OP Malhotra Class 11 Maths Solutions Chapter 24 Ellipse Chapter Test

Question 1. Find the eccentricity of the ellipse \( \frac{(x-3)^2}{8} + \frac{(y-4)^2}{6} = 1 \)
Answer: The given equation of the ellipse is \( \frac{(x-3)^2}{8} + \frac{(y-4)^2}{6} = 1 \).
Comparing this with the standard form \( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \), we find:
\( a^2 = 8 \)
\( b^2 = 6 \)
We use the formula that relates \( a^2 \), \( b^2 \), and eccentricity \( e \):
\( b^2 = a^2 (1 - e^2) \)
Substitute the values of \( a^2 \) and \( b^2 \):
\( 6 = 8 (1 - e^2) \)
Now, divide both sides by 8:
\( \frac{6}{8} = 1 - e^2 \)
\( \frac{3}{4} = 1 - e^2 \)
Rearrange to solve for \( e^2 \):
\( e^2 = 1 - \frac{3}{4} \)
\( e^2 = \frac{1}{4} \)
Take the square root to find \( e \). Since eccentricity must be positive:
\( e = \sqrt{\frac{1}{4}} \)
\( \implies \) \( e = \frac{1}{2} \)
The eccentricity of the given ellipse is \( \frac{1}{2} \). Eccentricity is a measure of how "stretched out" an ellipse is; a value of 0 means it's a perfect circle.
In simple words: We look at the numbers under the x and y terms in the ellipse equation. These give us \( a^2 \) and \( b^2 \). Then we use a special formula that links \( a^2 \), \( b^2 \), and eccentricity. We solve this formula to find the eccentricity, which tells us how round or flat the ellipse is.

🎯 Exam Tip: Always remember that \( a^2 \) is the larger of the two denominators for a horizontal ellipse, and \( b^2 \) is the smaller. For a vertical ellipse, \( a^2 \) is under the y-term and \( b^2 \) under the x-term.

Question 2. The distance between the foci of the ellipse \( 5x^2 + 9y^2 = 45 \) is
(a) 2
(b) 3
(c) 4
(d) 5
Answer: (c) 4
The given equation of the ellipse is \( 5x^2 + 9y^2 = 45 \).
To get it into standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we divide the entire equation by 45:
\( \frac{5x^2}{45} + \frac{9y^2}{45} = \frac{45}{45} \)
\( \implies \) \( \frac{x^2}{9} + \frac{y^2}{5} = 1 \)
By comparing this with the standard equation, we have:
\( a^2 = 9 \implies a = 3 \) (since \( a > 0 \))
\( b^2 = 5 \)
Next, we calculate the eccentricity \( e \) using the formula:
\( b^2 = a^2 (1 - e^2) \)
Substitute the values of \( a^2 \) and \( b^2 \):
\( 5 = 9 (1 - e^2) \)
Divide both sides by 9:
\( \frac{5}{9} = 1 - e^2 \)
Solve for \( e^2 \):
\( e^2 = 1 - \frac{5}{9} \)
\( e^2 = \frac{4}{9} \)
Take the square root for \( e \):
\( e = \sqrt{\frac{4}{9}} \)
\( \implies \) \( e = \frac{2}{3} \)
The distance between the foci of an ellipse is given by the formula \( 2ae \).
Substitute the values of \( a \) and \( e \):
Distance \( = 2 \times 3 \times \frac{2}{3} \)
Distance \( = 2 \times 2 \)
Distance \( = 4 \)
The foci are two fixed points inside an ellipse, and the sum of the distances from any point on the ellipse to these two foci is constant.
In simple words: First, we change the ellipse equation to its simple standard form to find values for \( a \) and \( b \). Then, we use these values to figure out the ellipse's eccentricity \( e \). Finally, we use a simple formula, \( 2ae \), to calculate the distance between the two special points called foci inside the ellipse.

🎯 Exam Tip: Always convert the given ellipse equation to its standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) or \( \frac{x^2}{b^2} + \frac{y^2}{a^2} = 1 \) before finding \( a^2 \) and \( b^2 \). This helps avoid mistakes in identifying \( a \) and \( b \).

