Class 11-NCERT Solutions-Chapter-10-Conic Sections

Answer:

The equation of a circle with centre \((h, k)\) and radius \(r\) is given by \((x - h)^2 + (y - k)^2 = r^2\).

Here, \(h = 0\), \(k = 2\), and \(r = 2\).

Substituting these values:

\((x - 0)^2 + (y - 2)^2 = 2^2\)

\(x^2 + y^2 - 4y + 4 = 4\)

\(x^2 + y^2 - 4y = 0\)

Answer:

Given: centre \((h, k) = (-2, 3)\) and radius \(r = 4\).

Using \((x - h)^2 + (y - k)^2 = r^2\):

\((x - (-2))^2 + (y - 3)^2 = 4^2\)

\((x + 2)^2 + (y - 3)^2 = 16\)

\(x^2 + 4x + 4 + y^2 - 6y + 9 = 16\)

\(x^2 + y^2 + 4x - 6y + 13 - 16 = 0\)

\(x^2 + y^2 + 4x - 6y - 3 = 0\)

Answer:

Given: centre \((h, k) = (\frac{1}{2}, \frac{1}{4})\) and radius \(r = \frac{1}{12}\).

Equation:

\((x - \frac{1}{2})^2 + (y - \frac{1}{4})^2 = (\frac{1}{12})^2\)

\(x^2 - x + \frac{1}{4} + y^2 - \frac{y}{2} + \frac{1}{16} = \frac{1}{144}\)

Multiplying by 144 to clear denominators:

\(144x^2 - 144x + 36 + 144y^2 - 72y + 9 = 1\)

\(144x^2 + 144y^2 - 144x - 72y + 45 - 1 = 0\)

\(144x^2 + 144y^2 - 144x - 72y + 44 = 0\)

Dividing by 4:

\(36x^2 + 36y^2 - 36x - 18y + 11 = 0\)

Answer:

Given: centre \((h, k) = (1, 1)\) and radius \(r = \sqrt{2}\).

Equation:

\((x - 1)^2 + (y - 1)^2 = (\sqrt{2})^2\)

\(x^2 - 2x + 1 + y^2 - 2y + 1 = 2\)

\(x^2 + y^2 - 2x - 2y + 2 = 2\)

\(x^2 + y^2 - 2x - 2y = 0\)

Answer:

Given: centre \((h, k) = (-a, -b)\) and radius \(r = \sqrt{a^2 - b^2}\).

Equation:

\((x - (-a))^2 + (y - (-b))^2 = (\sqrt{a^2 - b^2})^2\)

\((x + a)^2 + (y + b)^2 = a^2 - b^2\)

\(x^2 + 2ax + a^2 + y^2 + 2by + b^2 = a^2 - b^2\)

\(x^2 + y^2 + 2ax + 2by + b^2 + b^2 = 0\)

\(x^2 + y^2 + 2ax + 2by + 2b^2 = 0\)

Answer:

Comparing with the standard form \((x - h)^2 + (y - k)^2 = r^2\):

\((x - (-5))^2 + (y - 3)^2 = 6^2\)

Centre: \((-5, 3)\)

Radius: 6

Answer:

Method 1: Completing the square

\((x^2 - 4x) + (y^2 - 8y) = 45\)

\((x^2 - 4x + 4) + (y^2 - 8y + 16) = 45 + 4 + 16\)

\((x - 2)^2 + (y - 4)^2 = 65\)

Comparing with standard form:

Centre: \((2, 4)\)

Radius: \(\sqrt{65}\)

Answer:

Grouping x and y terms:

\((x^2 - 8x) + (y^2 + 10y) = 12\)

\((x^2 - 8x + 16) + (y^2 + 10y + 25) = 12 + 16 + 25\)

\((x - 4)^2 + (y + 5)^2 = 53\)

Comparing with \((x - h)^2 + (y - k)^2 = r^2\):

Centre: \((4, -5)\)

Radius: \(\sqrt{53}\)

Answer:

First, divide the equation by 2 to make coefficients of \(x^2\) and \(y^2\) equal to 1:

\(x^2 + y^2 - \frac{x}{2} = 0\)

\((x^2 - \frac{x}{2}) + y^2 = 0\)

Adding \((\frac{1}{4})^2 = \frac{1}{16}\) to both sides:

\((x^2 - \frac{x}{2} + \frac{1}{16}) + y^2 = \frac{1}{16}\)

\((x - \frac{1}{4})^2 + (y - 0)^2 = (\frac{1}{4})^2\)

Centre: \((\frac{1}{4}, 0)\)

Radius: \(\frac{1}{4}\)

Answer:

Let the equation of the circle be \((x - h)^2 + (y - k)^2 = r^2\).

