In geometry, a cone is one of the most fundamental three-dimensional figures. When this solid is intersected by a plane in different ways, the curves obtained are known as sections of a cone. These curves are collectively called conic sections.

A right circular cone is the solid generated by joining every point of a fixed circle to a fixed point not lying in the plane of the circle.
The fixed point is called the vertex, the circle is called the base, and the line joining the vertex to the centre of the base is known as the axis of the cone.

The cone is said to be double-napped when it is extended infinitely in both upward and downward directions, with the vertex as the common point of the two nappes.

A section of a cone is the curve obtained when a plane cuts the cone. The nature of the curve depends upon the inclination of the cutting plane with respect to the axis of the cone.

By changing the orientation of the cutting plane, different curves are produced, namely:

• Circle
• Ellipse
• Parabola
• Hyperbola

These curves are called conic sections.

A circle is obtained when the cutting plane is perpendicular to the axis of the cone.

Important Aspect

• The circle is a special case of an ellipse.
• It lies entirely on one nappe of the cone.
• The radius of the circle depends on the distance of the cutting plane from the vertex.

Standard Equation of a Circle with Centre at the Origin

Let the centre of the circle be at the origin \(O(0,0)\), and let its radius be \(r\).
Consider any point \(P(x,y)\) on the circle.
By definition of a circle, \[OP=r\] Using the distance formula, \[OP=\sqrt{x^2+y^2}\] Therefore, \[\sqrt{x^2+y^2}=r\] Squaring both sides, we obtain \[x^2+y^2=r^2\] This is the standard equation of a circle with centre at the origin and radius r.

Equation of a Circle with Centre at (h,k)

Let the centre of the circle be \(C(h,k)\), and let its radius be \(r\).
Let \(P(x,y)\) be any point on the circle.
By definition, \[CP=r\] Using the distance formula, \[\sqrt{\left(x-h)^2+(y-k)^2\right.}=r\] Squaring both sides, we get \[\left(x-h)^2+(y-k)^2=r^2\right.\] This is the general standard form of the equation of a circle.

General Equation of a Circle

Expanding the standard form, \[\left(x-h)^2+(y-k)^2=r^2\right.\] we obtain \[x^2+y^2+2gx+2fy+c=0\] where \[g=-h,\;f=-k,\;c=h^2+k^2-r^2\] This equation represents a circle provided \[r^2=g^2+f^2-c>0\]

The eccentricity \(e\) of a conic section is the ratio of the distance of any point on the curve from the focus to its distance from the corresponding directrix. \[e=\dfrac{PF}{PD}\]

Based on the value of \(e\):

• \(e=0 \Rightarrow\) Circle
• \(0\lt e\lt1 \Rightarrow\) Ellipse
• \(e=1 \Rightarrow\) Parabola
• \(e>1 \Rightarrow\) Hyperbola

This single parameter provides a unified way to study all conic sections.

Find an equation of the circle with centre at (0,0) and radius r.

Solution

We know that the general equation of a circle with centre \((h,k)\) and radius \(r\) is given by:

$$ \begin{aligned} \sqrt{(x-h)^2 + (y-k)^2} = r \end{aligned} $$

Since the centre of the circle is at \((0,0)\), we substitute \(h = 0\) and \(k = 0\) into the equation.

$$ \begin{aligned} h &= 0\\ k &= 0 \end{aligned} $$

Substituting these values, the equation becomes:

$$ \begin{aligned} \sqrt{x^2 + y^2} &= r\\ x^2 + y^2 &= r^2 \end{aligned} $$

Therefore, the required equation of the circle with centre at the origin and radius \(r\) is:

$$ \begin{aligned} x^2 + y^2 = r^2 \end{aligned} $$

Find the equation of the circle with centre (–3, 2) and radius 4.

Solution

The centre of the circle is given as \((-3, 2)\) and the radius is \(4\).