Question 3. Find the equation of an ellipse whose latus rectum is 8 and eccentricity is \( \frac{1}{3} \)
Answer: We are given two pieces of information about the ellipse:
1. The length of the latus rectum is 8.
2. The eccentricity \( e \) is \( \frac{1}{3} \).
The formula for the length of the latus rectum is \( \frac{2b^2}{a} \). So:
\( \frac{2b^2}{a} = 8 \)
\( \implies \) \( 2b^2 = 8a \)
\( \implies \) \( b^2 = 4a \) (Equation 1)
We also know the relationship between \( a^2 \), \( b^2 \), and \( e^2 \):
\( b^2 = a^2 (1 - e^2) \)
Substitute the value of \( e = \frac{1}{3} \):
\( b^2 = a^2 (1 - (\frac{1}{3})^2) \)
\( b^2 = a^2 (1 - \frac{1}{9}) \)
\( b^2 = a^2 (\frac{9-1}{9}) \)
\( b^2 = \frac{8}{9}a^2 \) (Equation 2)
Now, we can substitute \( b^2 = 4a \) from Equation 1 into Equation 2:
\( 4a = \frac{8}{9}a^2 \)
Since \( a \) cannot be zero for an ellipse, we can divide both sides by \( a \):
\( 4 = \frac{8}{9}a \)
To find \( a \), multiply by \( \frac{9}{8} \):
\( a = 4 \times \frac{9}{8} \)
\( a = \frac{36}{8} \)
\( \implies \) \( a = \frac{9}{2} \)
Now that we have \( a \), we can find \( b^2 \) using Equation 1:
\( b^2 = 4a \)
\( b^2 = 4 \times \frac{9}{2} \)
\( b^2 = 18 \)
The standard equation of an ellipse centered at the origin is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
Substitute \( a^2 = (\frac{9}{2})^2 = \frac{81}{4} \) and \( b^2 = 18 \):
\( \frac{x^2}{\frac{81}{4}} + \frac{y^2}{18} = 1 \)
\( \implies \) \( \frac{4x^2}{81} + \frac{y^2}{18} = 1 \)
To remove the denominators, find the least common multiple of 81 and 18, which is 162. Multiply the entire equation by 162:
\( 162 \left( \frac{4x^2}{81} \right) + 162 \left( \frac{y^2}{18} \right) = 162 \times 1 \)
\( \implies \) \( 2 \times 4x^2 + 9y^2 = 162 \)
\( \implies \) \( 8x^2 + 9y^2 = 162 \)
The latus rectum is a chord of the ellipse that passes through a focus and is perpendicular to the major axis.
In simple words: We used the given latus rectum length and eccentricity to set up two equations involving \( a \) and \( b \). By solving these equations together, we found the values for \( a \) and \( b \). Once we had \( a \) and \( b \), we could write the full equation of the ellipse.

🎯 Exam Tip: Remember the two key formulas for ellipses: \( b^2 = a^2(1-e^2) \) and the length of the latus rectum \( \frac{2b^2}{a} \). You will often need to solve them simultaneously to find \( a \) and \( b \).

Question 4. Find the equation of the ellipse whose foci are at \( (-2, 4) \) and \( (4, 4) \) and major and minor axis are 10 and 8 respectively. Also, find the eccentricity of the ellipse.
Answer: We are given the following information:
Foci: \( F_1(-2, 4) \) and \( F_2(4, 4) \).
Length of major axis \( 2a = 10 \).
Length of minor axis \( 2b = 8 \).
First, let's find \( a \) and \( b \):
From \( 2a = 10 \implies a = 5 \). So, \( a^2 = 25 \).
From \( 2b = 8 \implies b = 4 \). So, \( b^2 = 16 \).
Next, we find the distance between the foci. Using the distance formula between \( (-2, 4) \) and \( (4, 4) \):
Distance between foci \( = \sqrt{(4 - (-2))^2 + (4 - 4)^2} \)
\( = \sqrt{(4+2)^2 + 0^2} \)
\( = \sqrt{6^2} \)
\( = 6 \)
The distance between foci is also given by \( 2ae \). So:
\( 2ae = 6 \)
Since \( a = 5 \):
\( 2(5)e = 6 \)
\( 10e = 6 \)
\( \implies \) \( e = \frac{6}{10} = \frac{3}{5} \)
This is the eccentricity of the ellipse.
Now, let's find the center of the ellipse. The center is the midpoint of the segment connecting the two foci:
Center \( (h, k) = \left( \frac{-2+4}{2}, \frac{4+4}{2} \right) \)
\( (h, k) = \left( \frac{2}{2}, \frac{8}{2} \right) \)
\( \implies \) Center \( (h, k) = (1, 4) \)
Since the y-coordinates of the foci are the same (4), the major axis is parallel to the x-axis. The equation of an ellipse with its major axis parallel to the x-axis is:
\( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)
Substitute \( h=1, k=4, a^2=25, b^2=16 \):
\( \frac{(x-1)^2}{25} + \frac{(y-4)^2}{16} = 1 \)
To eliminate the denominators, multiply the entire equation by the LCM of 25 and 16, which is 400:
\( 400 \left( \frac{(x-1)^2}{25} \right) + 400 \left( \frac{(y-4)^2}{16} \right) = 400 \times 1 \)
\( \implies \) \( 16(x-1)^2 + 25(y-4)^2 = 400 \)
The center of an ellipse is the midpoint of both the major axis and the minor axis.
In simple words: We used the given lengths of the major and minor axes to find \( a \) and \( b \). The distance between the foci helped us calculate the eccentricity. By finding the midpoint of the foci, we located the center of the ellipse. Since the foci were on a horizontal line, we knew the major axis was horizontal. Then, we put all these values (\( a, b \), and the center) into the standard equation for an ellipse.

