10^{7} km from the sun.

II.

Question 1.

Find the locus of the points of trisection of double ordinate of a parabola y² = 4ax (a > 0).

Solution:

Equation of the parabola is y² = 4ax

P(x, y) and Q(x, -y) are the ends of the double ordinate.

T divides PQ in the ratio 1 : 2

Co-ordinates of T are (x, \(\frac{-y+2y}{3}\))

= (x, \(\frac{y}{3}\))

T divides PQ in the ratio 2 : 1

Co-ordinates of T are (x, \(\frac{-2y+y}{3}\))

= (x, –\(\frac{y}{3}\))

Suppose Co-ordinates of the required points

L and L’ be (x_{1}, y_{1})

y_{1} = ± \(\frac{y}{3}\) ⇒ y²_{1} = \(\frac{y^{2}}{9}\)

y² = 9y²_{1}

4ax_{1} = 9y²_{1}

Locus of (x_{1}, y_{1}) is 9y² = 4ax.

Question 2.

Find the equation of the parabola whose vertex and focus are on the positive X – axis at a distance of a and a’ from the origin respectively.

Solution:

Given Co-ordinates of A are (a, 0) and S are (a’, 0)

AS = a’ – a

Equation of the parabola is

y² = 4(a’ – a) (x – a).

Question 3.

If L and L‘ are the ends of the latus rectum of the parabola x² = 6y, find the equations of OL and OL’ where ‘O’ is the origin. Also find the angle between them.

Solution:

x² = 6y

Curve is symmetric about Y – axis

Extremities of latus rectum are

(2a, a), (-2a, a)

4a = 6 ⇒ a = \(\frac{3}{2}\)

Question 4.

Find the equation of the parabola whose axis is parallel to X-axis and which passes through these points. (-2, 1) (1, 2), and (-1, 3)

Solution:

Axis is parallel to X – axis

General equation be

x = ay² + by + c

Passes through (-2, 1) (1, 2) (-1, 3)

-2 = a + b + c ………. (i)

1 = 4a + 2b + c ……….. (ii)

-1 = 9a + 3b + c …………. (iii)

⇒ –\(\frac{5{2}\) = a

i.e., \(\frac{21}{2}\) = b

-10 = c

x = –\(\frac{5}{2}\)y² + \(\frac{21}{2}\)y – 10

5y² + 2x – 21y + 20 = 0

Question 5.

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).

Solution:

General equation be y = ax² + bx + c passes through (4, 5), (-2, 11), (-4, 21) we get

5 = 16a + 4b + c ……… (i)

11 = 4a – 2b + c ……….. (ii)

21 = 16a – 4b + c ……….. (iii)

(ii) – (i) we get

6 = -12a – 6b

(iii) – (ii) 10 = 12a – 2b

Solving we get

b = -2, a = 1/2; c = 5

y = \(\frac{1}{2}\)x² – 2x + 5

x² – 2y – 4x + 10 = 0

III.

Question 1.

Find the equation of the parabola whose focus is (-2, 3) and directrix is the line 2x + 3y – 4 = 0. Also find the length of the latus rectum and the equation of the axis of the parabola.

Solution:

Suppose P(x_{1}, y_{1}) is any point on the parabola.

S(-2, 3) is the focus.

SP² = (x_{1} + 2)² + (y_{1} – 3)²

Equation of the directrix is 2x + 3y – 4 = 0

PM is the perpendicular from P on the directrix

PM = \(\frac{2x_{1}+3y_{1}-4}{\sqrt{4+9}}\)

From Def. of parabola SP = PM ⇒ SP² = PM²

(x_{1} + 2)² + (y_{1} – 3)² = \(\frac{(2x_{1}+3y_{1}-4)^{2}}{13}\)

