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## Karnataka 2nd PUC Basic Maths Question Bank Chapter 16 Parabola

### 2nd PUC Basic Maths Parabola One Mark Questions and Answers

Question 1.

Find the equation of the parabola given that its.

Answer:

(a) Vertex is (0, 0) and focus is (4, 0)

The curve turn right side with a = 4

∴ Equation of parabola is y^{2} = 16x

(b) Vertex is (0, 0) and directrix is y = – 3

The curve turn upwords with a = 3

∴ Equation is x^{2} = 4ay = 4.3y =12y

x^{2} = 12y

(c) Focus is (1, 0) and the directrix is x = – 1

The curve turns right side with a = 1

And its equation is y^{2} = 4x

(d) Focus is (-4, 0) and directrix is x = 4

The curve turns left side and a = 4

The equation is y^{2} = – 16x

(e) Focus is (0, -3) and directrix is y = 3

The curve turns down words and its

Equation is x^{2} = -12y

(f) Focus is (0, 6) and vertex is (0, 0)

The curve turn upwords and a = 6 its

Equation is x^{2} = 24y.

Question 2.

Find the latus rectum of the parabola y^{2} = 16x.

Answer:

Length of L.R = 4a = 16.

Question 3.

Find the focus of the parabola y^{2} = -8x.

Answer:

Comparing with y^{2} = -4ax we get

a = 2, ∴ Focus – (-a, 0) = (-2, 0).

Question 4.

Find the equation of the directrix of the parabolay2 = 8x.

Answer:

Comparing with y^{2} = 4ax, 4a = 8 ⇒ a = 2

∴ Equation of directrix x = – 2.

Question 5.

Find the directrix of the parabola 3x^{2} = -8y.

Answer:

x^{2} = \(\frac{-8}{3} y\)

Comparing with x^{2} = -4ay

4a = \(\frac{8}{3}\) ⇒ a = \(\frac{2}{3}\)

Equation of dinectrix is y = \(\frac{2}{3}\) or 3y – 2 = 0.

Question 6.

In the parabola y^{2} = 8kx, if the length of L.R. is 4 find k.

Answer:

Comparing with y^{2} = 4ax we get 4a = 8k

But 4a = 4 (given) ⇒ 4 = 8k

⇒ k = \(\frac { 1 }{ 2 }\)

Question 7.

Find the length of L.R of 4y^{2} + 16x = 0.

Answer:

Given y^{2} = -4x

Comparing with y^{2} = 4ax we get 4a =4

∴ Length of L.R. = 4.

Question 8.

If the length of the L.R. of x^{2} = 4ky Is 8 find k.

Answer:

Comparing with x^{2} = 4ay

4a = 4k

8 = 4k ⇒ k = 2.

Question 9.

Find the length of the L.R of 3x^{2} + 4y = 0.

Answer:

x^{2} = \(\frac { -4 }{ 2 }y\)

Comparing with x^{2} = -4ay

We get length of L.R. = 4a = \(\frac { 4 }{ 3 }\)

Question 10.

Find the focus of the parabola x^{2} + 16y = 0.

Answer:

x^{2} = 16y

Comparing x^{2} = -4ay

⇒ 4a = 16 ⇒ a = 4

∴ focus = S(0, -4).

Question 11.

Find the co-ordinates of focus and length of L.R of x^{4} + 4y = 0.

Answer:

x^{2} = -4y

Comparing x^{2} – 4ay ⇒ 4a = 4 ⇒ a = 1

LT. of L.R = 4, and focus = (0, – 1).

Question 12.

Find focus and LT of L.R if y^{2} = 12x.

Answer

Comparing with y^{2} = 4ax

We get 4a = 12 ⇒ a = 3

∴ Lt of L.R= 12 and focus = (3, 0).

Question 13.

Find the equation of the parabola given that its vertex (0,0), axis is y-axis and passers through \(\left(\frac{1}{2}, 2\right)\)

Answer:

Given V = (0, 0) and axis is y-axis

∴ The parabola is of the form x^{2} = 4ay or x^{2} = -4ay

But this equation passes through the point \(\left(\frac{1}{2}, 2\right)\)

∴\(\left(\frac{1}{2}\right)^{2}\) = 4a.2 ⇒ \(\frac { 1 }{ 4 }\) = 8a a = \(\frac { 1 }{ 32 }\)

∴ Required equation of the parabola is x^{2} = \(\frac { 4 }{ 32 }y\)

x^{2} = \(\frac { 1 }{ 8 }y\)

8x^{2} = y = 0.

Question 14.

Find the equation of the parabola with vertex at (0, 0), axis is y-axis and it passes through (-1,-3).

