GSEB Class 11 Maths Solutions Chapter 11 શાંકવો 11.3

Question 1. \( \frac{x^2}{36}+\frac{y^2}{16} = 1 \)
Answer: Here, the denominator of \( \frac{x^2}{36} \) is larger than the denominator of \( \frac{y^2}{16} \), so the major axis is on the X-axis. Comparing the given equation with \( \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \), we get:
\( a^2 = 36 \), \( b^2 = 16 \)
\( a = 6 \), \( b = 4 \)
Now, calculate \( c \):
\( c = \sqrt{a^2-b^2} = \sqrt{36-16} = \sqrt{20} = 2\sqrt{5} \)
Thus, we find the following properties:
(1) Foci: \( (\pm c, 0) = (\pm 2\sqrt{5}, 0) \)
(2) Vertices: \( (\pm a, 0) = (\pm 6, 0) \)
(3) Length of major axis: \( 2a = 2 \times 6 = 12 \)
(4) Length of minor axis: \( 2b = 2 \times 4 = 8 \)
(5) Eccentricity: \( e = \frac{c}{a} = \frac{2\sqrt{5}}{6} = \frac{\sqrt{5}}{3} \)
(6) Length of latus rectum: \( \frac{2b^2}{a} = \frac{2 \times 16}{6} = \frac{16}{3} \)
In simple words: To analyze the ellipse, we first compare its equation to the standard form. This comparison helps us find 'a' and 'b'. Then, we compute 'c' using the relationship between 'a', 'b', and 'c'. Once 'a', 'b', and 'c' are known, we can easily determine the foci, vertices, axis lengths, eccentricity, and latus rectum by applying their specific formulas.

Exam Tip: Always identify whether the major axis is on the X or Y axis first, as this affects the coordinates of foci and vertices and the standard form of the equation.

Question 2. \( \frac{x^2}{4}+\frac{y^2}{25} = 1 \)
Answer: Here, the denominator of \( \frac{y^2}{25} \) is larger than the denominator of \( \frac{x^2}{4} \), so the major axis is on the Y-axis. Comparing the given equation with \( \frac{x^2}{b^2}+\frac{y^2}{a^2} = 1 \), we get:
\( b^2 = 4 \), \( a^2 = 25 \)
\( b = 2 \), \( a = 5 \)
Now, calculate \( c \):
\( c = \sqrt{a^2-b^2} = \sqrt{25-4} = \sqrt{21} \)
Thus, we find the following properties:
(1) Foci: \( (0, \pm c) = (0, \pm \sqrt{21}) \)
(2) Vertices: \( (0, \pm a) = (0, \pm 5) \)
(3) Length of major axis: \( 2a = 2 \times 5 = 10 \)
(4) Length of minor axis: \( 2b = 2 \times 2 = 4 \)
(5) Eccentricity: \( e = \frac{c}{a} = \frac{\sqrt{21}}{5} \)
(6) Length of latus rectum: \( \frac{2b^2}{a} = \frac{2 \times 4}{5} = \frac{8}{5} \)
In simple words: When the larger number is under \( y^2 \), the major axis is vertical. Identify 'a' and 'b' from the equation. Use these values to calculate 'c'. With 'a', 'b', and 'c', you can then determine all other properties like foci and vertices by using their specific formulas for a Y-axis oriented ellipse.

Exam Tip: Be careful to set up the general equation correctly, based on whether \(a^2\) is under \(x^2\) or \(y^2\).

Question 3. \( \frac{x^2}{16}+\frac{y^2}{9} = 1 \)
Answer: Here, the denominator of \( \frac{x^2}{16} \) is larger than the denominator of \( \frac{y^2}{9} \), so the major axis is on the X-axis. Comparing the given equation with \( \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \), we get:
\( a^2 = 16 \), \( b^2 = 9 \)
\( a = 4 \), \( b = 3 \)
Now, calculate \( c \):
\( c = \sqrt{a^2-b^2} = \sqrt{16-9} = \sqrt{7} \)
Thus, we find the following properties:
(1) Foci: \( (\pm c, 0) = (\pm \sqrt{7}, 0) \)
(2) Vertices: \( (\pm a, 0) = (\pm 4, 0) \)
(3) Length of major axis: \( 2a = 2 \times 4 = 8 \)
(4) Length of minor axis: \( 2b = 2 \times 3 = 6 \)
(5) Eccentricity: \( e = \frac{c}{a} = \frac{\sqrt{7}}{4} \)
(6) Length of latus rectum: \( \frac{2b^2}{a} = \frac{2 \times 9}{4} = \frac{9}{2} \)
In simple words: For an ellipse, always start by identifying the values of 'a' and 'b' from the equation. Then, use the formula to find 'c'. Once these primary values are known, you can directly calculate the foci, vertices, axis lengths, eccentricity, and latus rectum.

