Students must practice these Maths 2B Important Questions TS Inter Second Year Maths 2B Parabola Important Questions Very Short Answer Type to help strengthen their preparations for exams.

## TS Inter Second Year Maths 2B Parabola Important Questions Very Short Answer Type

Question 1.

Find the vertex and focus, and the equations of the directrix and axis of the parabola y^{2} = 16x.

Solution:

Given parabola is y^{2} = 16x

Compare with y^{2} = 4ax

we get 4a = 16 ⇒ a = 4

Vertex A = (0, 0)

Focus S = (a, 0) = (4, 0)

The equation of the directrix is x + a = 0

⇒ x + 4 = 0

The equation of the axis is y = 0

Question 2.

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

Solution:

Given equation of the parabola is y^{2} + 6y – 2x + 5 = 0

⇒ y^{2} + 6y = 2x – 5

⇒ (y)^{2} + 2 . y . 3 + (3)^{2} – (3)^{2} = 2x – 5

⇒ (y + 3)^{2} – 9 = 2x – 5

⇒ (y + 3)^{2} = 2x + 4

⇒ (y + 3)^{2} = 2(x + 2)

⇒ [y – (-3)]^{2} = 2[x – (-2)]

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

h = -2, k = -3,

4a = 2 ⇒ a = \(\frac{1}{2}\)

(i) Equation of the axis is y = k

⇒ y = -3

⇒ y + 3 = 0

(ii) Equation of the directrix is x = h – a

⇒ x = -2 – \(\frac{1}{2}\)

⇒ x = \(\frac{-5}{2}\)

⇒ 2x + 5 = 0

Question 3.

Find the coordinates of the vertex and focus, the equation of the directrix, and the axis of the parabola x^{2} – 2x + 4y – 3 = 0.

Solution:

Given the equation of the parabola is

x^{2} – 2x + 4y – 3 = 0

⇒ x^{2} – 2x = -4y + 3

⇒ (x)^{2} – 2 . x . (1) + (1)^{2} – (1)^{2} = -4y + 3

⇒ (x – 1)^{2} – 1 = -4y + 3

⇒ (x – 1)^{2} + 0 = -4y + 4

⇒ (x – 1)^{2} = -4(y – 1)

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

h = 1, k = 1, 4a = 4 ⇒ a = 1

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

(ii) Focus = (h, k – a) = (1, 1 – 1) = (1, 0)

(iii) Equation of the directrix is y = k + a

⇒ y = 1 + 1

⇒ y = 2

⇒ y – 2 = 0

(iv) Equation of the axis is x = h

⇒ x = 1

⇒ x – 1 = 0

Question 4.

Find the vertex and focus of x^{2} – 6x – 6y + 6 = 0.

Solution:

Given the equation of the parabola is

x^{2} – 6x – 6y + 6 = 0

⇒ x^{2} – 6x = 6y – 6

⇒ (x)^{2} – 2 . (x) . (3) + (3)^{2} – (3)^{2} = 6y – 6

⇒ (x – 3)^{2} – 9 = 6y – 6

⇒ (x – 3)^{2} = 6y + 3

⇒ (x – 3)^{2} = 6(y + \(\frac{1}{2}\))

⇒ (x – 3)^{2} = 6(y – (\(-\frac{1}{2}\)))

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

h = 3, k = \(\frac{-1}{2}\), 4a = 6 ⇒ a = \(\frac{3}{2}\)

(i) Vertex = (h, k) = (3, \(\frac{-1}{2}\))

(ii) Focus = (h, k + a) = (3, \(\frac{-1}{2}+\frac{3}{2}\)) = (3, 1)

Question 5.

Find the coordinates of the vertex and focus and the equations of the directrix and axis of the parabola x^{2} = -4y.

Solution:

Given parabola is x^{2} = -4y

Comparing with x^{2} = -4ay

we get 4a = 4 ⇒ a = -1

Vertex A = (0, 0)

Focus S = (0, -a) = (0, -1)

The equation of the directrix is y – a = 0

⇒ y – 1 = 0

The equation of the axis is x = 0.

Question 6.

Find the equation of the parabola whose vertex is (3, -2) and focus is (3, 1). [(AP) Mar. ’20, May ’17; (TS) May ’19, Mar. ’18]

Solution:

Given that Vertex = (3, -2)

Focus = (3, 1)

In abscissae of the vertex and focus are equal to 3.

Hence the axis of the parabola is x = 3, a line parallel to Y-axis.

a = distance between focus and vertex

= \(\sqrt{(3-3)^2+(-2-1)^2}\)

= 3

Since in vertex y-coordinate is less than the y-coordinate of focus.

