NCERT Class 11 Mathematics — Conic Sections, Exercise 10.2. All 12 questions solved.

A parabola is the set of points equidistant from a fixed point (the focus) and a fixed line (the directrix). With the vertex at the origin there are exactly four standard forms:

In every case the latus rectum has length $4a$ — which is simply the coefficient on the right-hand side, read without its sign.

Key insight. The variable that is squared names the other axis as the parabola’s axis. In $y^2 = 12x$ it is $y$ that is squared, so the axis is the $x$-axis; in $x^2 = 6y$ the axis is the $y$-axis. The sign of the other side then says which way it opens. Getting these two readings right settles four of the five things each of questions 1 to 6 asks for.

In questions 1 to 6, find the coordinates of the focus, the axis of the parabola, the equation of the directrix and the length of the latus rectum.

Question 1

$y^2 = 12x$

Solution. Comparing with $y^2 = 4ax$:

$$4a = 12 \quad\Longrightarrow\quad a = 3$$

Since $y$ is squared and the coefficient is positive, the parabola opens to the right along the $x$-axis.

Question 2

$x^2 = 6y$

Solution. Comparing with $x^2 = 4ay$:

$$4a = 6 \quad\Longrightarrow\quad a = \frac32$$

$x$ is squared with a positive coefficient, so the parabola opens upwards along the $y$-axis.

Question 3

$y^2 = -8x$

Solution. Comparing with $y^2 = -4ax$:

$$4a = 8 \quad\Longrightarrow\quad a = 2$$

The negative sign means it opens to the left, so the focus is at $(-a, 0)$ and the directrix on the positive side.

Question 4

$x^2 = -16y$

Solution. Comparing with $x^2 = -4ay$:

$$4a = 16 \quad\Longrightarrow\quad a = 4$$

Opens downwards.

Question 5

$y^2 = 10x$

Solution.

$$4a = 10 \quad\Longrightarrow\quad a = \frac52$$

Question 6

$x^2 = -9y$

Solution.

$$4a = 9 \quad\Longrightarrow\quad a = \frac94$$

Opens downwards, so the focus is below the origin.

In questions 7 to 12, find the equation of the parabola that satisfies the given conditions.

Question 7

Focus $(6, 0)$; directrix $x = -6$.

Solution. The focus is on the positive $x$-axis and the directrix is the vertical line on the opposite side, equidistant from the origin. So the vertex is at the origin and the form is $y^2 = 4ax$ with $a = 6$:

$$y^2 = 24x$$

Question 8

Focus $(0, -3)$; directrix $y = 3$.

Solution. The focus lies on the negative $y$-axis, so the parabola opens downwards: $x^2 = -4ay$ with $a = 3$:

$$x^2 = -12y$$

Question 9

Vertex $(0, 0)$; focus $(3, 0)$.

Solution. Vertex at the origin with the focus on the positive $x$-axis gives $y^2 = 4ax$ with $a = 3$:

$$y^2 = 12x$$

Question 10

Vertex $(0, 0)$; focus $(-2, 0)$.

Solution. The focus is on the negative $x$-axis, so the parabola opens to the left: $y^2 = -4ax$ with $a = 2$:

$$y^2 = -8x$$

Question 11

Vertex $(0, 0)$, passing through $(2, 3)$ and axis along the $x$-axis.

Solution. Axis along the $x$-axis means the form is $y^2 = 4ax$ (the point $(2,3)$ has a positive abscissa, so it opens right). Substituting the point:

$$3^2 = 4a(2) \quad\Longrightarrow\quad 9 = 8a \quad\Longrightarrow\quad a = \frac98$$

$$y^2 = 4\left(\frac98\right)x = \frac92 x \quad\Longrightarrow\quad 2y^2 = 9x$$

Question 12

Vertex $(0, 0)$, passing through $(5, 2)$ and symmetric with respect to the $y$-axis.

Solution. Symmetry about the $y$-axis means $x^2 = 4ay$ or $x^2 = -4ay$. The point $(5, 2)$ has a positive ordinate, so the parabola opens upwards:

$$5^2 = 4a(2) \quad\Longrightarrow\quad 25 = 8a \quad\Longrightarrow\quad a = \frac{25}{8}$$

$$x^2 = 4\left(\frac{25}{8}\right)y = \frac{25}{2}y \quad\Longrightarrow\quad 2x^2 = 25y$$

Common mistakes

• Confusing the axis with the squared variable. In $y^2 = 12x$ the axis is the $x$-axis, not the $y$-axis. The parabola is symmetric about the axis containing the focus, and the focus always sits on the axis named by the unsquared variable.
• Questions 3, 4 and 6, keeping the minus in $a$. Write $y^2 = -8x$ as $y^2 = -4ax$ with $a = 2$; the minus belongs to the form, not to $a$. Taking $a = -2$ then produces a focus at $(2,0)$ on the wrong side.
• Giving the latus rectum as $a$ or $2a$. It is $4a$ — that is, the whole coefficient of the linear term, without its sign. For $x^2 = -16y$ it is $16$.
• Question 8, choosing $x^2 = 4ay$. The focus at $(0,-3)$ is below the vertex, so the parabola opens downwards and the form carries a minus.
• Questions 11 and 12, guessing the direction. The point the parabola passes through decides it: $(2,3)$ has $x > 0$ so the curve opens right; $(5,2)$ has $y > 0$ so it opens up. Substituting into the wrong form gives a negative $a$, which is the signal that the choice was wrong.
• Assuming the vertex is at the origin without checking. In questions 7 and 8 it has to be deduced — the focus and directrix are equidistant from the origin and on opposite sides, which is exactly the condition for the vertex to be there.

Practise next

• Exercise 10.3 — ellipses, where two foci replace one and the eccentricity appears for the first time.
• Exercise 10.4 — hyperbolas, whose standard forms differ from the ellipse’s by a single sign.