OP Malhotra Class 11 Maths Solutions Chapter 25 Hyperbola Chapter Test

Question 1. Find the eccentricity and the coordinate of foci of the hyperbola \( 25x^2 - 9y^2 = 225 \).
Answer: First, we write the given equation of the hyperbola in its standard form. The equation is \( 25x^2 - 9y^2 = 225 \).
We divide by 225 to get:
\( \frac{25x^2}{225} - \frac{9y^2}{225} = \frac{225}{225} \)
\( \implies \frac{x^2}{9} - \frac{y^2}{25} = 1 \)
Now, we compare this with the standard equation of a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
From this, we find that \( a^2 = 9 \), so \( a = 3 \).
We also find that \( b^2 = 25 \), so \( b = 5 \).
Next, we find the eccentricity (e) using the formula \( b^2 = a^2(e^2 - 1) \).
Substitute the values of \( a^2 \) and \( b^2 \):
\( 25 = 9(e^2 - 1) \)
Divide both sides by 9:
\( \frac{25}{9} = e^2 - 1 \)
Add 1 to both sides:
\( \frac{25}{9} + 1 = e^2 \)
\( \frac{25+9}{9} = e^2 \)
\( \frac{34}{9} = e^2 \)
Take the square root of both sides (since \( e > 0 \)):
\( e = \sqrt{\frac{34}{9}} = \frac{\sqrt{34}}{3} \)
So, the eccentricity of the hyperbola is \( \frac{\sqrt{34}}{3} \).
Finally, we find the coordinates of the foci. For a hyperbola with its transverse axis along the x-axis, the foci are at \( (\pm ae, 0) \).
We calculate \( ae \):
\( ae = 3 \times \frac{\sqrt{34}}{3} = \sqrt{34} \)
Therefore, the coordinates of the foci are \( (\pm \sqrt{34}, 0) \). The eccentricity tells us how "stretched" the hyperbola is, and the foci are key points for its definition.
In simple words: First, change the hyperbola's equation into its simple form to find 'a' and 'b'. Then, use a special formula to calculate 'e', which is called eccentricity. After that, use 'a' and 'e' to find the positions of the two focus points, which are called foci.

🎯 Exam Tip: Always convert the hyperbola equation to its standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) or \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \) before finding eccentricity and foci. Remember the correct formula for eccentricity \( b^2 = a^2(e^2 - 1) \) and the foci coordinates \( (\pm ae, 0) \) or \( (0, \pm ae) \) based on the transverse axis.

Question 2. Find the value (s) of k so that the line \( 2x + y + k = 0 \) may touch the hyperbola \( 3x^2 - y^2 = 3 \).
Answer: We are given the equation of the hyperbola as \( 3x^2 - y^2 = 3 \).
To use the tangency condition, we first convert this to the standard form of a hyperbola:
Divide by 3: \( \frac{3x^2}{3} - \frac{y^2}{3} = \frac{3}{3} \)
\( \implies \frac{x^2}{1} - \frac{y^2}{3} = 1 \)
By comparing this with the standard equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), we get:
\( a^2 = 1 \)
\( b^2 = 3 \)
The given equation of the line is \( 2x + y + k = 0 \).
We can rewrite this in the slope-intercept form \( y = mx + c \):
\( y = -2x - k \)
So, for this line, the slope \( m = -2 \) and the y-intercept \( c = -k \).
For a line \( y = mx + c \) to touch a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), the tangency condition is \( c^2 = a^2m^2 - b^2 \). This condition ensures that the line intersects the hyperbola at exactly one point.
Now, we substitute the values of \( a^2, b^2, m \), and \( c \) into the tangency condition:
\( (-k)^2 = (1)(-2)^2 - 3 \)
\( k^2 = 1(4) - 3 \)
\( k^2 = 4 - 3 \)
\( k^2 = 1 \)
Take the square root of both sides:
\( k = \pm 1 \)
Thus, the possible values of k for which the line touches the hyperbola are \( 1 \) and \( -1 \).
In simple words: First, put the hyperbola equation into its basic form to find its special numbers 'a-squared' and 'b-squared'. Then, make the line equation look like 'y equals mx plus c'. Finally, use a rule that says 'c-squared' must equal 'a-squared times m-squared minus b-squared' to find the possible values for 'k'.