∴ The equation of the parabola is (x – h)^{2} = 4a(y – k)

⇒ (x – 3)^{2} = 4(3) (y + 2)

⇒ (x – 3)^{2} = 12(y + 2)

Question 7.

Find the equations of the parabola whose focus is S(1, -7) and vertex is A(1, -2). [(TS) May & Mar. ’15]

Solution:

Given that,

Focus, S = (1, -7)

Vertex, A(h, k) = (1, -2)

In vertex and focus, x-coordinates are equal to 1.

Hence, the axis of the parabola is x = 1, a line parallel to Y-axis,

a = distance between focus and vertex.

= \(\sqrt{(1-1)^2+(-7+2)^2}\)

= \(\sqrt{0+(-5)^2}\)

= √25

= 5

The y-coordinates of the vertex are greater than the y-coordinate of focus.

Focus is below the vertex.

∴ Equation of the parabola is (x – h)^{2} = -4a(y – k)

⇒ (x – 1)^{2} = -4(5) (y + 2)

⇒ (x – 1)^{2} = -20(y + 2)

Question 8.

Find the coordinates of the points on the parabola y^{2} = 2x whose focal distance is \(\frac{5}{2}\). [(AP) May & Mar. ’15]

Solution:

Given the equation of the parabola is y^{2} = 2x

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

4a = 2 ⇒ a = \(\frac{1}{2}\)

Let, P(x_{1}, y_{1}) be a point on the parabola y^{2} = 2x

Given that, focal distance = \(\frac{5}{2}\)

⇒ x_{1} + a = \(\frac{5}{2}\)

⇒ x_{1} + \(\frac{1}{2}\) = \(\frac{5}{2}\)

⇒ x_{1} = 2

Since P(x_{1}, y_{1}) lies on the parabola y^{2} = 2x then

⇒ \(\mathrm{y}_1^2\) = 2x_{1}

⇒ \(\mathrm{y}_1^2\) = 2(2)

⇒ \(\mathrm{y}_1^2\) = 4

⇒ \(\mathrm{y}_1^2\) = ±2

∴ The required points are (2, 2), (2, -2).

Question 9.

Find the coordinates of the point on the parabola y^{2} = 8x whose focal distance is 10. [(AP) May ’19, Mar. ’17, ’16; (TS) May & Mar. ’17; 14]

Solution:

Given the equation of the parabola is y^{2} = 8x

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

4a = 8 ⇒ a = 2

Let P(x1, y1) be a point on the parabola y^{2} = 8x

Given that focal distance = 10

⇒ x_{1} + a = 10

⇒ x_{1} + 2 = 10

⇒ x_{1} = 8

Since P(x_{1}, y_{1}) lies on the parabola then \(\mathrm{y}_1^2\) = 8x_{1}

⇒ \(\mathrm{y}_1^2\) = 8(8)

⇒ \(\mathrm{y}_1^2\) = 64

⇒ \(\mathrm{y}_1^2\) = ±8

∴ The required points are (8, 8), (8, -8).

Question 10.

If (\(\frac{1}{2}\), 2) is one extremity of a focal chord of the parabola y^{2} = 8x. Find the coordinates of the other extremity. [(AP) May ’18, (TS) ’16; ’14]

Solution:

Given the equation of the parabola is y^{2} = 8x

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

4a = 8 ⇒ a = 2

Given that one end of the focal chord is P = (\(\frac{1}{2}\), 2)

Let other end of the focal chord is Q(x, y) = \((\frac{1}{2}\), 2)

‘PSQ’ is a focal chord.

Since P, S, Q are collinear then

slope of \(\overline{\mathrm{SP}}\) = slope of \(\overline{\mathrm{SQ}}\)

⇒ \(\frac{2-0}{\frac{1}{2}-2}=\frac{y-0}{\frac{y^2}{8}-2}\)

⇒ \(\frac{2}{\frac{-3}{2}}=\frac{y}{\frac{y^2-16}{8}}\)

⇒ \(\frac{4}{-3}=\frac{8 y}{y^2-16}\)

⇒ y^{2} – 16 = -6y

⇒ y^{2} + 6y – 16 = 0

⇒ y^{2} + 8y – 2y – 16 = 0

⇒ y(y + 8) – 2(y + 8) = 0

⇒ (y + 8)(y – 2) = 0

⇒ y + 8 = 0 or y – 2 = 0

⇒ y = -8 or y = 2

If y = -8 then Q = (8, -8)

If y = 2 then Q = (\(\frac{1}{2}\), 2)

∴ Other end of the focal chord Q = (8, -8)

Question 11.