Since it passes through (4, 1): \((4 - h)^2 + (1 - k)^2 = r^2\) ... (i)

Since it passes through (6, 5): \((6 - h)^2 + (5 - k)^2 = r^2\) ... (ii)

Equating (i) and (ii):

\((4 - h)^2 + (1 - k)^2 = (6 - h)^2 + (5 - k)^2\)

\(16 - 8h + h^2 + 1 - 2k + k^2 = 36 - 12h + h^2 + 25 - 10k + k^2\)

\(17 - 8h - 2k = 61 - 12h - 10k\)

\(4h + 8k = 44 \Rightarrow h + 2k = 11\) ... (iii)

The centre \((h, k)\) lies on the line \(4x + y = 16\), so:

\(4h + k = 16\) ... (iv)

Multiplying (iv) by 2: \(8h + 2k = 32\). Subtracting (iii) from this:

\((8h + 2k) - (h + 2k) = 32 - 11 \Rightarrow 7h = 21 \Rightarrow h = 3\).

From (iv), \(k = 16 - 4(3) = 16 - 12 = 4\).

Centre is \((3, 4)\). Find radius using point (4, 1):

\(r^2 = (4 - 3)^2 + (1 - 4)^2 = 1^2 + (-3)^2 = 1 + 9 = 10\).

Equation: \((x - 3)^2 + (y - 4)^2 = 10\)

Answer:

Let centre be \((h, k)\). Since it passes through (2, 3) and (-1, 1), distances to centre are equal:

\((h - 2)^2 + (k - 3)^2 = (h + 1)^2 + (k - 1)^2\)

\(h^2 - 4h + 4 + k^2 - 6k + 9 = h^2 + 2h + 1 + k^2 - 2k + 1\)

\(-4h - 6k + 13 = 2h - 2k + 2\)

\(-6h - 4k = -11 \Rightarrow 6h + 4k = 11\) ... (i)

Centre lies on \(x - 3y - 11 = 0\), so \(h - 3k = 11\) ... (ii)

From (ii), \(h = 11 + 3k\). Substitute into (i):

\(6(11 + 3k) + 4k = 11 \Rightarrow 66 + 18k + 4k = 11\)

\(22k = -55 \Rightarrow k = -\frac{55}{22} = -\frac{5}{2}\).

\(h = 11 + 3(-\frac{5}{2}) = \frac{22 - 15}{2} = \frac{7}{2}\).

Radius square \(r^2 = (\frac{7}{2} - 2)^2 + (-\frac{5}{2} - 3)^2 = (\frac{3}{2})^2 + (-\frac{11}{2})^2 = \frac{9}{4} + \frac{121}{4} = \frac{130}{4}\).

Equation: \((x - \frac{7}{2})^2 + (y + \frac{5}{2})^2 = \frac{130}{4}\)

Simplify: \(x^2 - 7x + \frac{49}{4} + y^2 + 5y + \frac{25}{4} = \frac{130}{4}\)

\(x^2 + y^2 - 7x + 5y + \frac{74}{4} - \frac{130}{4} = 0\)

\(x^2 + y^2 - 7x + 5y - 14 = 0\)

Answer:

Since centre lies on x-axis, its coordinates are \((h, 0)\). Radius \(r = 5\).

Equation is \((x - h)^2 + (y - 0)^2 = 5^2\).

Passes through (2, 3): \((2 - h)^2 + 3^2 = 25\).

\((2 - h)^2 + 9 = 25 \Rightarrow (2 - h)^2 = 16\).

\(2 - h = \pm 4\).

If \(2 - h = 4 \Rightarrow h = -2\). Centre (-2, 0). Eq: \((x + 2)^2 + y^2 = 25\).

If \(2 - h = -4 \Rightarrow h = 6\). Centre (6, 0). Eq: \((x - 6)^2 + y^2 = 25\).

Answer:

The circle passes through origin O(0,0). Intercepts \(a\) and \(b\) on axes mean it passes through A(a, 0) and B(0, b).

Since \(\angle AOB = 90^\circ\), the chord AB is the diameter of the circle.

Midpoint of AB is the centre: \((\frac{a+0}{2}, \frac{0+b}{2}) = (\frac{a}{2}, \frac{b}{2})\).

Radius is distance from centre to origin: \(\sqrt{(\frac{a}{2})^2 + (\frac{b}{2})^2} = \sqrt{\frac{a^2+b^2}{4}}\).