🎯 Exam Tip: Remember the tangency condition \( c^2 = a^2m^2 - b^2 \) for a line \( y = mx + c \) to touch a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Incorrectly identifying \( a^2 \) or \( b^2 \) from the hyperbola's equation is a common mistake.

Question 3. From the following information, find the equation of the hyperbola and the equation of the transverse axis. Focus (-2, 1), Directrix : \( 2x - 3y + 1 = 0 \), \( e = \frac{2}{\sqrt{3}} \)
Answer: Let \( P(x, y) \) be any point on the hyperbola. According to the definition of a conic section, the distance from any point on the conic to the focus (PF) is equal to 'e' times the perpendicular distance from that point to the directrix (PM). This definition, \( PF = e \cdot PM \), is fundamental to understanding conic sections.
Given:
Focus \( F = (-2, 1) \)
Directrix equation: \( 2x - 3y + 1 = 0 \)
Eccentricity \( e = \frac{2}{\sqrt{3}} \)
Using the distance formula for PF and PM:
\( PF = \sqrt{(x - (-2))^2 + (y - 1)^2} = \sqrt{(x + 2)^2 + (y - 1)^2} \)
\( PM = \frac{|2x - 3y + 1|}{\sqrt{2^2 + (-3)^2}} = \frac{|2x - 3y + 1|}{\sqrt{4 + 9}} = \frac{|2x - 3y + 1|}{\sqrt{13}} \)
Now, apply the definition \( PF = e \cdot PM \):
\( \sqrt{(x + 2)^2 + (y - 1)^2} = \frac{2}{\sqrt{3}} \cdot \frac{|2x - 3y + 1|}{\sqrt{13}} \)
To remove the square root and absolute value, we square both sides:
\( (x + 2)^2 + (y - 1)^2 = \left(\frac{2}{\sqrt{3}}\right)^2 \cdot \left(\frac{2x - 3y + 1}{\sqrt{13}}\right)^2 \)
\( (x + 2)^2 + (y - 1)^2 = \frac{4}{3} \cdot \frac{(2x - 3y + 1)^2}{13} \)
\( (x + 2)^2 + (y - 1)^2 = \frac{4}{39} (2x - 3y + 1)^2 \)
Multiply both sides by 39:
\( 39[(x + 2)^2 + (y - 1)^2] = 4(2x - 3y + 1)^2 \)
Expand both sides:
\( 39[x^2 + 4x + 4 + y^2 - 2y + 1] = 4[ (2x)^2 + (-3y)^2 + 1^2 + 2(2x)(-3y) + 2(-3y)(1) + 2(1)(2x) ] \)
\( 39[x^2 + y^2 + 4x - 2y + 5] = 4[4x^2 + 9y^2 + 1 - 12xy - 6y + 4x] \)
\( 39x^2 + 39y^2 + 156x - 78y + 195 = 16x^2 + 36y^2 + 4 - 48xy - 24y + 16x \)
Rearrange all terms to one side to get the equation of the hyperbola:
\( (39x^2 - 16x^2) + (39y^2 - 36y^2) + 48xy + (156x - 16x) + (-78y + 24y) + (195 - 4) = 0 \)
\( 23x^2 + 3y^2 + 48xy + 140x - 54y + 191 = 0 \)
This is the required equation of the hyperbola.

Now, for the equation of the transverse axis:
The transverse axis of a hyperbola is a line that passes through the focus and is perpendicular to the directrix.
The directrix equation is \( 2x - 3y + 1 = 0 \). The slope of this line is \( m_D = -\frac{2}{-3} = \frac{2}{3} \).
Since the transverse axis is perpendicular to the directrix, its slope \( m_{TA} \) will be the negative reciprocal of \( m_D \).
\( m_{TA} = -\frac{1}{m_D} = -\frac{1}{2/3} = -\frac{3}{2} \)
The transverse axis passes through the focus \( (-2, 1) \). We use the point-slope form of a line \( y - y_1 = m(x - x_1) \):
\( y - 1 = -\frac{3}{2}(x - (-2)) \)
\( y - 1 = -\frac{3}{2}(x + 2) \)
Multiply by 2:
\( 2(y - 1) = -3(x + 2) \)
\( 2y - 2 = -3x - 6 \)
Rearrange into the general form \( Ax + By + C = 0 \):
\( 3x + 2y + 4 = 0 \)
This is the required equation of the transverse axis.