Exam Tip: Ensure that the square root of 'c' is kept in its simplest radical form when writing the coordinates of the foci.

Question 4. \( \frac{x^2}{25}+\frac{y^2}{100} = 1 \)
Answer: Here, the denominator of \( \frac{y^2}{100} \) is larger than the denominator of \( \frac{x^2}{25} \), so the major axis is on the Y-axis. Comparing the given equation with \( \frac{x^2}{b^2}+\frac{y^2}{a^2} = 1 \), we get:
\( b^2 = 25 \), \( a^2 = 100 \)
\( b = 5 \), \( a = 10 \)
Now, calculate \( c \):
\( c = \sqrt{a^2-b^2} = \sqrt{100-25} = \sqrt{75} = 5\sqrt{3} \)
Thus, we find the following properties:
(1) Foci: \( (0, \pm c) = (0, \pm 5\sqrt{3}) \)
(2) Vertices: \( (0, \pm a) = (0, \pm 10) \)
(3) Length of major axis: \( 2a = 2 \times 10 = 20 \)
(4) Length of minor axis: \( 2b = 2 \times 5 = 10 \)
(5) Eccentricity: \( e = \frac{c}{a} = \frac{5\sqrt{3}}{10} = \frac{\sqrt{3}}{2} \)
(6) Length of latus rectum: \( \frac{2b^2}{a} = \frac{2 \times 25}{10} = 5 \)
In simple words: When the major axis is vertical, the foci and vertices are on the Y-axis. Identify 'a' and 'b' by finding the square roots of the denominators. Then compute 'c' and apply the standard formulas to get all the requested features of the ellipse.

Exam Tip: Remember to simplify square roots like \( \sqrt{75} \) into \( 5\sqrt{3} \) for a clearer presentation of results.

Question 5. \( \frac{x^2}{49}+\frac{y^2}{36} = 1 \)
Answer: Here, the denominator of \( \frac{x^2}{49} \) is larger than the denominator of \( \frac{y^2}{36} \), so the major axis is on the X-axis. Comparing the given equation with \( \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \), we get:
\( a^2 = 49 \), \( b^2 = 36 \)
\( a = 7 \), \( b = 6 \)
Now, calculate \( c \):
\( c = \sqrt{a^2-b^2} = \sqrt{49-36} = \sqrt{13} \)
Thus, we find the following properties:
(1) Foci: \( (\pm c, 0) = (\pm \sqrt{13}, 0) \)
(2) Vertices: \( (\pm a, 0) = (\pm 7, 0) \)
(3) Length of major axis: \( 2a = 2 \times 7 = 14 \)
(4) Length of minor axis: \( 2b = 2 \times 6 = 12 \)
(5) Eccentricity: \( e = \frac{c}{a} = \frac{\sqrt{13}}{7} \)
(6) Length of latus rectum: \( \frac{2b^2}{a} = \frac{2 \times 36}{7} = \frac{72}{7} \)
In simple words: Always determine 'a' and 'b' from the given ellipse equation. Then calculate 'c' using the fundamental relationship \( c^2 = a^2 - b^2 \). With these values, you can use the appropriate formulas to calculate the foci, vertices, lengths of axes, eccentricity, and latus rectum.

Exam Tip: Keep fractions in their simplest form for the latus rectum and eccentricity, but do not convert to decimals unless specified.