Equation: \((x - \frac{a}{2})^2 + (y - \frac{b}{2})^2 = \frac{a^2+b^2}{4}\)

\(x^2 - ax + \frac{a^2}{4} + y^2 - by + \frac{b^2}{4} = \frac{a^2}{4} + \frac{b^2}{4}\)

Equation: \(x^2 + y^2 - ax - by = 0\)

Answer:

Centre \((h, k) = (2, 2)\).

Radius \(r\) is the distance between (2, 2) and (4, 5).

\(r^2 = (4 - 2)^2 + (5 - 2)^2 = 2^2 + 3^2 = 4 + 9 = 13\).

Equation: \((x - 2)^2 + (y - 2)^2 = 13\).

\(x^2 - 4x + 4 + y^2 - 4y + 4 = 13\)

\(x^2 + y^2 - 4x - 4y - 5 = 0\)

Answer:

The equation is \(x^2 + y^2 = 25\), so radius squared \(r^2 = 25\) (radius = 5) and centre is (0,0).

Calculate the distance squared of the point (-2.5, 3.5) from origin:

\(d^2 = (-2.5)^2 + (3.5)^2 = 6.25 + 12.25 = 18.5\).

Since \(18.5 < 25\) (i.e., \(d^2 < r^2\)), the point lies inside the circle.

Answer:

The equation is of the form \(y^2 = 4ax\). Here \(4a = 12 \Rightarrow a = 3\).

• Focus (a, 0): (3, 0)
• Axis: x-axis (y = 0)
• Directrix (x = -a): \(x = -3\)
• Latus Rectum (4a): 12

Answer:

The equation is of the form \(x^2 = 4ay\). Here \(4a = 6 \Rightarrow a = \frac{6}{4} = \frac{3}{2}\).

• Focus (0, a): \((0, \frac{3}{2})\)
• Axis: y-axis (x = 0)
• Directrix (y = -a): \(y = -\frac{3}{2}\)
• Latus Rectum (4a): 6

Answer:

The equation is of the form \(y^2 = -4ax\). Here \(4a = 8 \Rightarrow a = 2\).

• Focus (-a, 0): (-2, 0)
• Axis: x-axis
• Directrix (x = a): \(x = 2\)
• Latus Rectum (4a): 8

Answer:

The equation is of the form \(x^2 = -4ay\). Here \(4a = 16 \Rightarrow a = 4\).

• Focus (0, -a): (0, -4)
• Axis: y-axis
• Directrix (y = a): \(y = 4\)
• Latus Rectum (4a): 16

Answer:

Form \(y^2 = 4ax\). \(4a = 10 \Rightarrow a = 2.5\).

• Focus (a, 0): (2.5, 0)
• Axis: x-axis
• Directrix: \(x = -2.5\)
• Latus Rectum: 10

Answer:

Form \(x^2 = -4ay\). \(4a = 9 \Rightarrow a = \frac{9}{4}\).

• Focus (0, -a): \((0, -\frac{9}{4})\)
• Axis: y-axis
• Directrix: \(y = \frac{9}{4}\)
• Latus Rectum: 9

Answer:

Focus is on the positive x-axis \((a, 0)\) and directrix is \(x = -a\). This corresponds to \(y^2 = 4ax\).

Here \(a = 6\).

Equation: \(y^2 = 4(6)x \Rightarrow y^2 = 24x\).

Answer:

Focus is on the negative y-axis \((0, -a)\) and directrix is \(y = a\). This corresponds to \(x^2 = -4ay\).

Here \(a = 3\).

Equation: \(x^2 = -4(3)y \Rightarrow x^2 = -12y\).

Answer:

Vertex at origin and focus at (3, 0) (positive x-axis). Standard form \(y^2 = 4ax\).

Here \(a = 3\).

Equation: \(y^2 = 4(3)x \Rightarrow y^2 = 12x\).

Answer:

Vertex at origin and focus at (-2, 0) (negative x-axis). Standard form \(y^2 = -4ax\).

Here \(a = 2\).

Equation: \(y^2 = -4(2)x \Rightarrow y^2 = -8x\).

Answer:

Axis is along x-axis, so equation is either \(y^2 = 4ax\) or \(y^2 = -4ax\).

Point (2, 3) is in the first quadrant, so the parabola opens to the right: \(y^2 = 4ax\).

Substitute (2, 3): \(3^2 = 4a(2) \Rightarrow 9 = 8a \Rightarrow a = \frac{9}{8}\).