🎯 Exam Tip: When foci have the same y-coordinate, the major axis is horizontal. If they have the same x-coordinate, the major axis is vertical. This helps determine which standard equation form to use and which value corresponds to \( a^2 \) or \( b^2 \).

Question 5. Find the equation of the ellipse whose eccentricity is \( \frac { 1 }{ 2 } \) and whose foci are at the points \( (\pm 2, 0) \).
Answer: We are given:
Eccentricity \( e = \frac{1}{2} \).
Foci are at \( (\pm 2, 0) \).
For an ellipse with foci at \( (\pm ae, 0) \), we have \( ae = 2 \).
Substitute the value of \( e \):
\( a \left( \frac{1}{2} \right) = 2 \)
\( \implies \) \( a = 4 \)
So, \( a^2 = 4^2 = 16 \).
Now we use the relationship \( b^2 = a^2 (1 - e^2) \) to find \( b^2 \):
\( b^2 = 16 \left( 1 - \left( \frac{1}{2} \right)^2 \right) \)
\( b^2 = 16 \left( 1 - \frac{1}{4} \right) \)
\( b^2 = 16 \left( \frac{4-1}{4} \right) \)
\( b^2 = 16 \left( \frac{3}{4} \right) \)
\( \implies \) \( b^2 = 12 \)
Since the foci are on the x-axis and the center is at the origin \( (0,0) \), the equation of the ellipse is in the form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
Substitute \( a^2 = 16 \) and \( b^2 = 12 \):
\( \frac{x^2}{16} + \frac{y^2}{12} = 1 \)
When the foci are on the x-axis and the center is at the origin, the major axis of the ellipse lies along the x-axis.
In simple words: We used the given eccentricity and the position of the foci to find the values of \( a \) and \( b^2 \). The foci tell us that the ellipse is centered at the origin and stretched along the x-axis. Once we had \( a^2 \) and \( b^2 \), we just plugged them into the standard equation for an ellipse.

🎯 Exam Tip: If the foci are \( (\pm c, 0) \) and the center is at the origin, then \( ae = c \). Similarly, if they are \( (0, \pm c) \), then \( ae = c \). This shortcut helps quickly find \( a \) if \( e \) is known.