Question 6. \( \frac{x^2}{100}+\frac{y^2}{400} = 1 \)
Answer: Here, the denominator of \( \frac{y^2}{400} \) is larger than the denominator of \( \frac{x^2}{100} \), so the major axis is on the Y-axis. Comparing the given equation with \( \frac{x^2}{b^2}+\frac{y^2}{a^2} = 1 \), we get:
\( b^2 = 100 \), \( a^2 = 400 \)
\( b = 10 \), \( a = 20 \)
Now, calculate \( c \):
\( c = \sqrt{a^2-b^2} = \sqrt{400-100} = \sqrt{300} = 10\sqrt{3} \)
Thus, we find the following properties:
(1) Foci: \( (0, \pm c) = (0, \pm 10\sqrt{3}) \)
(2) Vertices: \( (0, \pm a) = (0, \pm 20) \)
(3) Length of major axis: \( 2a = 2 \times 20 = 40 \)
(4) Length of minor axis: \( 2b = 2 \times 10 = 20 \)
(5) Eccentricity: \( e = \frac{c}{a} = \frac{10\sqrt{3}}{20} = \frac{\sqrt{3}}{2} \)
(6) Length of latus rectum: \( \frac{2b^2}{a} = \frac{2 \times 100}{20} = 10 \)
In simple words: For a Y-axis oriented ellipse, identify 'a' and 'b' from the equation. Then, use these values to compute 'c'. With these key values, you can then find all the important features like foci, vertices, and lengths of axes by applying the relevant formulas.

Exam Tip: Remember that if the major axis is on the Y-axis, the coordinates of the foci and vertices will have zero for the x-component.

Question 7. \( 36x^2 + 4y^2 = 144 \)
Answer: First, convert the given equation into standard form by dividing by 144:
\( \frac{36x^2}{144} + \frac{4y^2}{144} = 1 \)
\( \frac{x^2}{4} + \frac{y^2}{36} = 1 \)
Here, the denominator of \( \frac{y^2}{36} \) is larger than the denominator of \( \frac{x^2}{4} \), so the major axis is on the Y-axis. Comparing the equation with \( \frac{x^2}{b^2}+\frac{y^2}{a^2} = 1 \), we get:
\( b^2 = 4 \), \( a^2 = 36 \)
\( b = 2 \), \( a = 6 \)
Now, calculate \( c \):
\( c = \sqrt{a^2-b^2} = \sqrt{36-4} = \sqrt{32} = 4\sqrt{2} \)
Thus, we find the following properties:
(1) Foci: \( (0, \pm c) = (0, \pm 4\sqrt{2}) \)
(2) Vertices: \( (0, \pm a) = (0, \pm 6) \)
(3) Length of major axis: \( 2a = 2 \times 6 = 12 \)
(4) Length of minor axis: \( 2b = 2 \times 2 = 4 \)
(5) Eccentricity: \( e = \frac{c}{a} = \frac{4\sqrt{2}}{6} = \frac{2\sqrt{2}}{3} \)
(6) Length of latus rectum: \( \frac{2b^2}{a} = \frac{2 \times 4}{6} = \frac{4}{3} \)
In simple words: Always begin by converting the general equation to its standard form, which means making the right side equal to 1. After that, identify whether the major axis is horizontal or vertical, and then proceed with calculating 'a', 'b', 'c', and all other ellipse properties using the standard formulas.

Exam Tip: The first step for equations not in standard form is always to divide by the constant on the right side to make it 1.

Question 8. \( 16x^2 + y^2 = 16 \)
Answer: First, convert the given equation into standard form by dividing by 16:
\( \frac{16x^2}{16} + \frac{y^2}{16} = 1 \)
\( \frac{x^2}{1} + \frac{y^2}{16} = 1 \)
Here, the denominator of \( \frac{y^2}{16} \) is larger than the denominator of \( \frac{x^2}{1} \), so the major axis is on the Y-axis. Comparing the equation with \( \frac{x^2}{b^2}+\frac{y^2}{a^2} = 1 \), we get:
\( b^2 = 1 \), \( a^2 = 16 \)
\( b = 1 \), \( a = 4 \)
Now, calculate \( c \):
\( c = \sqrt{a^2-b^2} = \sqrt{16-1} = \sqrt{15} \)
Thus, we find the following properties:
(1) Foci: \( (0, \pm c) = (0, \pm \sqrt{15}) \)
(2) Vertices: \( (0, \pm a) = (0, \pm 4) \)
(3) Length of major axis: \( 2a = 2 \times 4 = 8 \)
(4) Length of minor axis: \( 2b = 2 \times 1 = 2 \)
(5) Eccentricity: \( e = \frac{c}{a} = \frac{\sqrt{15}}{4} \)
(6) Length of latus rectum: \( \frac{2b^2}{a} = \frac{2 \times 1}{4} = \frac{1}{2} \)
In simple words: First, rewrite the given ellipse equation in standard form by ensuring the right side is 1. Then, identify 'a' and 'b' and determine if the major axis is horizontal or vertical. Finally, compute 'c' and apply the correct formulas to find all the requested characteristics of the ellipse.