Equation: \(y^2 = 4(\frac{9}{8})x \Rightarrow y^2 = \frac{9}{2}x \Rightarrow 2y^2 = 9x\).

Answer:

Symmetric w.r.t y-axis means equation is \(x^2 = 4ay\) or \(x^2 = -4ay\).

Point (5, 2) has positive y, so it opens upwards: \(x^2 = 4ay\).

Substitute (5, 2): \(5^2 = 4a(2) \Rightarrow 25 = 8a \Rightarrow a = \frac{25}{8}\).

Equation: \(x^2 = 4(\frac{25}{8})y \Rightarrow x^2 = \frac{25}{2}y \Rightarrow 2x^2 = 25y\).

Answer:

Since the denominator of \(x^2\) (36) is greater than the denominator of \(y^2\) (16), the major axis is along the x-axis.

Comparing with \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\), we have \(a^2 = 36\) and \(b^2 = 16\).

So, \(a = 6\) and \(b = 4\).

\(c = \sqrt{a^2 - b^2} = \sqrt{36 - 16} = \sqrt{20} = 2\sqrt{5}\).

Foci \((\pm c, 0)\): \((\pm 2\sqrt{5}, 0)\)

Vertices \((\pm a, 0)\): \((\pm 6, 0)\)

Length of major axis \((2a)\): \(12\)

Length of minor axis \((2b)\): \(8\)

Eccentricity \(e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3}\)

Length of Latus Rectum \(\frac{2b^2}{a} = \frac{2(16)}{6} = \frac{16}{3}\)

Answer:

Here \(25 > 4\), so the major axis is along the y-axis.

\(a^2 = 25 \Rightarrow a = 5\), \(b^2 = 4 \Rightarrow b = 2\).

\(c = \sqrt{a^2 - b^2} = \sqrt{25 - 4} = \sqrt{21}\).

Foci \((0, \pm c)\): \((0, \pm \sqrt{21})\)

Vertices \((0, \pm a)\): \((0, \pm 5)\)

Length of major axis \((2a)\): \(10\)

Length of minor axis \((2b)\): \(4\)

Eccentricity \(e = \frac{c}{a} = \frac{\sqrt{21}}{5}\)

Length of Latus Rectum \(\frac{2b^2}{a} = \frac{2(4)}{5} = \frac{8}{5}\)

Answer:

Major axis is along the x-axis (\(16 > 9\)).

\(a = 4\), \(b = 3\). \(c = \sqrt{16-9} = \sqrt{7}\).

Answer:

Major axis is along the y-axis (\(100 > 25\)).

\(a = 10\), \(b = 5\). \(c = \sqrt{100-25} = \sqrt{75} = 5\sqrt{3}\).

Answer:

Major axis is along the x-axis.

\(a = 7\), \(b = 6\). \(c = \sqrt{49-36} = \sqrt{13}\).

Foci: \((\pm \sqrt{13}, 0)\), Vertices: \((\pm 7, 0)\)

Axes: Major = 14, Minor = 12

Eccentricity: \(\frac{\sqrt{13}}{7}\), LR: \(\frac{2(36)}{7} = \frac{72}{7}\)

Answer:

Major axis is along the y-axis.

\(a = 20\), \(b = 10\). \(c = \sqrt{400-100} = \sqrt{300} = 10\sqrt{3}\).

Foci: \((0, \pm 10\sqrt{3})\), Vertices: \((0, \pm 20)\)

Axes: Major = 40, Minor = 20

Eccentricity: \(\frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2}\), LR: \(\frac{2(100)}{20} = 10\)

Answer:

Divide by 144: \(\frac{x^2}{4} + \frac{y^2}{36} = 1\).

Major axis is y-axis. \(a = 6\), \(b = 2\). \(c = \sqrt{36-4} = \sqrt{32} = 4\sqrt{2}\).

Foci: \((0, \pm 4\sqrt{2})\), Vertices: \((0, \pm 6)\)

Axes: Major = 12, Minor = 4

Eccentricity: \(\frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3}\), LR: \(\frac{2(4)}{6} = \frac{4}{3}\)

Answer:

Divide by 16: \(\frac{x^2}{1} + \frac{y^2}{16} = 1\).

Major axis is y-axis. \(a = 4\), \(b = 1\). \(c = \sqrt{16-1} = \sqrt{15}\).

Foci: \((0, \pm \sqrt{15})\), Vertices: \((0, \pm 4)\)

Axes: Major = 8, Minor = 2

Eccentricity: \(\frac{\sqrt{15}}{4}\), LR: \(\frac{2(1)}{4} = \frac{1}{2}\)

Answer:

Divide by 36: \(\frac{x^2}{9} + \frac{y^2}{4} = 1\).