The standard equation of a circle with centre \((h, k)\) and radius \(r\) is:

$$ \begin{aligned} (x - h)^2 + (y - k)^2 = r^2 \end{aligned} $$

Substituting \(h = -3\), \(k = 2\), and \(r = 4\) into the formula, we get:

$$ \begin{aligned} h &= -3\\ k &= 2\\ r &= 4\\ (x - (-3))^2 + (y - 2)^2 &= 4^2\\ (x + 3)^2 + (y - 2)^2 &= 16 \end{aligned} $$

Hence, the required equation of the circle is:

$$ \begin{aligned} (x + 3)^2 + (y - 2)^2 = 16 \end{aligned} $$

Find the centre and the radius of the circle x2 + y2 + 8x + 10y – 8 = 0

Solution

Given equation of the circle is:

$$ \begin{aligned} x^2 + y^2 + 8x + 10y - 8 = 0 \end{aligned} $$

To find the centre and radius, we complete the square for the \(x\) and \(y\) terms.

$$ \begin{aligned} x^2 + 8x + y^2 + 10y - 8 &= 0\\ x^2 + 8x + 16 - 16 + y^2 + 10y + 25 - 25 - 8 &= 0 \end{aligned} $$

Grouping perfect square terms, we rewrite the equation as:

$$ \begin{aligned} (x + 4)^2 + (y + 5)^2 - 16 - 25 - 8 &= 0\\ (x + 4)^2 + (y + 5)^2 &= 49\\ (x + 4)^2 + (y + 5)^2 &= 7^2 \end{aligned} $$

From the standard form of a circle, the centre is \((-4, -5)\) and the radius is \(7\).

Find the equation of the circle which passes through the points (2, – 2), and (3,4) and whose centre lies on the line x + y = 2.

Solution

The circle passes through \((2, -2)\) and \((3, 4)\), and its centre \((h, k)\) lies on the line \(x + y = 2\).

$$ \begin{aligned} (2 - h)^2 + (-2 - k)^2 &= r^2\\ (3 - h)^2 + (4 - k)^2 &= r^2 \end{aligned} $$

Equating both expressions to eliminate \(r^2\):

$$ \begin{aligned} (2 - h)^2 + (-2 - k)^2 &= (3 - h)^2 + (4 - k)^2\\ (h - 2)^2 + (k + 2)^2 &= (h - 3)^2 + (k - 4)^2 \end{aligned} $$

Expanding and simplifying:

$$ \begin{aligned} h^2 - 4h + 4 + k^2 + 4k + 4 &= h^2 - 6h + 9 + k^2 - 8k + 16\\ -4h + 4k + 8 &= -6h - 8k + 25\\ 2h + 12k - 17 &= 0\\ h + 6k &= 8.5 \end{aligned} $$

Since the centre lies on the line \(x + y = 2\):

$$ \begin{aligned} h + k &= 2\\ h &= 2 - k \end{aligned} $$

Substituting into \(h + 6k = 8.5\):

$$ \begin{aligned} 2 - k + 6k &= 8.5\\ 5k &= 6.5\\ k &= 1.3\\ h &= 2 - 1.3\\ h &= 0.7 \end{aligned} $$

Now calculating the radius, which is simply the distance from the centre to any point on its circumference:

$$ \begin{aligned} r^2 &= (2 - 0.7)^2 + (-2 - 1.3)^2\\ r^2 &= (1.3)^2 + (-3.3)^2\\ r^2 &= 1.69 + 10.89\\ r^2 &= 12.58 \end{aligned} $$

Therefore, the corrected equation of the circle is:

$$ \begin{aligned} (x - 0.7)^2 + (y - 1.3)^2 = 12.58 \end{aligned} $$

Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola \(y^2 = 8x\).

Solution

Given parabola:

$$ \begin{aligned} y^2 = 8x\\ y^2 = 4 \cdot 2x \end{aligned} $$

Comparing with the standard form \(y^2 = 4ax\), we get \(a = 2\)

$$ \begin{aligned} a = 2\\ \text{Focus } f(2, 0) \end{aligned} $$

Since the parabola opens towards the positive \(x\)-axis, the axis of the parabola is the \(x\)-axis

Equation of the directrix:

$$ \begin{aligned} x = -2 \end{aligned} $$

Length of the latus rectum:

$$ \begin{aligned} 4a = 4 \cdot 2 = 8 \end{aligned} $$

Thus, the focus is \((2, 0)\), the axis is the \(x\)-axis, the directrix is \(x = -2\), and the length of the latus rectum is \(8\)

Find the equation of the parabola with focus (2,0) and directrix x = – 2.

Solution

The focus of the parabola is given as \((2, 0)\) and the directrix is the line \(x = -2\)

For a parabola with focus \((a, 0)\) and directrix \(x = -a\), the standard equation is \(y^2 = 4ax\)

$$ \begin{aligned} F &= (2, 0)\\ \Rightarrow a &= 2\\ y^2 &= 4ax\\ y^2 &= 4 \cdot 2x\\ y^2 &= 8x \end{aligned} $$

Therefore, the required equation of the parabola is \(y^2 = 8x\)

Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2).