Exam Tip: Be careful when identifying \(a^2\) and \(b^2\) after normalization, especially if one denominator is 1.

Question 9. \( 4x^2 + 9y^2 = 36 \)
Answer: First, convert the given equation into standard form by dividing by 36:
\( \frac{4x^2}{36} + \frac{9y^2}{36} = 1 \)
\( \frac{x^2}{9} + \frac{y^2}{4} = 1 \)
Here, the denominator of \( \frac{x^2}{9} \) is larger than the denominator of \( \frac{y^2}{4} \), so the major axis is on the X-axis. Comparing the equation with \( \frac{x^2}{a^2}+\frac{y^2}{b^2} = 1 \), we get:
\( a^2 = 9 \), \( b^2 = 4 \)
\( a = 3 \), \( b = 2 \)
Now, calculate \( c \):
\( c = \sqrt{a^2-b^2} = \sqrt{9-4} = \sqrt{5} \)
Thus, we find the following properties:
(1) Foci: \( (\pm c, 0) = (\pm \sqrt{5}, 0) \)
(2) Vertices: \( (\pm a, 0) = (\pm 3, 0) \)
(3) Length of major axis: \( 2a = 2 \times 3 = 6 \)
(4) Length of minor axis: \( 2b = 2 \times 2 = 4 \)
(5) Eccentricity: \( e = \frac{c}{a} = \frac{\sqrt{5}}{3} \)
(6) Length of latus rectum: \( \frac{2b^2}{a} = \frac{2 \times 4}{3} = \frac{8}{3} \)
In simple words: Start by simplifying the equation into its standard form. Determine the values of 'a' and 'b' and identify the orientation of the major axis. Next, calculate 'c' using the relationship \( c^2 = a^2 - b^2 \). Finally, use 'a', 'b', and 'c' to compute all the necessary ellipse characteristics.

Exam Tip: Be vigilant for potential arithmetic errors during the normalization step and while simplifying square roots.

Find the Equation of Each Ellipse Satisfying the Given Conditions in the Following Questions 10 to 20:

Question 10. Vertices \( (\pm5, 0) \), Foci \( (\pm4, 0) \)
Answer: Since the vertices are \( (\pm5, 0) \) and the foci are \( (\pm4, 0) \), the major axis is on the X-axis. Therefore, the equation of the ellipse will be of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
From the vertices, \( a = 5 \).
From the foci, \( c = 4 \).
We know the relationship \( c^2 = a^2 - b^2 \).
Substituting the values: \( 4^2 = 5^2 - b^2 \)
\( 16 = 25 - b^2 \)
\( b^2 = 25 - 16 \)
\( b^2 = 9 \)
So, the required equation of the ellipse is \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \).
In simple words: When vertices and foci are on the X-axis, the ellipse is horizontal. Use the vertex to find 'a' and the focus to find 'c'. Then, calculate \( b^2 \) using the formula \( c^2 = a^2 - b^2 \). Finally, substitute these \( a^2 \) and \( b^2 \) values into the standard horizontal ellipse equation.

Exam Tip: The orientation of the major axis (X or Y) is critical; it dictates the form of the ellipse equation to use.

Question 11. Vertices \( (0, \pm 13) \), Foci \( (0, \pm 5) \)
Answer: Since the vertices are \( (0, \pm 13) \) and the foci are \( (0, \pm 5) \), the major axis is on the Y-axis. Therefore, the equation of the ellipse will be of the form \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \).
From the vertices, \( a = 13 \).
From the foci, \( c = 5 \).
We know the relationship \( c^2 = a^2 - b^2 \).
Substituting the values: \( 5^2 = 13^2 - b^2 \)
\( 25 = 169 - b^2 \)
\( b^2 = 169 - 25 \)
\( b^2 = 144 \)
So, the required equation of the ellipse is \( \frac{x^2}{144} + \frac{y^2}{169} = 1 \).
In simple words: If the vertices and foci lie on the Y-axis, the ellipse is vertical. Identify 'a' from the vertex coordinates and 'c' from the focus coordinates. Then, use these values to find \( b^2 \) using the relation \( c^2 = a^2 - b^2 \). Lastly, plug \( a^2 \) and \( b^2 \) into the standard vertical ellipse equation.