Major axis is x-axis. \(a = 3\), \(b = 2\). \(c = \sqrt{9-4} = \sqrt{5}\).

Foci: \((\pm \sqrt{5}, 0)\), Vertices: \((\pm 3, 0)\)

Axes: Major = 6, Minor = 4

Eccentricity: \(\frac{\sqrt{5}}{3}\), LR: \(\frac{2(4)}{3} = \frac{8}{3}\)

Answer:

Vertices are on the x-axis, so the equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).

\(a = 5\), \(c = 4\).

We know \(c^2 = a^2 - b^2 \Rightarrow 16 = 25 - b^2 \Rightarrow b^2 = 9\).

Equation: \(\frac{x^2}{25} + \frac{y^2}{9} = 1\)

Answer:

Vertices are on the y-axis, so the equation is \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\).

\(a = 13\), \(c = 5\).

\(b^2 = a^2 - c^2 = 169 - 25 = 144\).

Equation: \(\frac{x^2}{144} + \frac{y^2}{169} = 1\)

Answer:

Vertices on x-axis. \(a = 6\), \(c = 4\).

\(b^2 = a^2 - c^2 = 36 - 16 = 20\).

Equation: \(\frac{x^2}{36} + \frac{y^2}{20} = 1\)

Answer:

Major axis on x-axis. \(a = 3\). Minor axis on y-axis. \(b = 2\).

\(a^2 = 9\), \(b^2 = 4\).

Equation: \(\frac{x^2}{9} + \frac{y^2}{4} = 1\)

Answer:

Major axis on y-axis. \(a = \sqrt{5}\). Minor axis on x-axis. \(b = 1\).

\(a^2 = 5\), \(b^2 = 1\).

Equation: \(\frac{x^2}{1} + \frac{y^2}{5} = 1\)

Answer:

Foci on x-axis. \(c = 5\). Length of major axis \(2a = 26 \Rightarrow a = 13\).

\(b^2 = a^2 - c^2 = 169 - 25 = 144\).

Equation: \(\frac{x^2}{169} + \frac{y^2}{144} = 1\)

Answer:

Foci on y-axis. \(c = 6\). Length of minor axis \(2b = 16 \Rightarrow b = 8\).

\(a^2 = b^2 + c^2 = 64 + 36 = 100\).

Equation: \(\frac{x^2}{64} + \frac{y^2}{100} = 1\)

Answer:

Foci on x-axis. \(c = 3\), \(a = 4\).

\(b^2 = a^2 - c^2 = 16 - 9 = 7\).

Equation: \(\frac{x^2}{16} + \frac{y^2}{7} = 1\)

Answer:

Foci on x-axis means horizontal ellipse.

\(a^2 = b^2 + c^2 = 9 + 16 = 25\).

Equation: \(\frac{x^2}{25} + \frac{y^2}{9} = 1\)

Answer:

Since the major axis is the y-axis, equation is \(\frac{x^2}{b^2} + \frac{y^2}{a^2} = 1\).

Passes through (3, 2): \(\frac{9}{b^2} + \frac{4}{a^2} = 1\) ... (i)

Passes through (1, 6): \(\frac{1}{b^2} + \frac{36}{a^2} = 1\) ... (ii)

Multiply (ii) by 9: \(\frac{9}{b^2} + \frac{324}{a^2} = 9\) ... (iii)

Subtract (i) from (iii): \(\frac{320}{a^2} = 8 \Rightarrow a^2 = 40\).

Substitute \(a^2 = 40\) in (ii): \(\frac{1}{b^2} + \frac{36}{40} = 1 \Rightarrow \frac{1}{b^2} = 1 - \frac{9}{10} = \frac{1}{10} \Rightarrow b^2 = 10\).

Equation: \(\frac{x^2}{10} + \frac{y^2}{40} = 1\)

Answer:

Equation is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).

Passes through (4, 3): \(\frac{16}{a^2} + \frac{9}{b^2} = 1\) ... (i)

Passes through (6, 2): \(\frac{36}{a^2} + \frac{4}{b^2} = 1\) ... (ii)

Multiply (i) by 4 and (ii) by 9:
\(\frac{64}{a^2} + \frac{36}{b^2} = 4\)
\(\frac{324}{a^2} + \frac{36}{b^2} = 9\)

Subtracting: \(\frac{260}{a^2} = 5 \Rightarrow a^2 = 52\).