Question 6. Find the equation of the ellipse whose centre is the origin, length of major axis \( \frac{9}{2} \) and eccentricity \( \frac{1}{\sqrt{3}} \) where the major axis is the horizontal axis.
Answer: We are given:
Centre: Origin \( (0,0) \).
Length of major axis \( = \frac{9}{2} \).
Eccentricity \( e = \frac{1}{\sqrt{3}} \).
Major axis is horizontal.
For an ellipse with a horizontal major axis, the standard equation is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
The length of the major axis is \( 2a \). So:
\( 2a = \frac{9}{2} \)
\( \implies \) \( a = \frac{9}{4} \)
Thus, \( a^2 = \left( \frac{9}{4} \right)^2 = \frac{81}{16} \).
Now we find \( b^2 \) using the eccentricity formula:
\( b^2 = a^2 (1 - e^2) \)
Substitute \( a^2 = \frac{81}{16} \) and \( e = \frac{1}{\sqrt{3}} \):
\( b^2 = \frac{81}{16} \left( 1 - \left( \frac{1}{\sqrt{3}} \right)^2 \right) \)
\( b^2 = \frac{81}{16} \left( 1 - \frac{1}{3} \right) \)
\( b^2 = \frac{81}{16} \left( \frac{3-1}{3} \right) \)
\( b^2 = \frac{81}{16} \left( \frac{2}{3} \right) \)
\( b^2 = \frac{27 \times 3}{16} \times \frac{2}{3} \)
\( \implies \) \( b^2 = \frac{27 \times 2}{16} = \frac{27}{8} \)
Now, substitute \( a^2 = \frac{81}{16} \) and \( b^2 = \frac{27}{8} \) into the standard equation of the ellipse:
\( \frac{x^2}{\frac{81}{16}} + \frac{y^2}{\frac{27}{8}} = 1 \)
\( \implies \) \( \frac{16x^2}{81} + \frac{8y^2}{27} = 1 \)
To eliminate the denominators, multiply the entire equation by the LCM of 81 and 27, which is 81:
\( 81 \left( \frac{16x^2}{81} \right) + 81 \left( \frac{8y^2}{27} \right) = 81 \times 1 \)
\( \implies \) \( 16x^2 + 3 \times 8y^2 = 81 \)
\( \implies \) \( 16x^2 + 24y^2 = 81 \)
The major axis is the longest diameter of the ellipse, while the minor axis is the shortest, and they are perpendicular to each other.
In simple words: We are told the ellipse is centered at the origin and stretched horizontally. We used the length of the major axis to find \( a \), and then used \( a \) and the given eccentricity to find \( b^2 \). With \( a^2 \) and \( b^2 \), we wrote the equation of the ellipse in its standard form.

🎯 Exam Tip: Always pay attention to whether the major axis is horizontal or vertical, as this determines which term \( a^2 \) goes under (x-term for horizontal, y-term for vertical) in the standard equation.

Question 7. Find the equation of the ellipse whose minor axis is 4 and which has a distance of 6 units between foci.
Answer: We are given:
Length of minor axis \( 2b = 4 \).
Distance between foci \( 2ae = 6 \).
From the length of the minor axis:
\( 2b = 4 \implies b = 2 \). So, \( b^2 = 4 \).
From the distance between foci:
\( 2ae = 6 \implies ae = 3 \). So, \( (ae)^2 = 9 \).
We know the relationship for an ellipse: \( b^2 = a^2 - (ae)^2 \). This formula directly uses the distance from the center to a focus and the semi-minor axis length.
Substitute the values we found:
\( 4 = a^2 - 9 \)
Solve for \( a^2 \):
\( a^2 = 4 + 9 \)
\( \implies \) \( a^2 = 13 \)
Assuming the ellipse is centered at the origin and its major axis is along the x-axis (since no other information is given, this is the simplest standard form), the equation is:
\( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \)
Substitute \( a^2 = 13 \) and \( b^2 = 4 \):
\( \frac{x^2}{13} + \frac{y^2}{4} = 1 \)
The major and minor axes are symmetrical lines that pass through the center of the ellipse, and their lengths determine its shape.
In simple words: We used the length of the minor axis to find \( b^2 \) and the distance between the foci to find \( (ae)^2 \). Then, we used a special formula relating \( b^2 \), \( a^2 \), and \( (ae)^2 \) to figure out \( a^2 \). Finally, we put these \( a^2 \) and \( b^2 \) values into the simple standard equation of an ellipse.

🎯 Exam Tip: The formula \( b^2 = a^2 - (ae)^2 \) (or \( b^2 = a^2 - c^2 \) where \( c=ae \)) is a very useful shortcut to find \( a^2 \) or \( b^2 \) when you have the other two values. It comes directly from the definition of eccentricity.