Solution

The vertex of the parabola is \((0, 0)\) and the focus is \((0, 2)\)

Since the focus lies on the positive \(y\)-axis, the parabola opens upward

The distance between the vertex and focus is \(a = 2\), measured along the \(y\)-axis

The standard equation of a parabola opening upward with vertex at the origin is \(x^2 = 4ay\)

$$ \begin{aligned} x^2 &= 4ay\\ x^2 &= 4 \cdot 2 \cdot y\\ x^2 &= 8y \end{aligned} $$

Therefore, the required equation of the parabola is \(x^2 = 8y\)

Find the equation of the parabola which is symmetric about the y-axis, and passes through the point (2,–3).

Solution

The parabola is symmetric about the \(y\)-axis, which implies that it either opens upward or downward and has the standard form \(x^2 = 4ay\)

Since the parabola passes through the point \((2, -3)\), we substitute these coordinates into the equation to find the value of \(a\)

$$ \begin{aligned} x^2 &= 4ay\\ 2^2 &= 4 \cdot a \cdot (-3)\\ 4 &= -12a\\ a &= \frac{-4}{12}\\ a &= -\frac{1}{3} \end{aligned} $$

Substituting \(a = -\frac{1}{3}\) into the standard equation:

$$ \begin{aligned} x^2 &= 4 \cdot \left(-\frac{1}{3}\right) y\\ x^2 &= -\frac{4}{3}y\\ 3x^2 &= -4y \end{aligned} $$

Therefore, the required equation of the parabola is \(3x^2 = -4y\)

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse \[\dfrac{x^2}{25}+ \dfrac{y^2}{9}=1\]

Solution

Given ellipse:

$$ \begin{aligned} \frac{x^2}{25} + \frac{y^2}{9} = 1\\ a^2 &= 25\\ a &= 5\\ b^2 &= 9\\ b &= 3 \end{aligned} $$

Since \(a^2 > b^2\), the major axis lies along the \(x\)-axis

Eccentricity:

$$ \begin{aligned} c^2 &= a^2 - b^2\\ c &= \sqrt{25 - 9}\\ &= \sqrt{16}\\ &= 4\\ e &= \frac{c}{a} = \frac{4}{5} \end{aligned} $$

Foci:

$$ \begin{aligned} (\pm c, 0)\\ (\pm 4, 0) \end{aligned} $$

Vertices on the major axis:

$$ \begin{aligned} (\pm a, 0) = (\pm 5, 0) \end{aligned} $$

Length of major axis:

$$ \begin{aligned} 2a = 2 \times 5 = 10 \end{aligned} $$

Length of minor axis:

$$ \begin{aligned} 2b = 2 \times 3 = 6 \end{aligned} $$

Length of latus rectum:

$$ \begin{aligned} \frac{2b^2}{a} = \frac{2 \times 9}{5}\\ = \frac{18}{5} \end{aligned} $$

Thus, the foci are \((\pm 4, 0)\), the vertices are \((\pm 5, 0)\), the length of the major axis is \(10\), the length of the minor axis is \(6\), the eccentricity is \(\frac{4}{5}\), and the length of the latus rectum is \(\frac{18}{5}\)

Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse \(9x^2 + 4y^2 = 36\).

Solution

Given ellipse:

$$ \begin{aligned} 9x^2 + 4y^2 &= 36\\ \frac{9x^2}{36} + \frac{4y^2}{36} &= 1\\ \frac{x^2}{4} + \frac{y^2}{9} &= 1\\ \frac{x^2}{2^2} + \frac{y^2}{3^2} &= 1\\ a &= 3\\ b &= 2 \end{aligned} $$

Since the larger denominator is under \(y^2\), the major axis lies along the \(y\)-axis

Eccentricity:

$$ \begin{aligned} c &= \sqrt{a^2 - b^2}\\ &= \sqrt{9 - 4}\\ &= \sqrt{5}\\ e &= \frac{c}{a} = \frac{\sqrt{5}}{3} \end{aligned} $$

Major axis:

$$ \begin{aligned} 2a &= 2 \times 3\\ &= 6 \end{aligned} $$

Minor axis:

$$ \begin{aligned} 2b &= 2 \times 2\\ &= 4 \end{aligned} $$

Latus rectum:

$$ \begin{aligned} \frac{2b^2}{a} &= \frac{2 \times 2^2}{3}\\ &= \frac{8}{3} \end{aligned} $$

Foci:

$$ \begin{aligned} (0, \pm c)\\ (0, \pm \sqrt{5}) \end{aligned} $$

Vertices:

$$ \begin{aligned} (0, \pm a)\\ (0, \pm 3) \end{aligned} $$

Find the equation of the ellipse whose vertices are (± 13, 0) and foci are (± 5, 0).

