Anand Classes provides NCERT Solutions for Class 11 Maths Chapter 10 – Conic Sections (Exercise 10.2: Parabola) as per the latest NCERT and CBSE syllabus (2025–2026). This exercise focuses on the concept and derivation of the parabola, its standard equation, geometrical properties, and latus rectum. Students will learn how to find the focus, directrix, and axis of a parabola and solve problems based on equations of parabolas in different forms. Each question is solved step-by-step to strengthen understanding of coordinate geometry and prepare for CBSE exams, JEE Main, JEE Advanced, NDA, and CUET. Click the print button to download study material and notes.
NCERT Question 1 : Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola y2 = 12x
Solution :
Compare with standard form y2 = 4ax.
4a = 12 ⇒ a = 3.
Focus = (a, 0) = (3, 0).
Axis = x-axis.
Directrix = x = −a ⇒ x = −3.
Length of latus rectum = 4a = 12.
NCERT Question 2 : Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola x2 = 6y
Solution
Compare with standard form x2 = 4ay.
4a = 6 ⇒ a = 6/4 = 3/2.
Focus = (0, a) = (0, 3/2).
Axis = y-axis.
Directrix = y = −a ⇒ y = −3/2.
Length of latus rectum = 4a = 6.
NCERT Question 3 : Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola y2 = −8x
Solution
Compare with standard form y2 = −4ax (left-opening).
−4a = −8 ⇒ a = 2.
Focus = (−a, 0) = (−2, 0).
Axis = x-axis.
Directrix = x = a ⇒ x = 2.
Length of latus rectum = 4a = 8.
NCERT Question 4 : Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola x2 = −16y
Solution
Compare with standard form x2 = −4ay (downward-opening).
−4a = −16 ⇒ a = 4.
Focus = (0, −a) = (0, −4).
Axis = y-axis.
Directrix = y = a ⇒ y = 4.
Length of latus rectum = 4a = 16.
NCERT Question 5 : Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola y2 = 10x
Solution
Compare with standard form y2 = 4ax.
4a = 10 ⇒ a = 10/4 = 5/2.
Focus = (a, 0) = (5/2, 0).
Axis = x-axis.
Directrix = x = −a ⇒ x = −5/2.
Length of latus rectum = 4a = 10.
NCERT Question 6 : Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum for the parabola x2 = −9y
Solution
Compare with standard form x2 = −4ay (downward-opening).
−4a = −9 ⇒ a = 9/4.
Focus = (0, −a) = (0, −9/4).
Axis = y-axis.
Directrix = y = a ⇒ y = 9/4.
Length of latus rectum = 4a = 9.
For more worked examples on parabolas and downloadable practice sheets for board and competitive exam prep, check out the parabola revision notes and problem sets from Anand Classes.
NCERT Question.7 : Find the equation of parabola with
focus (6, 0) and directrix x = −6.
Solution:
Given focus $(6, 0)$ and directrix $x = -6$.
Since the focus lies on the x-axis, the x-axis is the axis of the parabola.
As the directrix $x = -6$ is to the left of the y-axis, the parabola opens towards the right.
Equation of a parabola with vertex at origin and axis along the x-axis is $$y^2 = 4ax$$
Here, $a = 6$.
Therefore,
$$y^2 = 4(6)x$$
$$y^2 = 24x$$
NCERT Question.8 : Find the equation of parabola with
Focus (0, –3) and Directrix $y = 3$.
Solution:
Given focus $(0, -3)$ and directrix $y = 3$.
Since the focus lies on the y-axis, the y-axis is the axis of the parabola.
As the directrix $y = 3$ is above the x-axis, the parabola opens downwards.
Equation of a downward opening parabola is $x^2 = -4ay$.
Here, $a = 3$.
Therefore,
$$x^2 = -4(3)y$$
$$x^2 = -12y$$
NCERT Question.9 : Find the equation of parabola with
Vertex (0, 0) and Focus (3, 0)
Solution:
Given vertex $(0, 0)$ and focus $(3, 0)$.
Since the vertex is at the origin and the focus lies on the positive x-axis, the parabola opens to the right.
Equation of the parabola is $y^2 = 4ax$.
Here, $a = 3$.
Therefore,
$$y^2 = 4(3)x$$
$$y^2 = 12x$$
NCERT Question.10 : Find the equation of parabola with
Vertex (0, 0) and Focus (–2, 0).
Solution:
Given vertex $(0, 0)$ and focus $(-2, 0)$.
Since the focus lies on the negative x-axis, the parabola opens towards the left.
Equation of a left-opening parabola is $y^2 = -4ax$.
Here, $a = 2$.
Therefore,
$$y^2 = -4(2)x$$
$$y^2 = -8x$$
NCERT Question.11 : Find the equation of parabola with
Vertex (0, 0), Passes through (2, 3) and Axis along x-axis
Solution:
Equation of the parabola is $y^2 = 4ax$.
Substituting $(x, y) = (2, 3)$,
$$3^2 = 4a(2)$$
$$9 = 8a$$
$$a = \frac{9}{8}$$
Therefore,
$$y^2 = 4\left(\frac{9}{8}\right)x$$
$$y^2 = \frac{9x}{2}$$
Multiplying both sides by 2,
$$2y^2 = 9x$$
Hence, the equation of the parabola is $2y^2 = 9x$.
NCERT Question.12 : Find the equation of parabola with
Vertex (0, 0); Passes through (5, 2) and Symmetric with respect to y-axis.
Solution:
Equation of the parabola is $x^2 = 4ay$.
Substituting $(x, y) = (5, 2)$,
$$5^2 = 4a(2)$$
$$25 = 8a$$
$$a = \frac{25}{8}$$
Therefore,
$$x^2 = 4\left(\frac{25}{8}\right)y$$
$$x^2 = \frac{25y}{2}$$
Multiplying both sides by 2,
$$2x^2 = 25y$$
Hence, the equation of the parabola is $2x^2 = 25y$.
These parabola examples are excellent for Class 11 Maths, JEE, and NDA practice — covering standard forms, focus–directrix relationships, and symmetry properties.