Exam Tip: Remember that for a vertical ellipse, 'a' corresponds to the y-coordinate of the vertex and 'c' to the y-coordinate of the focus.

Question 12. Vertices \( (\pm6, 0) \), Foci \( (\pm4, 0) \)
Answer: Since the vertices are \( (\pm6, 0) \) and the foci are \( (\pm4, 0) \), the major axis is on the X-axis. Therefore, the equation of the ellipse will be of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
From the vertices, \( a = 6 \).
From the foci, \( c = 4 \).
We know the relationship \( c^2 = a^2 - b^2 \).
Substituting the values: \( 4^2 = 6^2 - b^2 \)
\( 16 = 36 - b^2 \)
\( b^2 = 36 - 16 \)
\( b^2 = 20 \)
So, the required equation of the ellipse is \( \frac{x^2}{36} + \frac{y^2}{20} = 1 \).
In simple words: When the vertices and foci are on the X-axis, the ellipse is horizontal. Extract 'a' from the vertices and 'c' from the foci. Use these to compute \( b^2 \) via \( c^2 = a^2 - b^2 \). Finally, write the ellipse equation using the obtained \( a^2 \) and \( b^2 \).

Exam Tip: Confirm that the given vertices and foci consistently indicate the same major axis orientation before proceeding with calculations.

Question 13. Endpoints of Major Axis \( (\pm3, 0) \), Endpoints of Minor Axis \( (0, \pm2) \)
Answer: Since the endpoints of the major axis are \( (\pm3, 0) \), the major axis is on the X-axis. This means \( a = 3 \).
Since the endpoints of the minor axis are \( (0, \pm2) \), the minor axis is on the Y-axis. This means \( b = 2 \).
Therefore, the equation of the ellipse will be of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
Substituting the values of \( a \) and \( b \):
\( \frac{x^2}{3^2} + \frac{y^2}{2^2} = 1 \)
So, the required equation of the ellipse is \( \frac{x^2}{9} + \frac{y^2}{4} = 1 \).
In simple words: If the major axis ends are on the X-axis and minor axis ends are on the Y-axis, the ellipse is horizontal. The x-coordinate of the major axis endpoint gives 'a', and the y-coordinate of the minor axis endpoint gives 'b'. Simply substitute these \( a^2 \) and \( b^2 \) values into the standard horizontal ellipse equation.

Exam Tip: For problems involving endpoints, 'a' is directly the length from the center to a major axis endpoint, and 'b' is the length from the center to a minor axis endpoint.

Question 14. Endpoints of Major Axis \( (0, \pm \sqrt{5}) \), Endpoints of Minor Axis \( (\pm 1, 0) \)
Answer: Since the endpoints of the major axis are \( (0, \pm \sqrt{5}) \), the major axis is on the Y-axis. This means \( a = \sqrt{5} \).
Since the endpoints of the minor axis are \( (\pm 1, 0) \), the minor axis is on the X-axis. This means \( b = 1 \).
Therefore, the equation of the ellipse will be of the form \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \).
Substituting the values of \( a \) and \( b \):
\( \frac{x^2}{1^2} + \frac{y^2}{(\sqrt{5})^2} = 1 \)
So, the required equation of the ellipse is \( \frac{x^2}{1} + \frac{y^2}{5} = 1 \).
In simple words: If the major axis ends are on the Y-axis and minor axis ends are on the X-axis, the ellipse is vertical. The y-coordinate of the major axis endpoint gives 'a', and the x-coordinate of the minor axis endpoint gives 'b'. Just substitute these \( a^2 \) and \( b^2 \) values into the standard vertical ellipse equation.

Exam Tip: Remember to square the values of 'a' and 'b' before substituting them into the ellipse equation.

Question 15. Length of Major Axis 26, Foci \( (\pm5, 0) \)
Answer: Given that the length of the major axis is 26, we have \( 2a = 26 \), which means \( a = 13 \).
Since the foci are \( (\pm5, 0) \), the major axis is on the X-axis. This means \( c = 5 \).
Therefore, the equation of the ellipse will be of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
We know the relationship \( c^2 = a^2 - b^2 \).
Substituting the values: \( 5^2 = 13^2 - b^2 \)
\( 25 = 169 - b^2 \)
\( b^2 = 169 - 25 \)
\( b^2 = 144 \)
So, the required equation of the ellipse is \( \frac{x^2}{169} + \frac{y^2}{144} = 1 \).
In simple words: The length of the major axis helps us find 'a'. The foci tell us 'c' and the orientation of the ellipse (X-axis here). With 'a' and 'c', we can calculate \( b^2 \). Once \( a^2 \) and \( b^2 \) are known, we can write the ellipse's equation.