Substitute \(a^2 = 52\) in (i): \(\frac{16}{52} + \frac{9}{b^2} = 1 \Rightarrow \frac{4}{13} + \frac{9}{b^2} = 1\).

\(\frac{9}{b^2} = 1 - \frac{4}{13} = \frac{9}{13} \Rightarrow b^2 = 13\).

Equation: \(\frac{x^2}{52} + \frac{y^2}{13} = 1\)

Answer:

Form: \(\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1\) (Horizontal Transverse Axis).

\(a^2 = 16 \Rightarrow a = 4\), \(b^2 = 9 \Rightarrow b = 3\).

\(c^2 = a^2 + b^2 = 16 + 9 = 25 \Rightarrow c = 5\).

Foci \((\pm c, 0)\): \((\pm 5, 0)\)

Vertices \((\pm a, 0)\): \((\pm 4, 0)\)

Eccentricity \(e = \frac{c}{a} = \frac{5}{4}\)

Latus Rectum \(\frac{2b^2}{a} = \frac{2(9)}{4} = \frac{9}{2}\)

Answer:

Form: \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\) (Vertical Transverse Axis).

\(a^2 = 9 \Rightarrow a = 3\), \(b^2 = 27\).

\(c^2 = 9 + 27 = 36 \Rightarrow c = 6\).

Foci \((0, \pm c)\): \((0, \pm 6)\)

Vertices \((0, \pm a)\): \((0, \pm 3)\)

Eccentricity \(e = \frac{6}{3} = 2\)

Latus Rectum \(\frac{2(27)}{3} = 18\)

Answer:

Divide by 36: \(\frac{y^2}{4} - \frac{x^2}{9} = 1\).

Vertical transverse axis. \(a = 2\), \(b = 3\).

\(c = \sqrt{4+9} = \sqrt{13}\).

Foci: \((0, \pm \sqrt{13})\), Vertices: \((0, \pm 2)\)

Eccentricity: \(\frac{\sqrt{13}}{2}\), LR: \(\frac{2(9)}{2} = 9\)

Answer:

Divide by 576: \(\frac{x^2}{36} - \frac{y^2}{64} = 1\).

Horizontal transverse axis. \(a = 6\), \(b = 8\).

\(c = \sqrt{36+64} = \sqrt{100} = 10\).

Foci: \((\pm 10, 0)\), Vertices: \((\pm 6, 0)\)

Eccentricity: \(\frac{10}{6} = \frac{5}{3}\), LR: \(\frac{2(64)}{6} = \frac{64}{3}\)

Answer:

Divide by 36: \(\frac{5y^2}{36} - \frac{x^2}{4} = 1 \Rightarrow \frac{y^2}{36/5} - \frac{x^2}{4} = 1\).

Vertical transverse axis. \(a^2 = \frac{36}{5}\), \(b^2 = 4\).

\(c^2 = \frac{36}{5} + 4 = \frac{56}{5} \Rightarrow c = \sqrt{\frac{56}{5}} = 2\sqrt{\frac{14}{5}}\).

Foci: \((0, \pm 2\sqrt{\frac{14}{5}})\)

Vertices: \((0, \pm \frac{6}{\sqrt{5}})\)

Eccentricity: \(\frac{\sqrt{56/5}}{\sqrt{36/5}} = \frac{\sqrt{56}}{6} = \frac{2\sqrt{14}}{6} = \frac{\sqrt{14}}{3}\)

LR: \(\frac{2(4)}{6/\sqrt{5}} = \frac{8\sqrt{5}}{6} = \frac{4\sqrt{5}}{3}\)

Answer:

Divide by 784: \(\frac{y^2}{16} - \frac{x^2}{49} = 1\).

Vertical. \(a = 4\), \(b = 7\). \(c = \sqrt{16+49} = \sqrt{65}\).

Foci: \((0, \pm \sqrt{65})\), Vertices: \((0, \pm 4)\)

Eccentricity: \(\frac{\sqrt{65}}{4}\), LR: \(\frac{2(49)}{4} = \frac{49}{2}\)

Answer:

Horizontal hyperbola. \(a = 2, c = 3\).

\(b^2 = c^2 - a^2 = 9 - 4 = 5\).

Equation: \(\frac{x^2}{4} - \frac{y^2}{5} = 1\)

Answer:

Vertical hyperbola. \(a = 5, c = 8\).

\(b^2 = c^2 - a^2 = 64 - 25 = 39\).

Equation: \(\frac{y^2}{25} - \frac{x^2}{39} = 1\)

Answer:

Vertical hyperbola. \(a = 3, c = 5\).