⇒ 13(x²_{1} + 4x_{1} + 4 + y²_{1} – 6y_{1} + 9) = (2x_{1} + 3y_{1} – 4)²

⇒ 13x²_{1} + 13y²_{1} + 52x_{1} – 78y_{1} + 169

= 4x²_{1} + 9y²_{1} + 16 + 12x_{1}y_{1} – 16x_{1} – 24y_{1}

⇒ 9×2_{1} – 12x_{1}y_{1} + 4y²_{1} + 68x_{1} – 54y_{1} + 153 = 0

Locus of P(x_{1}, y_{1}) is

9x² – 12xy + 4y² + 68x- 54y + 153 = 0

Length of the latus rectum = 4a

Perpendicular distance from S on directrix

Length of the latus rectum = 4a = \(\frac{2}{\sqrt{3}}\)

The axis is perpendicular to the .directrix Equation of the directrix can be taken as

3x – 2y + k = 0

This line passes through S (-2, 3)

– 6 – 6 + k = 0 ⇒ k = 12

Equation of the axis is 3x – 2y + 12 = 0

Question 2.

Prove that the area of the triangle inscribed in the parabola y² = 4ax is \(\frac{1}{8a}\)|(y_{1} – y_{2}) (y_{2} – y_{3}) (y_{3} – y_{1})| sq. units where y_{1}, y_{2}, y_{3} are the ordinates of its vertices.

Solution:

Suppose P(at²_{1}, 2at_{1}), Q(at²_{2}, 2at_{2}),

R(at²_{3}, 2at_{3}) are the vertices of ∆PQR.

Area of ∆PQR = \(\frac{1}{2}\) | at²_{1} (2at_{2} – 2at_{3}) + at²_{2} (2at_{3} – 2at_{1}) + at²_{3}(2at_{1} – 2at_{2})|

= \(\frac{1}{2}\) . 2a² |t²_{1} (t_{2} – t_{3}) + t²_{2}(t_{3} – t_{1}) + t²_{3}(t_{1} – t_{2})|

= a² |(t_{1} – t_{2}) (t_{2} – t_{3}) (t_{3} – t_{1})|

= \(\frac{1}{8a}\) |(2at_{1} – 2at_{2}) (2at_{2} – 2at_{3}) (2at_{3} – 2at_{1}) |

= \(\frac{1}{8a}\) |(y_{1} – y_{2}) (y_{2} – y_{3}) (y_{3} – y_{1})|

Where P(x_{1}, y_{1}), Q(x_{2}, y_{2}), R(x_{3}, y_{3}) are the vertices of ∆PQR.

Question 3.

Find the co-ordinates of the vertex and focus, the equation of the directrix and axis of the -following parabolas.

(i) y² + 4x + 4y – 3 = 0

(ii) x² – 2x + 4y – 3 = 0

Solution:

i) y² + 4x + 4y – 3 = 0

⇒ y² + 4y = – 4x + 3

⇒ y² + 4y + 4 = – 4x +. 3 + 4

⇒ (y + 2)² = – 4x + 7

⇒ [y – (-2)]² = -4[x – \(\frac{7}{4}\)]

h = \(\frac{7}{4}\), k = -2, a = 1

Vertex A(h, k) = (\(\frac{7}{4}\), -2)

Focus (h-a, k) = (\(\frac{7}{4}\) – 1, -2)

= (\(\frac{3}{4}\), -2)

Equation of the directrix x – h – a = 0

x – \(\frac{7}{4}\) – 1 = 0

4x – 11 = 0

Equation of the axis is y – k = 0

y + 2 = 0

ii) x² – 2x + 4y – 3 = 0

⇒ x² – 2x = – 4y + 3

⇒ x – 2x + 1 = – 4y + 3 + 1

(x – 1)² = – 4y + 4

= – 4[y – 1 ]

(x – 1)² = – 4[y – 1]

h = 1; k = 1; a = 1

Vertex A(h, k) = (1, 1)

Focus (h, k – a) =(1, 1 -1)

= (1, 0)

Equation of the directrix

y – k – a = 0

y – 1 – 1 = 0

y – 2 = 0

Equation of the axis is,

x – h = 0

x – 1 = 0