Question 8. Find the equation of the ellipse whose centre is at \( (4, -1) \), focus at \( (1, -1) \) and given that it passes through \( (8, 0) \).
Answer: We are given:
Center \( C(h, k) = (4, -1) \).
One focus \( S_1(1, -1) \).
The ellipse passes through the point \( P(8, 0) \).
Since the y-coordinates of the center and the focus are the same \( (-1) \), the major axis of the ellipse is parallel to the x-axis. The distance from the center to a focus is \( ae \):
\( ae = |h - x_{focus}| = |4 - 1| = 3 \). So, \( (ae)^2 = 9 \).
For an ellipse with its major axis parallel to the x-axis and center \( (h,k) \), the equation is:
\( \frac{(x-h)^2}{a^2} + \frac{(y-k)^2}{b^2} = 1 \)
Substitute the center \( (h,k) = (4, -1) \):
\( \frac{(x-4)^2}{a^2} + \frac{(y-(-1))^2}{b^2} = 1 \)
\( \implies \) \( \frac{(x-4)^2}{a^2} + \frac{(y+1)^2}{b^2} = 1 \)
We also know the relationship \( (ae)^2 = a^2 - b^2 \). Using \( (ae)^2 = 9 \):
\( 9 = a^2 - b^2 \)
\( \implies \) \( b^2 = a^2 - 9 \)
Substitute this expression for \( b^2 \) into the ellipse equation:
\( \frac{(x-4)^2}{a^2} + \frac{(y+1)^2}{a^2-9} = 1 \) (Equation 1)
The ellipse passes through the point \( (8, 0) \). Substitute \( x=8 \) and \( y=0 \) into Equation 1:
\( \frac{(8-4)^2}{a^2} + \frac{(0+1)^2}{a^2-9} = 1 \)
\( \implies \) \( \frac{4^2}{a^2} + \frac{1^2}{a^2-9} = 1 \)
\( \implies \) \( \frac{16}{a^2} + \frac{1}{a^2-9} = 1 \)
Multiply the entire equation by \( a^2(a^2-9) \) to clear the denominators:
\( 16(a^2-9) + a^2 = a^2(a^2-9) \)
\( 16a^2 - 144 + a^2 = a^4 - 9a^2 \)
\( 17a^2 - 144 = a^4 - 9a^2 \)
Rearrange the terms to form a quadratic equation in \( a^2 \):
\( a^4 - 26a^2 + 144 = 0 \)
Let \( A = a^2 \). The equation becomes:
\( A^2 - 26A + 144 = 0 \)
Factor this quadratic equation:
\( A^2 - 18A - 8A + 144 = 0 \)
\( A(A-18) - 8(A-18) = 0 \)
\( (A-18)(A-8) = 0 \)
So, \( A = 18 \) or \( A = 8 \). This means \( a^2 = 18 \) or \( a^2 = 8 \).
Now we check which value of \( a^2 \) is valid by finding \( b^2 = a^2 - 9 \):
Case 1: If \( a^2 = 8 \), then \( b^2 = 8 - 9 = -1 \). This is not possible because \( b^2 \) must be positive.
Case 2: If \( a^2 = 18 \), then \( b^2 = 18 - 9 = 9 \). This is a valid value for \( b^2 \).
So, we have \( a^2 = 18 \) and \( b^2 = 9 \).
Substitute these values back into the ellipse equation:
\( \frac{(x-4)^2}{18} + \frac{(y+1)^2}{9} = 1 \)
Multiply by the LCM of 18 and 9, which is 18:
\( 18 \left( \frac{(x-4)^2}{18} \right) + 18 \left( \frac{(y+1)^2}{9} \right) = 18 \times 1 \)
\( \implies \) \( (x-4)^2 + 2(y+1)^2 = 18 \)
An ellipse can be thought of as a conic section formed by intersecting a cone with a plane that is not parallel to the base, side, or top.
In simple words: We used the center and focus points to figure out that the ellipse is horizontal and to find \( ae \). This allowed us to write a general equation with \( a^2 \) as the unknown. We then plugged in the coordinates of the point the ellipse passes through to form an equation for \( a^2 \). Solving this equation gave us \( a^2 \), and from there we found \( b^2 \), allowing us to write the final equation of the ellipse.

🎯 Exam Tip: When the ellipse passes through a specific point, substitute that point's coordinates into the general ellipse equation to create an additional condition. This often leads to solving a quadratic equation for \( a^2 \) or \( b^2 \), so always check for valid positive solutions.