Exam Tip: Distinguish between the length of the major axis (\(2a\)) and the value of 'a' itself. A common mistake is using \(2a\) directly instead of \(a\).

Question 16. Length of Minor Axis 16, Foci \( (0, \pm6) \)
Answer: Given that the length of the minor axis is 16, we have \( 2b = 16 \), which means \( b = 8 \).
Since the foci are \( (0, \pm6) \), the major axis is on the Y-axis. This means \( c = 6 \).
Therefore, the equation of the ellipse will be of the form \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \).
We know the relationship \( c^2 = a^2 - b^2 \).
Substituting the values: \( 6^2 = a^2 - 8^2 \)
\( 36 = a^2 - 64 \)
\( a^2 = 36 + 64 \)
\( a^2 = 100 \)
So, the required equation of the ellipse is \( \frac{x^2}{64} + \frac{y^2}{100} = 1 \).
In simple words: The length of the minor axis helps us find 'b'. The foci tell us 'c' and that the ellipse is vertical. With 'b' and 'c', we calculate \( a^2 \). Once \( a^2 \) and \( b^2 \) are known, we can write the ellipse's equation.

Exam Tip: For a vertical ellipse, the \(a^2\) value is always under the \(y^2\) term, and \(b^2\) is under the \(x^2\) term.

Question 17. Foci \( (\pm3, 0) \), \(a = 4\)
Answer: Since the foci are \( (\pm3, 0) \), the major axis is on the X-axis. This means \( c = 3 \).
Given \( a = 4 \).
Therefore, the equation of the ellipse will be of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
We know the relationship \( c^2 = a^2 - b^2 \).
Substituting the values: \( 3^2 = 4^2 - b^2 \)
\( 9 = 16 - b^2 \)
\( b^2 = 16 - 9 \)
\( b^2 = 7 \)
So, the required equation of the ellipse is \( \frac{x^2}{16} + \frac{y^2}{7} = 1 \).
In simple words: The foci provide 'c' and the ellipse's orientation (X-axis). Since 'a' is already given, we can directly find \( b^2 \) using the formula \( c^2 = a^2 - b^2 \). Then, substitute the values of \( a^2 \) and \( b^2 \) into the standard equation for a horizontal ellipse.

Exam Tip: When 'a' is given directly, it simplifies the initial steps, allowing direct calculation of \(b^2\).

Question 18. \( b = 3 \), \( c = 4 \), Center at origin and foci on X-axis.
Answer: Given \( b = 3 \) and \( c = 4 \).
Since the foci are on the X-axis, the major axis is on the X-axis. Therefore, the equation of the ellipse will be of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
We know the relationship \( c^2 = a^2 - b^2 \).
Substituting the values: \( 4^2 = a^2 - 3^2 \)
\( 16 = a^2 - 9 \)
\( a^2 = 16 + 9 \)
\( a^2 = 25 \)
So, the required equation of the ellipse is \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \).
In simple words: When 'b' and 'c' are given, and the foci are on the X-axis (meaning a horizontal ellipse), we can find \( a^2 \) using \( c^2 = a^2 - b^2 \). After finding \( a^2 \), substitute it along with the given \( b^2 \) into the standard horizontal ellipse equation.

Exam Tip: The center being at the origin is a standard assumption unless stated otherwise, allowing use of simplified standard ellipse equations.