Solution

The vertices of the ellipse are \((\pm 13, 0)\) and the foci are \((\pm 5, 0)\), so the major axis lies along the \(x\)-axis

$$ \begin{aligned} a &= 13\\ c &= 5\\ c^2 &= a^2 - b^2\\ b^2 &= a^2 - c^2\\ &= 13^2 - 5^2\\ &= 169 - 25\\ &= 144\\ b &= \sqrt{144}\\ &= 12 \end{aligned} $$

Using the standard form of an ellipse with major axis along the \(x\)-axis:

$$ \begin{aligned} \frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1\\ \frac{x^2}{13^2} + \frac{y^2}{12^2} &= 1\\ \frac{x^2}{169} + \frac{y^2}{144} &= 1 \end{aligned} $$

Therefore, the required equation of the ellipse is \(\dfrac{x^2}{169} + \dfrac{y^2}{144} = 1\)

Find the equation of the ellipse, whose length of the major axis is 20 and foci are (0, ± 5).

Solution

The length of the major axis of the ellipse is \(20\), so \(2a = 20\)

$$ \begin{aligned} 2a &= 20\\ a &= \frac{20}{2}\\ a &= 10 \end{aligned} $$

The foci are given as \((0, \pm 5)\), hence \(c = 5\)

Using the relation \(c^2 = a^2 - b^2\), we find \(b\)

$$ \begin{aligned} b^2 &= a^2 - c^2\\ &= 10^2 - 5^2\\ &= 100 - 25\\ &= 75\\ b &= 5\sqrt{3} \end{aligned} $$

Since the foci lie on the \(y\)-axis, the major axis is along the \(y\)-axis, so the standard equation is:

$$ \begin{aligned} \frac{x^2}{b^2} + \frac{y^2}{a^2} &= 1\\ \frac{x^2}{(5\sqrt{3})^2} + \frac{y^2}{10^2} &= 1\\ \frac{x^2}{75} + \frac{y^2}{100} &= 1 \end{aligned} $$

Therefore, the required equation of the ellipse is \(\dfrac{x^2}{75} + \dfrac{y^2}{100} = 1\)

Find the equation of the ellipse, with major axis along the x-axis and passing through the points (4, 3) and (– 1,4).

Solution

Since the major axis is along the \(x\)-axis, the standard equation of the ellipse is:

$$ \begin{aligned} \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \end{aligned} $$

The ellipse passes through the points \((4, 3)\) and \((-1, 4)\), so we substitute these coordinates into the equation

$$ \begin{aligned} \frac{(4)^2}{a^2} + \frac{(3)^2}{b^2} &= 1\\ \frac{16}{a^2} + \frac{9}{b^2} &= 1\\ \frac{(-1)^2}{a^2} + \frac{(4)^2}{b^2} &= 1\\ \frac{1}{a^2} + \frac{16}{b^2} &= 1 \end{aligned} $$

From the second equation, express \(\frac{1}{a^2}\) in terms of \(b^2\)

$$ \begin{aligned} \frac{1}{a^2} &= 1 - \frac{16}{b^2} \end{aligned} $$

Substitute this into the first equation

$$ \begin{aligned} \frac{16}{a^2} + \frac{9}{b^2} &= 1\\ 16\left(1 - \frac{16}{b^2}\right) + \frac{9}{b^2} &= 1\\ 16 - \frac{256}{b^2} + \frac{9}{b^2} &= 1\\ 16 - \frac{247}{b^2} &= 1\\ 15 &= \frac{247}{b^2}\\ b^2 &= \frac{247}{15} \end{aligned} $$

Now substitute \(b^2\) back to find \(a^2\)

$$ \begin{aligned} \frac{1}{a^2} &= 1 - \frac{16}{b^2}\\ &= 1 - \frac{16 \times 15}{247}\\ &= \frac{247 - 240}{247}\\ &= \frac{7}{247}\\ a^2 &= \frac{247}{7} \end{aligned} $$

