NCERT Class 11 Mathematics — Conic Sections, Miscellaneous Exercise on Chapter 10. All 8 questions solved.

Almost every question here is an applied one, and each is solved the same way:

1. Set up axes so that the conic takes its simplest standard form — usually with the vertex or centre at the origin.
2. Use the given measurements as a point on the curve to find the unknown constant.
3. Answer the question asked, which is normally the value of $y$ (or $x$) at some other stated position.

Key insight. The choice of origin is yours, and it decides how much algebra follows. In question 3 the cable’s lowest point is $6$ m above the roadway, but putting the origin at the cable’s vertex rather than on the roadway turns the parabola into a bare $x^2 = 4ay$ with no shifted terms — and the $6$ m is simply added back at the end.

Question 1

If a parabolic reflector is $20$ cm in diameter and $5$ cm deep, find the focus.

Solution. Place the vertex at the origin with the axis along the positive $x$-axis, so the reflector’s cross-section is

$$y^2 = 4ax$$

“$20$ cm in diameter and $5$ cm deep” means that at depth $x = 5$ the reflector is $20$ cm across, so it reaches $y = 10$ on each side. The point $(5, 10)$ therefore lies on the parabola:

$$10^2 = 4a(5) \quad\Longrightarrow\quad 100 = 20a \quad\Longrightarrow\quad a = 5$$

The focus of $y^2 = 4ax$ is at $(a, 0)$:

$$\text{focus} = (5, 0)$$

That is $5$ cm from the vertex along the axis — exactly the depth of the reflector, so the focus sits at the midpoint of the diameter across the open end.

Question 2

An arch is in the form of a parabola with its axis vertical. The arch is $10$ m high and $5$ m wide at the base. How wide is it $2$ m from the vertex of the parabola?

Solution. The axis is vertical and the arch opens downwards, so put the vertex at the origin and use

$$x^2 = -4ay$$

The base is $10$ m below the vertex and $5$ m wide, so the point $\left(\tfrac52,\ -10\right)$ lies on the curve:

$$\left(\frac52\right)^2 = -4a(-10) \quad\Longrightarrow\quad \frac{25}{4} = 40a \quad\Longrightarrow\quad 4a = \frac{25}{40} = \frac58$$

So the equation is $x^2 = -\tfrac58 y$.

“$2$ m from the vertex” means $y = -2$:

$$x^2 = -\frac58(-2) = \frac54 \quad\Longrightarrow\quad x = \frac{\sqrt5}{2}$$

The width is twice this:

$$2x = \sqrt5 \approx 2.23 \text{ m}$$

Question 3

The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway, which is horizontal and $100$ m long, is supported by vertical wires attached to the cable, the longest wire being $30$ m and the shortest being $6$ m. Find the length of a supporting wire attached to the roadway $18$ m from the middle.

Solution. Put the origin at the lowest point of the cable, which is $6$ m above the roadway. The cable opens upwards:

$$x^2 = 4ay$$

At the ends of the bridge, $x = \pm50$ (half of $100$ m), and there the cable is $30$ m above the roadway — that is, $30 – 6 = 24$ m above the vertex. So $(50, 24)$ lies on the curve:

$$50^2 = 4a(24) \quad\Longrightarrow\quad 2500 = 96a \quad\Longrightarrow\quad 4a = \frac{2500}{24} = \frac{625}{6}$$

The equation is $x^2 = \tfrac{625}{6}y$, that is $y = \tfrac{6x^2}{625}$.

At $x = 18$:

$$y = \frac{6(324)}{625} = \frac{1944}{625} = 3.1104 \text{ m above the vertex}$$

The wire runs from the roadway to the cable, so add back the $6$ m:

$$6 + 3.1104 = 9.1104 \approx 9.11 \text{ m}$$

Question 4

An arch is in the form of a semi-ellipse. It is $8$ m wide and $2$ m high at the centre. Find the height of the arch at a point $1.5$ m from one end.

Solution. Put the centre of the full ellipse at the origin with the $8$ m span along the $x$-axis. Then the semi-major axis is $a = 4$ and the semi-minor axis is $b = 2$:

$$\frac{x^2}{16} + \frac{y^2}{4} = 1$$

A point $1.5$ m from one end is $4 – 1.5 = 2.5$ m from the centre, so $x = 2.5$:

$$\frac{6.25}{16} + \frac{y^2}{4} = 1 \quad\Longrightarrow\quad \frac{y^2}{4} = 1 – 0.390625 = 0.609375$$

$$y^2 = 2.4375 \quad\Longrightarrow\quad y = 1.5612\ldots \approx 1.56 \text{ m}$$

Measuring from the end rather than the centre is the step that most often goes wrong; converting to a distance from the centre first keeps the ellipse equation usable as it stands.

Question 5

A rod of length $12$ cm moves with its ends always touching the coordinate axes. Determine the equation of the locus of a point $\mathrm P$ on the rod which is $3$ cm from the end in contact with the $x$-axis.