Prove that the point on the parabola y^{2} = 4ax (a > 0) nearest to the focus is its vertex.

Solution:

Given equation of the parabola is y^{2} = 4ax (a > 0)

Let P(at^{2}, 2at) be a point on the parabola y^{2} = 4ax

Which is nearest to the focus, S(a, 0), then

SP = \(\sqrt{\left(a t^2-a\right)^2+(2 a t-0)^2}\)

⇒ SP = \(\sqrt{\left(a t^2-a\right)^2+4 a^2 t^2}\)

⇒ SP^{2} = (at^{2} – a)^{2} + 4a^{2}t^{2}

⇒ SP^{2} = a^{2}(t^{2} – 1)^{2} + 4a^{2}t^{2}

Let f(t) = a^{2}(t^{2} – 1)^{2} + 4a^{2}t^{2}

f'(t) = a^{2} . 2(t^{2} – 1) . 2t + 4a^{2} . 2t

= 4a^{2}t(t^{2} – 1) + 8a^{2}t

= 4a^{2}t(t^{2} – 1 + 2)

= 4a^{2}t (t^{2} + 1)

= 4a^{2}(t^{3} + 1)

for minimum value of f(t), then f'(t) = 0

4a^{2}t(t^{2} + 1) = 0 ⇒ t = 0

f”(t) = 4a^{2}(3t^{2} + 1)

If t = 0, then f”(0) = 4a^{2} > 0

∴ At t = 0, f(t) is minimum

Then P = [a(0)^{2}, 2a(0)] = (0, 0)

∴ The point on the parabola y^{2} = 4ax, which is nearest to the focus is its vertex A(0, 0).

Question 12.

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

Solution:

Since the vertex and focus are on the positive X-axis

∴ The equation of the parabola is

(y – k)^{2} = 4a(x – h) ………..(1)

Now, given that OA = a

vertex, A(h, k) = (a, 0)

a = SA = a’ – a

∴ The equation of the parabola is

(y – 0)^{2} = 4(a’ – a) (x – a)

⇒ y^{2} = 4(a’ – a) (x – a)

Question 13.

Find the value of k if the line 2y = 5x + k is a tangent to the parabola y^{2} = 6x. [(TS) May ’18, Mar. ’16, (AP) ’18]

Solution:

Given the equation of the parabola is y^{2} = 6x

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

4a = 6 ⇒ a = \(\frac{3}{2}\)

Given the equation of the straight line is

2y = 5x + k

⇒ y = \(\frac{5}{2} x+\frac{k}{2}\)

Comparing with y = mx + c, we get

m = \(\frac{5}{2}\), c = \(\frac{k}{2}\)

Given that the line y = \(\frac{5}{2} x+\frac{k}{2}\) is tangent to the parabola y^{2} = 6x then c = \(\frac{a}{m}\)

⇒ \(\frac{k}{2}=\frac{3 / 2}{5 / 2}\)

⇒ \(\frac{k}{2}=\frac{3}{5}\)

⇒ k = \(\frac{6}{5}\)

Question 14.

Find the equation of a tangent to the parabola y^{2} = 16x inclined at an angle of 60° with its axis and the point of contact. [(AP) May ’16]

Solution:

Given the equation of the parabola is y^{2} = 16x

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

Given that inclination of a tangent θ = 60°

The slope of the tangent, m = tan θ

= tan 60°

= √3

∴ The equation of the tangent

Question 15.

Find the condition for the straight line lx + my + n = 0 to be a tangent to the parabola y^{2} = 4ax and find the coordinates of the point of contact. (Mar. ’99, ’93)

Solution:

Suppose lx + my + n = 0 …….(1)

be a tangent to the parabola y^{2} = 4ax

Let P(x_{1}, y_{1}) be the point of contact.

The equation of the tangent at ‘P’ is S_{1} = 0

yy_{1} – 2a(x + x_{1}) = 0

yy_{1} – 2ax – 2ax_{1} = 0

2ax – y_{1}y + 2ax_{1} = 0 ……….(2)

Now (1) & (2) represent the same line

∴ Point of contact P(x_{1}, y_{1}) = \(\left(\frac{\mathbf{n}}{l}, \frac{-2 \mathrm{am}}{l}\right)\)

Suppose ‘P’ lies on the lx + my + n = 0, lx_{1} + my_{1} + n = 0

⇒ \(l\left(\frac{\mathrm{n}}{l}\right)+\mathrm{m}\left(\frac{-2 \mathrm{am}}{l}\right)+\mathrm{n}=0\)

⇒ ln – 2am^{2} + ln = 0

⇒ 2ln – 2am^{2} = 0

⇒ ln – am^{2} = 0

⇒ am^{2} = ln

Question 16.