\(b^2 = 25 - 9 = 16\).

Equation: \(\frac{y^2}{9} - \frac{x^2}{16} = 1\)

Answer:

Horizontal hyperbola. \(c = 5\). Transverse axis \(2a = 8 \Rightarrow a = 4\).

\(b^2 = c^2 - a^2 = 25 - 16 = 9\).

Equation: \(\frac{x^2}{16} - \frac{y^2}{9} = 1\)

Answer:

Vertical hyperbola. \(c = 13\). Conjugate axis \(2b = 24 \Rightarrow b = 12\).

\(a^2 = c^2 - b^2 = 169 - 144 = 25\).

Equation: \(\frac{y^2}{25} - \frac{x^2}{144} = 1\)

Answer:

Horizontal hyperbola. \(c = 3\sqrt{5}\).

LR: \(\frac{2b^2}{a} = 8 \Rightarrow b^2 = 4a\).

\(c^2 = a^2 + b^2 \Rightarrow 45 = a^2 + 4a \Rightarrow a^2 + 4a - 45 = 0\).

\((a + 9)(a - 5) = 0 \Rightarrow a = 5\) (since \(a>0\)).

\(b^2 = 4(5) = 20\).

Equation: \(\frac{x^2}{25} - \frac{y^2}{20} = 1\)

Answer:

Horizontal hyperbola. \(c = 4\).

\(\frac{2b^2}{a} = 12 \Rightarrow b^2 = 6a\).

\(16 = a^2 + 6a \Rightarrow a^2 + 6a - 16 = 0 \Rightarrow (a+8)(a-2) = 0 \Rightarrow a = 2\).

\(b^2 = 12\).

Equation: \(\frac{x^2}{4} - \frac{y^2}{12} = 1\)

Answer:

Horizontal hyperbola. \(a = 7\).

\(e = \frac{c}{a} = \frac{4}{3} \Rightarrow c = \frac{28}{3}\).

\(b^2 = c^2 - a^2 = \frac{784}{9} - 49 = \frac{784 - 441}{9} = \frac{343}{9}\).

Equation: \(\frac{x^2}{49} - \frac{9y^2}{343} = 1\)

Answer:

Vertical hyperbola: \(\frac{y^2}{a^2} - \frac{x^2}{b^2} = 1\). \(c = \sqrt{10}\).

\(a^2 + b^2 = 10 \Rightarrow b^2 = 10 - a^2\).

Passes through (2, 3): \(\frac{9}{a^2} - \frac{4}{10 - a^2} = 1\).

Let \(u = a^2\). \(9(10-u) - 4u = u(10-u) \Rightarrow 90 - 13u = 10u - u^2\).

\(u^2 - 23u + 90 = 0 \Rightarrow (u-18)(u-5) = 0\).

If \(a^2 = 18\), \(b^2 = -8\) (impossible). So \(a^2 = 5\), \(b^2 = 5\).

Equation: \(\frac{y^2}{5} - \frac{x^2}{5} = 1\) or \(y^2 - x^2 = 5\).

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Answer:

Let the vertex of the parabola be at the origin (0, 0) and the axis be along the x-axis.

Equation of the parabola: \(y^2 = 4ax\).

Since the diameter is 20 cm and depth is 5 cm, the coordinates of the end point of the diameter are \((5, 10)\).

Substituting \(x = 5\) and \(y = 10\) in the equation:

\((10)^2 = 4a(5) \Rightarrow 100 = 20a \Rightarrow a = 5\).

The focus is at \((a, 0)\). Thus, the focus is at (5, 0), which is the mid-point of the diameter.

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Answer:

Let the vertex of the parabola be at the origin (0, 0). Since the axis is vertical and it opens downwards (as it's an arch), the equation is \(x^2 = -4ay\).

The base is 5 m wide and height is 10 m. The coordinates of the end points of the base are \((\pm 2.5, -10)\).

Substitute \((2.5, -10)\) into the equation:

\((2.5)^2 = -4a(-10) \Rightarrow 6.25 = 40a \Rightarrow a = \frac{6.25}{40}\).

We need the width at 2 m from the vertex. Let the point be \((x, -2)\).

\(x^2 = -4(\frac{6.25}{40})(-2) = \frac{8 \times 6.25}{40} = \frac{50}{40} = 1.25\).

\(x = \sqrt{1.25} \approx 1.118\) m.

Total width = \(2x = 2(1.118) \approx \mathbf{2.23}\) m.

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Answer:

Let the lowest point of the cable be the vertex. Let vertex be at \((0, 6)\) because the shortest wire is 6 m.