In simple words: First, use the definition of a conic section, which says the distance from any point on the curve to the focus is 'e' times the distance to the directrix. Write this as a math equation and square both sides to remove square roots. Expand and simplify to get the hyperbola's equation. Then, find the line for the transverse axis by making it pass through the focus and be at a right angle to the directrix.

🎯 Exam Tip: The definition \( PF = e \cdot PM \) is crucial for finding the equation of a conic section when given the focus, directrix, and eccentricity. Remember that the transverse axis is perpendicular to the directrix and passes through the focus.

Question 4. Find the equation of the hyperbola whose eccentricity is \( \sqrt{5} \) and the sum of whose semi-axes is 9.
Answer: Let the equation of the hyperbola be in the standard form \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \). Here, 'a' is the length of the semi-transverse axis and 'b' is the length of the semi-conjugate axis.
We are given the eccentricity \( e = \sqrt{5} \).
We are also given that the sum of the semi-axes is 9, so \( a + b = 9 \).
From this, we can write \( b = 9 - a \).
The relationship between \( a, b \), and \( e \) for a hyperbola is \( b^2 = a^2(e^2 - 1) \). This formula connects the lengths of the axes to the eccentricity.
Substitute the values of \( b \) and \( e \) into this formula:
\( (9 - a)^2 = a^2((\sqrt{5})^2 - 1) \)
\( (9 - a)^2 = a^2(5 - 1) \)
\( (9 - a)^2 = 4a^2 \)
Now, expand the left side and move all terms to one side:
\( 81 - 18a + a^2 = 4a^2 \)
\( 0 = 4a^2 - a^2 + 18a - 81 \)
\( 0 = 3a^2 + 18a - 81 \)
Divide the entire equation by 3 to simplify:
\( a^2 + 6a - 27 = 0 \)
We can factor this quadratic equation:
\( (a + 9)(a - 3) = 0 \)
This gives two possible values for \( a \): \( a = -9 \) or \( a = 3 \).
Since 'a' represents a length, it must be a positive value. So, \( a = 3 \).
Now, we find 'b' using \( b = 9 - a \):
\( b = 9 - 3 = 6 \)
So, we have \( a^2 = 3^2 = 9 \) and \( b^2 = 6^2 = 36 \).
Substitute these values back into the standard equation of the hyperbola:
\( \frac{x^2}{9} - \frac{y^2}{36} = 1 \)
This is the required equation of the hyperbola.
In simple words: Use the given eccentricity and the sum of the semi-axes to set up equations. Solve these equations to find the lengths 'a' and 'b' of the semi-axes. Remember that 'a' must be a positive number. Once you have 'a' and 'b', put them into the standard hyperbola equation.

🎯 Exam Tip: When solving for lengths like 'a' or 'b', always choose the positive value since lengths cannot be negative. Carefully use the relationship \( b^2 = a^2(e^2 - 1) \) or \( a^2 = b^2(e^2 - 1) \) depending on the transverse axis.