Find the equation of the tangent and normal to the parabola x^{2} – 4x – 8y + 12 = 0 at (4, \(\frac{3}{2}\)).

Solution:

Given equation of the parabola is x^{2} – 4x – 8y + 12 = 0

given point P(x_{1}, y_{1}) = (4, \(\frac{3}{2}\))

∴ The equation of the tangent is S_{1} = 0

⇒ xx_{1} – 2(x + x_{1}) – 4(y + y_{1}) + 12 = 0

⇒ x(4) – 2(x + 4) – 4(y + \(\frac{3}{2}\)) + 12 = 0

⇒ 4x – 2x – 8 – 4y – 6 + 12 = 0

⇒ 2x – 4y – 2 = 0

⇒ x – 2y – 1 = 0

Slope of x – 2y – 1 = 0 is m = \(\frac{-1}{-2}=\frac{1}{2}\)

Slope of normal is \(\frac{-1}{m}=\frac{-1}{\frac{1}{2}}\) = -2

∴ Equation of normal at (4, \(\frac{3}{2}\)) is

⇒ y – \(\frac{3}{2}\) = -2(x – 4)

⇒ \(\frac{2 y-3}{2}\) = -2x + 8

⇒ 2y – 3 = -4x + 16

⇒ 4x + 2y – 19 = 0

Question 17.

Find the equation of the normal to the parabola y^{2} = 4x which is parallel to y- 2x + 5 = 0. [(TS) Mar. ’19]

Solution:

Given the equation of the parabola is y^{2} = 4x

∴ a = 1

Given the equation of the straight line is y – 2x + 5 = 0

slope m = \(\frac{-(-2)}{1}\) = 2

slope of normal = 2

equation of the normal at ‘t’ is y + xt = 2at + at^{3}

∴ slope = -t = 2

⇒ t = -2

equation of the normal is

y – 2x = 2 . 1(-2) + 1(-2)

⇒ y – 2x = -4 – 8

⇒ y – 2x = -12

⇒ 2x – y – 12 = 0

Question 18.

Show that the equation of the tangent to the parabola y^{2} = 4ax at the point ‘t’ is yt = x + at^{2}.

Solution:

Given the equation of the parabola is y2 = 4ax

Given a point on the parabola is P(at^{2}, 2at)

The equation of the tangent at P(at^{2}, 2at) is S_{1} = 0

⇒ yy_{1} – 2a(x + x_{1}) = 0

⇒ y(2at) – 2a(x + at^{2}) = 0

⇒ yt – (x + at^{2}) = 0

⇒ yt = x + at^{2}

Question 19.

Find the equations of the tangent and normal to the parabola y^{2} = 6x at the positive end of the latus rectum. [(TS) Mar. ’20]

Solution:

Given the equation of the parabola is y^{2} = 6x

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

4a = 6 ⇒ a = \(\frac{3}{2}\)

The positive end of the latus rectum,

L = (a, 2a) = (\(\frac{3}{2}\), 3)

The equation of the tangent at L(\(\frac{3}{2}\), 3) to the parabola y^{2} = 6x is S_{1} = 0

⇒ yy_{1} – 2a(x + x_{1}) = 0

⇒ \(y(3)-2\left(\frac{3}{2}\right)\left(x+\frac{3}{2}\right)=0\)

⇒ \(3 y-3\left(x+\frac{3}{2}\right)=0\)

⇒ 3y – 3x – \(\frac{9}{2}\) = 0

⇒ 6y – 6x – 9 = 0

⇒ 2x – 2y + 3 = 0

The equation of the normal at L(\(\frac{3}{2}\), 3) to the parabola y^{2} = 6x is

y – y_{1} = \(\frac{-y_1}{2 a}\) (x – x_{1})

⇒ y – 3 = \(\frac{-3}{2\left(\frac{3}{2}\right)}\left(x-\frac{3}{2}\right)\)

⇒ y – 3 = \(-1\left(x-\frac{3}{2}\right)\)

⇒ y – 3 = \(\frac{-2 x+3}{2}\)

⇒ 2y – 6 = -2x + 3

⇒ 2x + 2y – 9 = 0

Question 20.

Show that the line 2x – y + 2 = 0 is a tangent to the parabola y^{2} = 16x. Find the point of contact also.