Substituting \(a^2\) and \(b^2\) into the ellipse equation

$$ \begin{aligned} \frac{x^2}{a^2} + \frac{y^2}{b^2} &= 1\\ \frac{7x^2}{247} + \frac{15y^2}{247} &= 1\\ 7x^2 + 15y^2 &= 247 \end{aligned} $$

Therefore, the required equation of the ellipse is \(7x^2 + 15y^2 = 247\)

Find the coordinates of the foci and the vertices, the eccentricity,the length of the latus rectum of the hyperbolas:
(i) \(\dfrac{x^2}{9}-\dfrac{y^2}{16}=1\)
(ii) \(y^2 – 16x^2 = 16\)

Solution

(i) For the hyperbola:

$$ \begin{aligned} \frac{x^2}{9} - \frac{y^2}{16} = 1\\ a^2 &= 9\\ a &= 3\\ b^2 &= 16\\ b &= 4\\ c &= \sqrt{a^2 + b^2}\\ &= \sqrt{9 + 16}\\ &= \sqrt{25}\\ &= 5 \end{aligned} $$

Foci:

$$ \begin{aligned} (\pm c, 0)\\ (\pm 5, 0) \end{aligned} $$

Vertices:

$$ \begin{aligned} (\pm a, 0)\\ (\pm 3, 0) \end{aligned} $$

Eccentricity:

$$ \begin{aligned} e = \frac{c}{a}\\ = \frac{5}{3} \end{aligned} $$

Length of latus rectum:

$$ \begin{aligned} \frac{2b^2}{a}\\ = \frac{2 \times 4^2}{3}\\ = \frac{32}{3} \end{aligned} $$

(ii) For the hyperbola:

$$ \begin{aligned} y^2 - 16x^2 &= 16\\ \frac{y^2}{16} - x^2 &= 1\\ a^2 &= 16\\ a &= 4\\ b^2 &= 1\\ b &= 1\\ c &= \sqrt{a^2 + b^2}\\ &= \sqrt{16 + 1}\\ &= \sqrt{17} \end{aligned} $$

Foci:

$$ \begin{aligned} (0, \pm c)\\ (0, \pm \sqrt{17}) \end{aligned} $$

Vertices:

$$ \begin{aligned} (0, \pm a)\\ (0, \pm 4) \end{aligned} $$

Eccentricity:

$$ \begin{aligned} e = \frac{c}{a}\\ = \frac{\sqrt{17}}{4} \end{aligned} $$

Length of latus rectum:

$$ \begin{aligned} \frac{2b^2}{a}\\ = \frac{2 \times 1}{4}\\ = \frac{1}{2} \end{aligned} $$

Find the equation of the hyperbola with foci (0, ± 3) and vertices \(\left(0,\; \pm\dfrac{\sqrt{11}}{2}\right)\)

Solution

The foci of the hyperbola are \((0, \pm 3)\) and the vertices are \(\left(0, \pm \dfrac{\sqrt{11}}{2}\right)\), so the transverse axis lies along the \(y\)-axis

$$ \begin{aligned} a &= \frac{\sqrt{11}}{2}\\ c &= 3\\ b^2 &= c^2 - a^2\\ &= 9 - \left(\frac{\sqrt{11}}{2}\right)^2\\ &= 9 - \frac{11}{4}\\ b^2 &= \frac{36 - 11}{4}\\ &= \frac{25}{4} \end{aligned} $$

Since the transverse axis is along the \(y\)-axis, the standard equation of the hyperbola is:

$$ \begin{aligned} \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\\ \frac{y^2}{\frac{11}{4}} - \frac{x^2}{\frac{25}{4}} = 1\\ \frac{4y^2}{11} - \frac{4x^2}{25} = 1 \end{aligned} $$

Multiplying throughout by \(275\) to remove denominators:

$$ \begin{aligned} 100y^2 - 44x^2 = 275 \end{aligned} $$

Therefore, the required equation of the hyperbola is \(100y^2 - 44x^2 = 275\)

Find the equation of the hyperbola where foci are (0, ±12) and the length of the latus rectum is 36.