Question 5. Find the equation of the hyperbola whose foci are \( (4, 1),(8, 1) \) and whose eccentricity is 2.
Answer: We are given the foci of the hyperbola as \( F_1(4, 1) \) and \( F_2(8, 1) \).
Since the y-coordinates of the foci are the same, the transverse axis of the hyperbola is parallel to the x-axis.
The center of the hyperbola is the midpoint of the two foci. The center point is a key reference for the hyperbola's position.
Center \( (h, k) = \left(\frac{4+8}{2}, \frac{1+1}{2}\right) = \left(\frac{12}{2}, \frac{2}{2}\right) = (6, 1) \).
The distance between the foci is \( 2ae \). We can calculate this distance:
\( 2ae = \sqrt{(8-4)^2 + (1-1)^2} = \sqrt{4^2 + 0^2} = \sqrt{16} = 4 \)
So, \( 2ae = 4 \implies ae = 2 \).
We are given the eccentricity \( e = 2 \).
Substitute \( e = 2 \) into \( ae = 2 \):
\( a(2) = 2 \implies a = 1 \).
Now we find \( b^2 \) using the formula \( b^2 = a^2(e^2 - 1) \).
\( b^2 = (1)^2((2)^2 - 1) \)
\( b^2 = 1(4 - 1) \)
\( b^2 = 3 \).
The standard equation of a hyperbola with its center at \( (h, k) \) and transverse axis parallel to the x-axis is \( \frac{(x-h)^2}{a^2} - \frac{(y-k)^2}{b^2} = 1 \).
Substitute the values of \( h=6, k=1, a^2=1 \), and \( b^2=3 \):
\( \frac{(x-6)^2}{1} - \frac{(y-1)^2}{3} = 1 \)
This can also be written as:
\( 3(x-6)^2 - (y-1)^2 = 3 \)
This is the required equation of the hyperbola.
In simple words: First, find the middle point of the two focus points to get the center of the hyperbola. Then, use the distance between the focus points and the given eccentricity to find 'a'. Use 'a' and 'e' to find 'b-squared'. Finally, put these values along with the center coordinates into the hyperbola's equation.

🎯 Exam Tip: If the foci have the same y-coordinate, the transverse axis is parallel to the x-axis. If they have the same x-coordinate, it's parallel to the y-axis. This determines the form of the hyperbola's standard equation.

Question 6. The line \( y = x + \sqrt{7} \) touches the hyperbola \( 9x^2 - 16y^2 = 144 \).
Answer: We need to determine if the given line touches the hyperbola. We will use the method of substituting the line equation into the hyperbola equation and checking the discriminant.
The equation of the hyperbola is \( 9x^2 - 16y^2 = 144 \).
The equation of the line is \( y = x + \sqrt{7} \).
Substitute the expression for \( y \) from the line equation into the hyperbola equation:
\( 9x^2 - 16(x + \sqrt{7})^2 = 144 \)
Expand \( (x + \sqrt{7})^2 \):
\( 9x^2 - 16(x^2 + 2\sqrt{7}x + (\sqrt{7})^2) = 144 \)
\( 9x^2 - 16(x^2 + 2\sqrt{7}x + 7) = 144 \)
Distribute the -16:
\( 9x^2 - 16x^2 - 32\sqrt{7}x - 112 = 144 \)
Combine like terms:
\( -7x^2 - 32\sqrt{7}x - 112 - 144 = 0 \)
\( -7x^2 - 32\sqrt{7}x - 256 = 0 \)
Multiply by -1 to make the leading coefficient positive:
\( 7x^2 + 32\sqrt{7}x + 256 = 0 \)
This is a quadratic equation in the form \( Ax^2 + Bx + C = 0 \), where \( A = 7 \), \( B = 32\sqrt{7} \), and \( C = 256 \).
For a line to touch a curve (i.e., be tangent to it), the quadratic equation formed by their intersection must have exactly one solution. This occurs when the discriminant \( D = B^2 - 4AC \) is equal to zero. This condition means there's only one point of intersection.
Let's calculate the discriminant:
\( D = (32\sqrt{7})^2 - 4(7)(256) \)
\( D = (32^2 \times (\sqrt{7})^2) - (28 \times 256) \)
\( D = (1024 \times 7) - 7168 \)
\( D = 7168 - 7168 \)
\( D = 0 \)
Since the discriminant is zero, the quadratic equation has exactly one real solution for \( x \). This means the line intersects the hyperbola at exactly one point, which implies the line touches the hyperbola. The equation can also be written as \( (\sqrt{7}x + 16)^2 = 0 \), showing a perfect square.
In simple words: Replace 'y' in the hyperbola equation with the expression from the line equation. This will give you a quadratic equation for 'x'. Calculate the discriminant of this quadratic equation. If the discriminant is zero, it means the line touches the hyperbola at just one point.

🎯 Exam Tip: To show that a line touches a conic section, substitute the linear equation into the conic's equation. If the resulting quadratic equation has a discriminant \( D = 0 \), then the line is tangent (touches) the conic.