Answer:

Given V = (0, 0) and axis is y-axis,

∴ The parabola is of the form x^{2 }= 4ay or x^{2} = 4ay,

But this equation passes through (-1, -3) which ¡s in III quadrant

∴ The curve turn down words and is of the fonn

x^{2} = -4ay, it passes through (-1, -3)

∴ (-1)^{2} = 4a(-3) ⇒ 1 = 12a ⇒ a = \(\frac { 1 }{ 12 }y\)

x^{2} = \(\frac { -1 }{ 3 }\) or 3x^{2} + y = 0

### 2nd PUC Basic Maths Parabola Three or Four Marks Questions and Answers

Question 1.

Write the characteristics of the following.

(i) y^{2} = 16x

Answer:

Comparing with y^{2} = 4ax we get 4a = 16 ⇒ a = 4

Vertex = V(0, 0)

Focus = S(4, 0)

Directrix is x = -4 or x + 4 = 0

Axis x-axis and y = 0

Tangent is y-axis and x = 0

latus rectum is x = 4 or x – 4 = 0

Ends of L.R L (4, 8) L'(4, -8)

Length of L.R = 4a = 4.4 = 16.

Question 2.

y^{2} + 4x = 0.

Answer:

y^{2} = -4x

Comparing this with yy^{2} = -4ax ⇒ 4a = + 4 ⇒ a = 1

Vertex = (0, 0), focus = S(-1, 0), Directrix is x = 1,or x – 1 = 0

Axis x-axis and y = 0, Tangent is y-axis and x = 0

Ends of L.R, L = (-1, 2) L = (-1; -2)

Equation of L.R is x = – 1 or x + 1 = 0

Length of L.R = 4.

Question 3.

x^{2} = 8y

Answer:

Comparing this with x^{2} = 4ay we get 4a = 8 ⇒ a = 2

Vertex = V(0, 0)

Focus = S(0, 2)

Directrix isy = -2 or y + 2 = 0

Axis is y-axis and x = 0,

Tangent is x-axis and y = 0

Ends of L.R., L(4, 2) and L'(-4, 2)

Equation of L.R is y = 2 or y – 2 = 0

Length of L.R = 8.

Question 4.

3x^{2} = -8y.

Answer:

⇒ x^{2} = \(\frac{-8}{3} y\)

Companng with x^{2} = -4ay we get 4a = \(\frac{+8}{3}\) ⇒ a = \(\frac{2}{3}\)

Vertex V (0, 0)

Focus = s\(\left(0, \frac{-2}{3}\right)\)

Directrix is y = or 3x – 2 = 0

Axis ¡s y-axis and x = 0

Tangent is x-axis and y = 0

Ends of L.R. L = \(\left(\frac{4}{3}, \frac{-2}{3}\right)\) and L’= \(\left(\frac{-4}{3}, \frac{-2}{3}\right)\)

EquationofL.R. ¡s y = \(\frac{-2}{3}\) = 3y + 2 = 0

Length of L.R. = \(\frac{8}{3}\)

Question 5.

Find the equation of a parabola with proper choice of co-ordinate axis is y^{2} = 4ax.

Answer:

Proof: Let S be the focus and l be the directrix.

Let SZ = 2 a

⇒ OZ = OS = a

Choose O as the origin and the line ZOX as x-axis and the line YOY’ through O as y-axis. With this choice of co-ordinate axes we have S(a, o) and Z(-a, o)

The equation of directrix is x = – a

Let P(x, y) be any point on the parabola then by the defmation of parabola we have

Distance of P from S = Distance of P fram the line l

i.e. PS = PM ∵ PM = ZN

PS^{2} = ZN^{2}

PS^{2} = (OZ + ON)^{2}

(x – a)^{2} + y^{2} = (a + x)^{2}

y^{2} = 4ax which is the equation of the parabola

Question 6.

Find the equation of parabola if the vertex and axis of parabola are given.

(a) Vertex is at the orgin and axis along x-aixs and passing through the point p(2,3)

Answer:

Given vertex = (0, 0) and x-axis So its equation is either y^{2} = 4ax or y^{2} = – 4ax, but the

parabola passes through the point P(2, 3) which’ is in 1st quadrant

∴ The parabola turns right side i.e., y^{2} = 4ax

It passes through (2, 3) ∴ 9 = 4a.2 ⇒ a = \(\frac{9}{8}\)

∴ Equation of parabola is y^{2} = \(4 . \frac{9}{8} x\)

⇒ 2y^{2} = 9 x.

(b) Vertex at the origin and passing through the point P(2, -3) and which is symmetric about x-axis.

Answer:

Given V = (0,0) and axis is y-axis So its equation is x^{2} = 4ay or x^{2} = -4ay But parabola passes through the point (2, -3). So it lies in the 4th quadrant,

∴ the parabola turns down words i.e. x^{2} = – 4ay

but it passes through the point (2, -3)

∴ (-2)^{2} = – 4a(-3)

4 = 12a ⇒ a = \(\frac{1}{3}\)

∴ Equation of parabola is x^{2} = \(4 . \frac{1}{3} y\)

⇒ 3x^{2} + 4y = 0.