NCERT Class 11 Maths Chapter 11 Exercise 11.2: Parabolas

Understanding the Standard Forms of a Parabola

To solve any question in conic sections ex 11 2 class 11 ncert, you must first understand the four standard forms of a parabola where the vertex is at the origin (0, 0). These equations are determined by the position of the axis of symmetry along the coordinate axes. When the parabola opens to the right, its equation is $y^2 = 4ax$ (with focus at $(a,0)$ and directrix $x = -a$). If it opens to the left, it is written as $y^2 = -4ax$ (with focus at $(-a,0)$ and directrix $x = a$). Similarly, vertical parabolas open upwards as $x^2 = 4ay$ (focus $(0,a)$, directrix $y = -a$) or downwards as $x^2 = -4ay$ (focus $(0,-a)$, directrix $y = a$). In all these standard cases, $a$ is assumed to be a positive constant representing the distance from the vertex to the focus. The line segment perpendicular to the axis of symmetry, passing through the focus with endpoints on the parabola, is called the latus rectum. Its length is always equal to $4a$, regardless of the direction the parabola opens.

Key Terminology for Exercise 11.2

Focus

The fixed point, usually denoted as F, from which distances to any point on the parabola are measured.

Directrix

The fixed straight line perpendicular to the axis of symmetry, situated opposite to the direction the parabola opens.

Axis of Symmetry

The line passing through the focus and vertex, dividing the parabola into two congruent, mirrored halves.

Latus Rectum

A chord passing through the focus, perpendicular to the axis of symmetry, with both endpoints lying on the parabola. Its length is 4a.

Step-by-Step Problem Solving Framework

1. Identify the Orientation — Look at the squared term in the given equation. If it is $y^2$, the axis of symmetry is the x-axis. If it is $x^2$, the axis of symmetry is the y-axis.
2. Compare with the Standard Equation — Compare your given equation with one of the four standard forms ($y^2 = \pm 4ax$ or $x^2 = \pm 4ay$) to isolate the value of $4a$ and compute $a$ (where $a > 0$).
3. Extract Coordinates and Equations — Using the calculated value of $a$, write down the focus coordinates, directrix equation, axis equation, and the length of the latus rectum ($4a$).
4. Formulate Equations from Given Constraints — If given geometric constraints like focus or directrix, identify the form of the parabola, determine $a$, and substitute it back into the standard equation.

Fully Worked Illustrative Examples

• Example 1: Find the focus, axis of the parabola, directrix, and length of the latus rectum for $y^2 = -12x$. Step 1: Identify the standard form. The equation is of the form $y^2 = -4ax$. Step 2: Compare coefficients. Here, $-4a = -12$, which gives $a = 3$. Step 3: State the parameters. - Since the equation is $y^2 = -4ax$, the focus lies on the negative x-axis at $(-a, 0)$, which is $(-3, 0)$. - The axis of symmetry is the x-axis ($y = 0$). - The equation of the directrix is $x = a$, which means $x = 3$. - The length of the latus rectum is $4a = 12$.
• Example 2: Find the equation of the parabola with vertex $(0,0)$ and focus $(0, -4)$. Step 1: Identify the orientation. The focus is at $(0, -4)$, which lies on the negative y-axis. This corresponds to the standard form $x^2 = -4ay$. Step 2: Determine the value of $a$. Comparing $(0, -a)$ with $(0, -4)$ gives $a = 4$. Step 3: Write the final equation. Substitute $a = 4$ into $x^2 = -4ay$: $x^2 = -4(4)y \Rightarrow x^2 = -16y$.

Common Mistakes & Board Exam Tips

1. Sign of 'a': Always keep $a > 0$ when determining coordinates. For example, in $y^2 = -8x$, do not set $a = -2$. Instead, match it directly to $-4ax$, giving $4a = 8$, so $a = 2$. Keep your coordinates sign-accurate manually (e.g., focus is at $(-a, 0) = (-2, 0)$).
2. Latus Rectum Length: Remember, length is a physical dimension and must always be positive. Even if the equation has a negative sign (like $y^2 = -16x$), the length of the latus rectum is positive $16$.
3. Coordinate Axes: Ensure you do not swap axes. A focus of $(0, 3)$ means a vertical parabola ($x^2 = 4ay$), whereas a focus of $(3, 0)$ means a horizontal parabola ($y^2 = 4ax$).

Practice Questions with Solutions

• Q: Find the focus, axis, directrix, and length of the latus rectum for the parabola $x^2 = 6y$. A: Step 1: Compare the given equation $x^2 = 6y$ with the standard form $x^2 = 4ay$. Step 2: Calculate the value of $a$. Here, $4a = 6 \Rightarrow a = \frac{6}{4} = \frac{3}{2}$. Step 3: Extract the parameters based on the upward standard form $x^2 = 4ay$: - Focus: $(0, a) = (0, \frac{3}{2})$ - Axis of symmetry: y-axis ($x = 0$) - Directrix: $y = -a \Rightarrow y = -\frac{3}{2}$ (or $2y + 3 = 0$) - Length of latus rectum: $4a = 6$ Final answer: Focus is $(0, \frac{3}{2})$, Axis is $x = 0$, Directrix is $y = -\frac{3}{2}$, Latus Rectum is $6$.
• Q: Find the equation of the parabola with vertex $(0,0)$ and focus $(3,0)$. A: Step 1: The focus is at $(3,0)$, which lies on the positive x-axis. Therefore, the standard form is $y^2 = 4ax$. Step 2: Determine $a$. Comparing $(a,0)$ with $(3,0)$, we get $a = 3$. Step 3: Substitute $a = 3$ into the standard equation: $y^2 = 4(3)x \Rightarrow y^2 = 12x$. Final answer: $y^2 = 12x$
• Q: Find the equation of the parabola with vertex $(0,0)$, directrix $y = 3$. A: Step 1: The directrix is the horizontal line $y = 3$. This indicates a vertical parabola symmetric about the y-axis. Step 2: Since the directrix is positive ($y = a$), the parabola opens downwards, matching the form $x^2 = -4ay$. Step 3: Comparing the directrix formula $y = a$ with $y = 3$, we find $a = 3$. Step 4: Substitute $a = 3$ into $x^2 = -4ay$: $x^2 = -4(3)y \Rightarrow x^2 = -12y$. Final answer: $x^2 = -12y$
• Q: Find the equation of the parabola with vertex $(0,0)$, symmetric about the y-axis, and passing through the point $(2, -3)$. A: Step 1: Since the parabola is symmetric about the y-axis, its standard equation is either $x^2 = 4ay$ or $x^2 = -4ay$. Step 2: The point $(2, -3)$ lies in the fourth quadrant (where $y$ is negative). Thus, the parabola must open downwards, corresponding to $x^2 = -4ay$. Step 3: Substitute the point $(2, -3)$ into the equation to find $a$: $(2)^2 = -4a(-3) \Rightarrow 4 = 12a \Rightarrow a = \frac{4}{12} = \frac{1}{3}$. Step 4: Write the equation using $a = \frac{1}{3}$: $x^2 = -4\left(\frac{1}{3}\right)y \Rightarrow x^2 = -\frac{4}{3}y$ or $3x^2 = -4y$. Final answer: $3x^2 = -4y$

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