Question 7. Find the equation of the hyperbola whose foci are \( (0, \pm 13) \) and the length of the conjugate axis is 20.
Answer: We are given the foci of the hyperbola as \( (0, \pm 13) \).
Since the x-coordinates of the foci are zero, the transverse axis of the hyperbola lies along the y-axis.
For such a hyperbola, the standard equation is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \).
From the foci \( (0, \pm 13) \), we know that \( ae = 13 \).
We are given that the length of the conjugate axis is 20. The length of the conjugate axis is \( 2b \).
So, \( 2b = 20 \implies b = 10 \).
This means \( b^2 = 10^2 = 100 \).
The relationship between \( a, b \), and \( e \) for a hyperbola is \( b^2 = a^2(e^2 - 1) \).
We can rewrite this as \( b^2 = (ae)^2 - a^2 \). This form is useful because we know \( ae \).
Substitute the known values:
\( 100 = (13)^2 - a^2 \)
\( 100 = 169 - a^2 \)
Now, solve for \( a^2 \):
\( a^2 = 169 - 100 \)
\( a^2 = 69 \).
Now that we have \( a^2 = 69 \) and \( b^2 = 100 \), we can write the equation of the hyperbola. Remember, the orientation means \( a^2 \) is under \( y^2 \).
Substitute these values into the standard equation \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \):
\( \frac{y^2}{69} - \frac{x^2}{100} = 1 \)
This is the required equation of the hyperbola.
In simple words: Look at the given focus points to figure out if the hyperbola opens up/down or left/right. Use the focus information to find 'ae' and the length of the conjugate axis to find 'b'. Then, use the formula that connects 'a', 'b', and 'e' to calculate 'a-squared'. Finally, put these values into the correct hyperbola equation.

🎯 Exam Tip: When foci are \( (0, \pm c) \), the transverse axis is along the y-axis, and the equation form is \( \frac{y^2}{a^2} - \frac{x^2}{b^2} = 1 \). Remember that \( c = ae \).

Question 8. Find the equation of the hyperbola whose transverse and conjugate axes are the x and y axes respectively, given that the length of conjugate axis is 5 and distance between the foci is 13.
Answer: We are given that the transverse axis is the x-axis and the conjugate axis is the y-axis. This means the standard form of the hyperbola equation is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
We are given the length of the conjugate axis is 5. The length of the conjugate axis is \( 2b \).
So, \( 2b = 5 \implies b = \frac{5}{2} \).
This means \( b^2 = \left(\frac{5}{2}\right)^2 = \frac{25}{4} \).
We are given that the distance between the foci is 13. The distance between the foci is \( 2ae \).
So, \( 2ae = 13 \implies ae = \frac{13}{2} \).
The relationship between \( a, b \), and \( e \) for a hyperbola is \( b^2 = a^2(e^2 - 1) \). This important formula helps link the semi-axes lengths to the eccentricity.
We can rearrange this to use \( ae \): \( b^2 = (ae)^2 - a^2 \).
Substitute the known values:
\( \frac{25}{4} = \left(\frac{13}{2}\right)^2 - a^2 \)
\( \frac{25}{4} = \frac{169}{4} - a^2 \)
Now, solve for \( a^2 \):
\( a^2 = \frac{169}{4} - \frac{25}{4} \)
\( a^2 = \frac{169 - 25}{4} \)
\( a^2 = \frac{144}{4} \)
\( a^2 = 36 \).
Now that we have \( a^2 = 36 \) and \( b^2 = \frac{25}{4} \), we can write the equation of the hyperbola.
Substitute these values into the standard equation \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \):
\( \frac{x^2}{36} - \frac{y^2}{\frac{25}{4}} = 1 \)
To simplify the fraction in the denominator, we can write \( \frac{y^2}{\frac{25}{4}} \) as \( \frac{4y^2}{25} \):
\( \frac{x^2}{36} - \frac{4y^2}{25} = 1 \)
To remove the denominators, multiply the entire equation by the least common multiple of 36 and 25, which is 900:
\( 25x^2 - 144y^2 = 900 \)
This is the required equation of the hyperbola.
In simple words: First, use the length of the conjugate axis to find 'b' and the distance between foci to find 'ae'. Then, use the relationship \( b^2 = (ae)^2 - a^2 \) to solve for 'a-squared'. Finally, put these 'a-squared' and 'b-squared' values into the standard hyperbola equation and simplify it.