Solution:

Given the equation of the parabola is y^{2} = 16x

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

4a = 16 ⇒ a = 4

Given the equation of the straight line is 2x – y + 2 = 0

⇒ y = 2x + 2

Comparing with y = mx + c we get

m = 2, c = 2

Now, c = 2

\(\frac{\mathrm{a}}{\mathrm{m}}=\frac{4}{2}\) = 2

⇒ c = \(\frac{\mathrm{a}}{\mathrm{m}}\)

∴ The line 2x – y + 2 = 0 is a tangent to the parabola y^{2} = 16x

∴ Point of contact, P = \(\left(\frac{a}{m^2}, \frac{2 a}{m}\right)\)

= \(\left(\frac{4}{(2)^2}, \frac{2 \cdot 4}{2}\right)\)

= (1, 4)

Question 21.

Find the position of the point (6, -6) with respect to the parabola y^{2} = 6x.

Solution:

Given the equation of the parabola is y^{2} = 6x

Comparing with y^{2} = 4ax,

we get 4a = 6 ⇒ a = \(\frac{3}{2}\)

Let the given point A(x_{1}, y_{1}) = (6, -6)

Now \(S_{11}=y_1{ }^2-4 a x_1\)

= (-6)^{2} – 4(\(\frac{3}{2}\))(6)

= 36 – 36

= 0

Since S_{11} = 0, then the point A(6, -6) lies on the parabola y^{2} = 6x.

Question 22.

Find the position of the point (0, 1) with respect to the parabola y^{2} = 6x.

Solution:

Given the equation of the parabola is y^{2} = 6x

Comparing with y^{2} = 4ax,

we get 4a = 6 ⇒ a = \(\frac{3}{2}\)

Let the given point A(x_{1}, y_{1}) = (0, 1)

Now \(S_{11}=y_1{ }^2-4 a x_1\)

= (1)^{2} – 4(\(\frac{3}{2}\))(0)

= 1 > 0

Since S_{11} > 0, then the point A(0, 1) lies outside the parabola y^{2} = 6x.

Question 23.

Find the position of the point (2, 3) with respect to the parabola y^{2} = 6x.

Solution:

Given the equation of the parabola is y^{2} = 6x

Comparing with y^{2} = 4ax,

we get 4a = 6 ⇒ a = \(\frac{3}{2}\)

Let the given point A(x_{1}, y_{1}) = (2, 3)

Now \(S_{11}=y_1{ }^2-4 a x_1\)

= (3)^{2} – 4(\(\frac{3}{2}\))(2)

= 9 – 12

= -3 > 0

Since S_{11} < 0, then the point A(2, 3) lies inside the parabola y^{2} = 6x.

Question 24.

A comet moves in a parabolic orbit with the sun in a focus. When the comet is 2 × 10^{7} km from the sun, the line from the sun to it makes an angle π/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 y^{2} = 4ax.

P is the position of the comet,

given ∠XSP = π/2

SP is perpendicular to the axis of the parabola

SP is the semi latus rectum

2a = 2 × 10^{7}

⇒ a = 10^{7} km

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

Question 25.

A double ordinate of the curve y^{2} = 4ax is of length 8a. Prove that the line from the vertex to it is at right angles.

Solution:

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

Let P = (at^{2}, 2at) and P’ = (at^{2}, -2at) be the ends of double ordinate PP’

given that, the length of the double ordinate = 8a

Question 26.

Find the length of the latus rectum of the parabola y^{2} = 4ax.

Solution:

The equation of the parabola is y^{2} = 4ax …….(1)

Let LSL’ be the latus rectum of the parabola.

Let LL’ be the length of the latus rectum of the parabola y^{2} = 4ax.

If SL = l, then L = (a, l)

Since L lies on the parabola, y^{2} = 4ax, then

l^{2} = 4a × a

⇒ l^{2} = 4a^{2}

⇒ l = 2a

∴ SL = 2a

Now LL’ = 2SL = 2(2a) = 4a

∴ Length of the latus rectum = 4a.

Question 27.

Prove that the equation of the normal to the parabola y^{2} = 4ax at P(x_{1}, y_{1}) is y – y_{1} = \(\frac{-\mathbf{y}_1}{2 a}\)(x – x_{1}).

Solution:

Let S = y^{2} – 4ax = 0 be the given parabola.

The slope of the tangent at P is

m = \(\left(\frac{\mathrm{dy}}{\mathrm{dx}}\right)_{\mathrm{P}}=\frac{2 \mathrm{a}}{\mathrm{y}_1}\)

The slope of the normal at

P = \(\frac{-1}{m}=\frac{-1}{\frac{2 a}{y_1}}=\frac{-y_1}{2 a}\)

∴ The equation of the normal at P is

y – y_{1} = \(\frac{-1}{m}\)(x – x_{1})

y – y_{1} = \(\frac{-y_1}{2 a}\)(x – x_{1})