Question 19. Center at origin, major axis on Y-axis, and passes through points \( (3, 2) \) and \( (1, 6) \).
Answer: Since the center is at the origin and the major axis is on the Y-axis, the equation of the ellipse will be of the form \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \).
The ellipse passes through the points \( (3, 2) \) and \( (1, 6) \).
Substituting \( (3, 2) \) into the equation:
\( \frac{3^2}{b^2} + \frac{2^2}{a^2} = 1 \implies \frac{9}{b^2} + \frac{4}{a^2} = 1 \) (Equation 1)
Substituting \( (1, 6) \) into the equation:
\( \frac{1^2}{b^2} + \frac{6^2}{a^2} = 1 \implies \frac{1}{b^2} + \frac{36}{a^2} = 1 \) (Equation 2)
Multiply Equation 2 by 9:
\( 9 \times (\frac{1}{b^2} + \frac{36}{a^2}) = 9 \times 1 \implies \frac{9}{b^2} + \frac{324}{a^2} = 9 \) (Equation 3)
Subtract Equation 1 from Equation 3:
\( (\frac{9}{b^2} + \frac{324}{a^2}) - (\frac{9}{b^2} + \frac{4}{a^2}) = 9 - 1 \)
\( \frac{324-4}{a^2} = 8 \)
\( \frac{320}{a^2} = 8 \)
\( 320 = 8a^2 \)
\( a^2 = \frac{320}{8} = 40 \)
Substitute \( a^2 = 40 \) into Equation 1:
\( \frac{9}{b^2} + \frac{4}{40} = 1 \)
\( \frac{9}{b^2} + \frac{1}{10} = 1 \)
\( \frac{9}{b^2} = 1 - \frac{1}{10} \)
\( \frac{9}{b^2} = \frac{9}{10} \)
\( b^2 = 10 \)
So, the required equation of the ellipse is \( \frac{x^2}{10} + \frac{y^2}{40} = 1 \).
In simple words: For an ellipse passing through two points with its major axis on the Y-axis, set up two equations by substituting each point's coordinates into the general equation. Solve these simultaneous equations for \( a^2 \) and \( b^2 \). Then, use these values to write the final ellipse equation.

Exam Tip: Solving simultaneous equations accurately is crucial when the ellipse passes through given points. Double-check your arithmetic steps.

Question 20. Major axis on X-axis and passes through points \( (4, 3) \) and \( (6, 2) \).
Answer: Since the major axis is on the X-axis, the equation of the ellipse will be of the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
The ellipse passes through the points \( (4, 3) \) and \( (6, 2) \).
Substituting \( (4, 3) \) into the equation:
\( \frac{4^2}{a^2} + \frac{3^2}{b^2} = 1 \implies \frac{16}{a^2} + \frac{9}{b^2} = 1 \) (Equation 1)
Substituting \( (6, 2) \) into the equation:
\( \frac{6^2}{a^2} + \frac{2^2}{b^2} = 1 \implies \frac{36}{a^2} + \frac{4}{b^2} = 1 \) (Equation 2)
To eliminate \( b^2 \), multiply Equation 1 by 4 and Equation 2 by 9:
Equation 1 \(\times\) 4: \( 4 \times (\frac{16}{a^2} + \frac{9}{b^2}) = 4 \times 1 \implies \frac{64}{a^2} + \frac{36}{b^2} = 4 \) (Equation 3)
Equation 2 \(\times\) 9: \( 9 \times (\frac{36}{a^2} + \frac{4}{b^2}) = 9 \times 1 \implies \frac{324}{a^2} + \frac{36}{b^2} = 9 \) (Equation 4)
Subtract Equation 3 from Equation 4:
\( (\frac{324}{a^2} + \frac{36}{b^2}) - (\frac{64}{a^2} + \frac{36}{b^2}) = 9 - 4 \)
\( \frac{324-64}{a^2} = 5 \)
\( \frac{260}{a^2} = 5 \)
\( 260 = 5a^2 \)
\( a^2 = \frac{260}{5} = 52 \)
Substitute \( a^2 = 52 \) into Equation 1:
\( \frac{16}{52} + \frac{9}{b^2} = 1 \)
\( \frac{4}{13} + \frac{9}{b^2} = 1 \)
\( \frac{9}{b^2} = 1 - \frac{4}{13} \)
\( \frac{9}{b^2} = \frac{13-4}{13} \)
\( \frac{9}{b^2} = \frac{9}{13} \)
\( b^2 = 13 \)
So, the required equation of the ellipse is \( \frac{x^2}{52} + \frac{y^2}{13} = 1 \).
This can also be written as \( x^2 + 4y^2 = 52 \) (by multiplying the equation by 52).
In simple words: For a horizontal ellipse passing through two points, set up two equations by substituting the coordinates. Multiply each equation to make one term (like \( \frac{1}{b^2} \)) eliminable. Subtract the equations to solve for \( a^2 \), then substitute \( a^2 \) back into one of the original equations to find \( b^2 \). Finally, write the ellipse equation using the calculated \( a^2 \) and \( b^2 \) values.

Exam Tip: Be mindful of algebraic manipulation when solving simultaneous equations, especially with fractions. Always cross-check the solution with the original given points.