Solution

The foci of the hyperbola are \((0, \pm 12)\), so the transverse axis lies along the \(y\)-axis and \(c = 12\)

The length of the latus rectum is given as \(36\), and for a hyperbola:

$$ \begin{aligned} \frac{2b^2}{a} &= 36\\ 2b^2 &= 36a\\ b^2 &= 18a \end{aligned} $$

Using the relation \(c^2 = a^2 + b^2\):

$$ \begin{aligned} 12^2 &= a^2 + 18a\\ 144 &= a^2 + 18a\\ a^2 + 18a - 144 &= 0\\ (a - 6)(a + 24) &= 0\\ a &= 6 \end{aligned} $$

Substituting \(a = 6\) to find \(b^2\):

$$ \begin{aligned} b^2 &= 18a\\ b^2 &= 18 \times 6\\ b^2 &= 108 \end{aligned} $$

Since the transverse axis is along the \(y\)-axis, the standard equation of the hyperbola is:

$$ \begin{aligned} \frac{y^2}{a^2} - \frac{x^2}{b^2} &= 1\\ \frac{y^2}{36} - \frac{x^2}{108} &= 1 \end{aligned} $$

Therefore, the required equation of the hyperbola is \(\dfrac{y^2}{36} - \dfrac{x^2}{108} = 1\)

A beam is supported at its ends by supports which are 12 metres apart. Since the load is concentrated at its centre, there is a deflection of 3 cm at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection 1 cm?

Solution

Let the beam be supported at points \(A\) and \(B\), which are \(12\) m apart. The centre of the beam is taken as the origin, and the axis of symmetry is vertical

Since the beam is deflected downward in the shape of a parabola, its equation can be written as:

$$ \begin{aligned} y = ax^2 \end{aligned} $$

At the centre, the deflection is \(3\) cm, so when \(x = 0\), we have:

$$ \begin{aligned} y &= -3 \end{aligned} $$

The supports are \(12\) m apart, so each support is \(6\) m from the centre. At the supports, there is no deflection, so when \(x = 6\), \(y = 0\)

$$ \begin{aligned} 0 &= a(6)^2 - 3\\ 36a &= 3\\ a &= \frac{1}{12} \end{aligned} $$

Thus, the equation of the deflected beam becomes:

$$ \begin{aligned} y = \frac{1}{12}x^2 - 3 \end{aligned} $$

We now find how far from the centre the deflection is \(1\) cm, so set \(y = -1\)

$$ \begin{aligned} -1 &= \frac{1}{12}x^2 - 3\\ \frac{1}{12}x^2 &= 2\\ x^2 &= 24\\ x &= 2\sqrt{6} \end{aligned} $$

Therefore, the deflection is \(1\) cm at a distance of \(2\sqrt{6}\) metres from the centre

A rod AB of length 15 cm rests in between two coordinate axes in such a way that the end point A lies on x-axis and end point B lies on y-axis. A point \(P(x, y)\) is taken on the rod in such a way that \(AP = 6\; cm\). Show that the locus of \(P\) is an ellipse.

Solution

Let the coordinate axes be mutually perpendicular with origin \(O\). The rod \(AB\) of length \(15\) cm rests between the axes such that \(A\) lies on the \(x\)-axis and \(B\) lies on the \(y\)-axis

Let the coordinates of \(A\) and \(B\) be \((a, 0)\) and \((0, b)\) respectively. Since the length of the rod is constant, we have:

$$ \begin{aligned} AB^2 &= a^2 + b^2\\ a^2 + b^2 &= 15^2\\ a^2 + b^2 &= 225 \end{aligned} $$

A point \(P(x, y)\) lies on the rod such that \(AP = 6\) cm. Since \(AP : PB = 6 : 9 = 2 : 3\), point \(P\) divides the segment \(AB\) internally in the ratio \(2 : 3\)

Using the section formula, the coordinates of \(P\) are:

$$ \begin{aligned} x &= \frac{3a + 2 \cdot 0}{5} = \frac{3a}{5}\\ y &= \frac{3 \cdot 0 + 2b}{5} = \frac{2b}{5} \end{aligned} $$

From these relations, express \(a\) and \(b\) in terms of \(x\) and \(y\)

$$ \begin{aligned} a &= \frac{5x}{3}\\ b &= \frac{5y}{2} \end{aligned} $$

Substituting into the equation \(a^2 + b^2 = 225\)

$$ \begin{aligned} \left(\frac{5x}{3}\right)^2 + \left(\frac{5y}{2}\right)^2 &= 225\\ \frac{25x^2}{9} + \frac{25y^2}{4} &= 225 \end{aligned} $$

Dividing throughout by \(225\)

$$ \begin{aligned} \frac{x^2}{81} + \frac{y^2}{36} &= 1 \end{aligned} $$

This is the standard equation of an ellipse. Hence, the locus of point \(P\) is an ellipse