Solution. Let the rod’s ends be $\mathrm A(p, 0)$ on the $x$-axis and $\mathrm B(0, q)$ on the $y$-axis. Since the rod has length $12$:

$$p^2 + q^2 = 144 \qquad (\star)$$

$\mathrm P$ is $3$ cm from $\mathrm A$ and therefore $9$ cm from $\mathrm B$, so it divides $\mathrm{AB}$ in the ratio $3 : 9 = 1 : 3$ measured from $\mathrm A$. By the section formula:

$$\mathrm P = \left(\frac{1 \cdot 0 + 3 \cdot p}{4},\ \frac{1 \cdot q + 3 \cdot 0}{4}\right) = \left(\frac{3p}{4},\ \frac q4\right)$$

Writing $\mathrm P = (x, y)$:

$$x = \frac{3p}{4} \ \Longrightarrow\ p = \frac{4x}{3}, \qquad y = \frac q4 \ \Longrightarrow\ q = 4y$$

Substituting into $(\star)$ — this is the step that eliminates the moving parameters:

$$\frac{16x^2}{9} + 16y^2 = 144$$

Dividing by $144$:

$$\frac{x^2}{81} + \frac{y^2}{9} = 1$$

The locus is an ellipse.

Question 6

Find the area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 12y$ to the ends of its latus rectum.

Solution. Comparing $x^2 = 12y$ with $x^2 = 4ay$ gives $a = 3$, so the focus is $(0, 3)$ and the latus rectum is the horizontal chord through it, $y = 3$.

Its endpoints satisfy $x^2 = 12(3) = 36$, so $x = \pm6$:

$$(-6, 3) \quad\text{and}\quad (6, 3)$$

The triangle has vertices $(0,0)$, $(-6,3)$ and $(6,3)$. Its base is the latus rectum, of length $12$, and its height is the distance from the vertex to the line $y = 3$, which is $3$:

$$\text{area} = \frac12 \times 12 \times 3 = 18$$

(The base is the latus rectum, whose length $4a = 12$ could have been quoted directly.)

Question 7

A man running a racecourse notes that the sum of the distances from the two flag posts from him is always $10$ m and the distance between the flag posts is $8$ m. Find the equation of the path traced by the man.

Solution. “The sum of the distances from two fixed points is constant” is the definition of an ellipse, with the flag posts as its foci.

$$2a = 10 \quad\Longrightarrow\quad a = 5$$ $$2c = 8 \quad\Longrightarrow\quad c = 4$$

$$b^2 = a^2 – c^2 = 25 – 16 = 9$$

Taking the centre at the origin with the foci on the $x$-axis:

$$\frac{x^2}{25} + \frac{y^2}{9} = 1$$

Note that $2a = 10 > 8 = 2c$, as it must be — the constant sum always exceeds the distance between the foci, otherwise no such point exists.

Question 8

An equilateral triangle is inscribed in the parabola $y^2 = 4ax$, where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.

Solution. By symmetry, the other two vertices are reflections of each other in the $x$-axis, say $\mathrm P(h, k)$ and $\mathrm Q(h, -k)$ with $k > 0$.

The side $\mathrm{PQ}$ is vertical with length $2k$, and the origin $\mathrm O$ is the third vertex. For the triangle to be equilateral, $\mathrm{OP}$ must make an angle of $30^\circ$ with the $x$-axis (half of the $60^\circ$ apex angle at $\mathrm O$):

$$\tan 30^\circ = \frac kh \quad\Longrightarrow\quad k = \frac{h}{\sqrt3}$$

Since $\mathrm P$ lies on the parabola, $k^2 = 4ah$:

$$\frac{h^2}{3} = 4ah \quad\Longrightarrow\quad h = 12a \qquad (h \ne 0)$$

Then

$$k = \frac{12a}{\sqrt3} = 4\sqrt3\,a$$

The side length is

$$\mathrm{PQ} = 2k = 8\sqrt3\,a$$

Check with $\mathrm{OP}$: $\sqrt{h^2 + k^2} = \sqrt{144a^2 + 48a^2} = \sqrt{192a^2} = 8\sqrt3\,a$ ✓ — all three sides agree, confirming the triangle really is equilateral.

Common mistakes

• Question 1, using the diameter as $y$. The reflector is $20$ cm across, so the rim reaches $y = 10$, not $y = 20$. Halving the width is the step to remember.
• Question 2, forgetting the parabola opens downwards. An arch’s vertex is at the top, so the base point has a negative ordinate and the form carries a minus.
• Question 3, measuring the cable height from the roadway. With the origin at the cable’s lowest point, the end of the cable is $24$ m above the origin, not $30$. The $6$ m is added back only at the very end.
• Question 4, taking $x = 1.5$. The point is $1.5$ m from the end, which is $2.5$ m from the centre. The ellipse equation is written about the centre, so the distance must be converted.
• Question 5, using the ratio $3 : 12$. $\mathrm P$ divides the rod into pieces of $3$ and $9$, so the ratio is $1 : 3$ — the two parts, not part and whole.
• Question 6, taking the base as $2a$. The latus rectum has length $4a = 12$. Its endpoints are found by substituting $y = a$ into the equation, which also gives the height $3$ directly.
• Question 8, assuming the apex angle at the origin is $60^\circ$ on one side. The full angle is $60^\circ$, so each half is $30^\circ$ — using $60^\circ$ for the half-angle gives $h = 4a$ and a triangle that is not equilateral.

Practise next

• Exercise 11.1 — the next chapter, Introduction to Three Dimensional Geometry, which extends coordinates to space.
• Exercise 10.3 and 10.4 — worth revising alongside questions 4, 5 and 7, all of which reduce to the standard ellipse form.