Equation: \(x^2 = 4a(y - 6)\).

The bridge is 100 m long, so the ends are at \(x = \pm 50\). The longest wire is 30 m, so \(y = 30\). Point is \((50, 30)\).

\(50^2 = 4a(30 - 6) \Rightarrow 2500 = 4a(24) \Rightarrow 4a = \frac{2500}{24}\).

We need the length of the wire at 18 m from the middle (\(x = 18\)). Let length be \(L\).

\(18^2 = \frac{2500}{24}(L - 6)\)

\(324 = \frac{2500}{24}(L - 6) \Rightarrow L - 6 = \frac{324 \times 24}{2500} = 3.11\).

\(L = 6 + 3.11 = \mathbf{9.11}\) m.

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Answer:

Let the centre of the base be the origin (0, 0).

Width \(2a = 8 \Rightarrow a = 4\). Height \(b = 2\).

Equation: \(\frac{x^2}{16} + \frac{y^2}{4} = 1\).

A point 1.5 m from one end (which is at \(x=4\)) corresponds to \(x = 4 - 1.5 = 2.5\).

Substitute \(x = 2.5\) or \(5/2\):

\(\frac{(2.5)^2}{16} + \frac{y^2}{4} = 1 \Rightarrow \frac{6.25}{16} + \frac{y^2}{4} = 1\).

\(\frac{y^2}{4} = 1 - \frac{6.25}{16} = \frac{9.75}{16}\).

\(y^2 = \frac{9.75 \times 4}{16} = \frac{9.75}{4} = 2.4375\).

\(y \approx \mathbf{1.56}\) m.

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Answer:

Let the rod AB of length 12 cm touch axes at A(a, 0) and B(0, b). Point P(x, y) is on the rod such that AP = 3 cm. Thus PB = 9 cm.

Using section formula or trigonometry:

\(x = \frac{1 \cdot 0 + 3 \cdot a}{4}\) (Wait, simplified using similar triangles is better).

Let \(\theta\) be the angle the rod makes with the x-axis.

\(x = (12-3)\cos\theta = 9\cos\theta \Rightarrow \cos\theta = \frac{x}{9}\).

\(y = 3\sin\theta \Rightarrow \sin\theta = \frac{y}{3}\).

Using \(\sin^2\theta + \cos^2\theta = 1\):

\(\frac{y^2}{9} + \frac{x^2}{81} = 1\) or \(\frac{x^2}{81} + \frac{y^2}{9} = 1\).

Equation: An ellipse \(\frac{x^2}{81} + \frac{y^2}{9} = 1\).

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Answer:

Equation: \(x^2 = 12y\). Compare with \(x^2 = 4ay \Rightarrow 4a = 12 \Rightarrow a = 3\).

Vertex is at (0, 0). Focus is at (0, 3).

The latus rectum is the horizontal line passing through the focus. Ends of LR are \((\pm 2a, a)\), which are \((\pm 6, 3)\).

The triangle is formed by points (0, 0), (6, 3), and (-6, 3).

Base of triangle = Distance between (6, 3) and (-6, 3) = 12.

Height of triangle = y-coordinate of LR = 3.

Area = \(\frac{1}{2} \times \text{Base} \times \text{Height} = \frac{1}{2} \times 12 \times 3 = \mathbf{18}\) sq units.

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Answer:

This describes an ellipse where the flag posts are the foci.

Sum of distances = \(2a = 10 \Rightarrow a = 5\).

Distance between foci = \(2c = 8 \Rightarrow c = 4\).

We know \(b^2 = a^2 - c^2 = 25 - 16 = 9\).

The equation of the path is \(\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1\).

Equation: \(\frac{x^2}{25} + \frac{y^2}{9} = 1\).

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Answer:

Let the side length be \(L\). One vertex is at (0, 0). Since the parabola is symmetric about the x-axis, the other two vertices make angles of \(30^\circ\) and \(-30^\circ\) with the axis.

Coordinates of one vertex: \((L \cos 30^\circ, L \sin 30^\circ) = (L\frac{\sqrt{3}}{2}, \frac{L}{2})\).

Since this point lies on \(y^2 = 4ax\):

\((\frac{L}{2})^2 = 4a(L\frac{\sqrt{3}}{2})\)

\(\frac{L^2}{4} = 2\sqrt{3}aL\)

Dividing by \(L\) (since \(L \ne 0\)): \(\frac{L}{4} = 2\sqrt{3}a\).

\(L = 8\sqrt{3}a\).

Length of side: \(8a\sqrt{3}\).