Question 9. Find the coordinates of the vertices and the foci and the length of the latus rectum of the ellipse \( 9x^2 + 25y^2 = 225 \).
Answer: The given equation of the ellipse is \( 9x^2 + 25y^2 = 225 \).
To convert it to the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), divide the entire equation by 225:
\( \frac{9x^2}{225} + \frac{25y^2}{225} = \frac{225}{225} \)
\( \implies \) \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \)
By comparing this with the standard equation of a horizontal ellipse centered at the origin, \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we get:
\( a^2 = 25 \implies a = 5 \) (since \( a > 0 \))
\( b^2 = 9 \implies b = 3 \) (since \( b > 0 \))
Now we can find the required properties:
1. **Coordinates of the Vertices:**
For a horizontal ellipse centered at the origin, the vertices are at \( (\pm a, 0) \).
\( \implies \) Vertices are \( (\pm 5, 0) \).
2. **Coordinates of the Foci:**
First, we need to find the eccentricity \( e \). We use the formula \( b^2 = a^2 (1 - e^2) \):
\( 9 = 25 (1 - e^2) \)
\( \implies \) \( \frac{9}{25} = 1 - e^2 \)
\( \implies \) \( e^2 = 1 - \frac{9}{25} \)
\( e^2 = \frac{25-9}{25} = \frac{16}{25} \)
\( \implies \) \( e = \sqrt{\frac{16}{25}} = \frac{4}{5} \)
For a horizontal ellipse centered at the origin, the foci are at \( (\pm ae, 0) \).
\( ae = 5 \times \frac{4}{5} = 4 \)
\( \implies \) Foci are \( (\pm 4, 0) \).
3. **Length of the Latus Rectum:**
The formula for the length of the latus rectum is \( \frac{2b^2}{a} \).
Length of latus rectum \( = \frac{2 \times 9}{5} = \frac{18}{5} \) units.
The latus rectum is an important feature for sketching the shape and curvature of the ellipse near its foci.
In simple words: First, we change the given equation into its standard form to easily find \( a^2 \) and \( b^2 \), which gives us \( a \) and \( b \). Then, we use \( a \) to find the vertices. To find the foci, we first need to calculate the eccentricity \( e \) using \( a \) and \( b \). Once we have \( e \), we can find the foci. Finally, we apply a specific formula to calculate the length of the latus rectum using \( a \) and \( b \).

🎯 Exam Tip: Remember to clearly list all the requested components (vertices, foci, latus rectum) and ensure you use the correct formulas for a horizontal or vertical ellipse as determined by \( a^2 \) and \( b^2 \).

Question 10. Find the eccentricity and the equations of the directrices of the ellipse \( 7x^2 + 16y^2 = 112 \).
Answer: The given equation of the ellipse is \( 7x^2 + 16y^2 = 112 \).
To convert it to the standard form \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), divide the entire equation by 112:
\( \frac{7x^2}{112} + \frac{16y^2}{112} = \frac{112}{112} \)
\( \implies \) \( \frac{x^2}{16} + \frac{y^2}{7} = 1 \)
By comparing this with the standard equation of a horizontal ellipse centered at the origin, \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), we get:
\( a^2 = 16 \implies a = 4 \) (since \( a > 0 \))
\( b^2 = 7 \)
Now, let's find the eccentricity \( e \):
We use the formula \( b^2 = a^2 (1 - e^2) \):
\( 7 = 16 (1 - e^2) \)
\( \implies \) \( \frac{7}{16} = 1 - e^2 \)
\( \implies \) \( e^2 = 1 - \frac{7}{16} \)
\( e^2 = \frac{16-7}{16} = \frac{9}{16} \)
\( \implies \) \( e = \sqrt{\frac{9}{16}} = \frac{3}{4} \)
Next, let's find the equations of the directrices:
For a horizontal ellipse centered at the origin, the equations of the directrices are \( x = \pm \frac{a}{e} \).
Substitute the values of \( a \) and \( e \):
\( \frac{a}{e} = \frac{4}{\frac{3}{4}} \)
\( = 4 \times \frac{4}{3} \)
\( = \frac{16}{3} \)
\( \implies \) The equations of the directrices are \( x = \pm \frac{16}{3} \).
These can also be written as \( 3x = 16 \) and \( 3x = -16 \), or \( 3x - 16 = 0 \) and \( 3x + 16 = 0 \).
The directrices are fixed lines used in the definition of an ellipse, where for any point on the ellipse, the ratio of its distance from a focus to its distance from a directrix is equal to the eccentricity.
In simple words: First, we rewrite the ellipse equation into a simpler standard form to find \( a^2 \) and \( b^2 \). Then, we use these values to calculate the eccentricity \( e \), which tells us how oval the ellipse is. Finally, we use \( a \) and \( e \) to find the equations of the directrices, which are two special lines outside the ellipse.

🎯 Exam Tip: The equations of directrices for a horizontal ellipse are \( x = \pm \frac{a}{e} \), and for a vertical ellipse are \( y = \pm \frac{a}{e} \). Make sure to distinguish between these based on the orientation of the major axis.