🎯 Exam Tip: Ensure you correctly assign 'a' and 'b' to the transverse and conjugate axes based on the given information. Remember that \( 2b \) is the length of the conjugate axis and \( 2ae \) is the distance between foci.

Question 9. Find the equation to the conic whose focus is \( (1, -1) \), eccentricity is \( \frac{1}{2} \) and the directrix is the line \( x - y = 3 \). Is the conic section an ellipse?
Answer: Let \( P(x, y) \) be any point on the conic section. The definition of a conic section states that the ratio of the distance from any point on the conic to the focus (PF) to the perpendicular distance from that point to the directrix (PM) is equal to the eccentricity (e). This relationship \( PF = e \cdot PM \) defines all conic sections.
Given:
Focus \( F = (1, -1) \)
Directrix equation: \( x - y = 3 \), which can be written as \( x - y - 3 = 0 \).
Eccentricity \( e = \frac{1}{2} \).
First, calculate the distance PF:
\( PF = \sqrt{(x - 1)^2 + (y - (-1))^2} = \sqrt{(x - 1)^2 + (y + 1)^2} \)
Next, calculate the perpendicular distance PM from \( P(x, y) \) to the directrix \( x - y - 3 = 0 \):
\( PM = \frac{|Ax + By + C|}{\sqrt{A^2 + B^2}} = \frac{|1x - 1y - 3|}{\sqrt{1^2 + (-1)^2}} = \frac{|x - y - 3|}{\sqrt{1 + 1}} = \frac{|x - y - 3|}{\sqrt{2}} \)
Now, apply the definition \( PF = e \cdot PM \):
\( \sqrt{(x - 1)^2 + (y + 1)^2} = \frac{1}{2} \cdot \frac{|x - y - 3|}{\sqrt{2}} \)
To eliminate the square roots and the absolute value, square both sides of the equation:
\( (x - 1)^2 + (y + 1)^2 = \left(\frac{1}{2}\right)^2 \cdot \left(\frac{x - y - 3}{\sqrt{2}}\right)^2 \)
\( (x - 1)^2 + (y + 1)^2 = \frac{1}{4} \cdot \frac{(x - y - 3)^2}{2} \)
\( (x - 1)^2 + (y + 1)^2 = \frac{1}{8} (x - y - 3)^2 \)
Multiply both sides by 8:
\( 8[(x - 1)^2 + (y + 1)^2] = (x - y - 3)^2 \)
Expand both sides:
Left side:
\( 8[ (x^2 - 2x + 1) + (y^2 + 2y + 1) ] \)
\( 8[ x^2 + y^2 - 2x + 2y + 2 ] \)
\( 8x^2 + 8y^2 - 16x + 16y + 16 \)
Right side (using \( (A - B - C)^2 = A^2 + B^2 + C^2 - 2AB + 2BC - 2CA \), or \( (X-3)^2 \)):
\( (x - y - 3)^2 = x^2 + (-y)^2 + (-3)^2 + 2(x)(-y) + 2(-y)(-3) + 2(-3)(x) \)
\( = x^2 + y^2 + 9 - 2xy + 6y - 6x \)
Now, set the expanded left side equal to the expanded right side:
\( 8x^2 + 8y^2 - 16x + 16y + 16 = x^2 + y^2 + 9 - 2xy + 6y - 6x \)
Move all terms to one side to form the general equation of the conic:
\( (8x^2 - x^2) + (8y^2 - y^2) + 2xy + (-16x + 6x) + (16y - 6y) + (16 - 9) = 0 \)
\( 7x^2 + 7y^2 + 2xy - 10x + 10y + 7 = 0 \)
This is the required equation of the conic.

To determine if the conic section is an ellipse, we look at the eccentricity \( e \).
Given \( e = \frac{1}{2} \).
Since \( 0 < e < 1 \), the conic section is indeed an ellipse. The value of eccentricity precisely tells us the shape of the curve.
In simple words: First, use the definition of a conic section (distance to focus equals 'e' times distance to directrix) to set up an equation. Square both sides to get rid of the square root, then expand and simplify to find the conic's equation. Finally, check the value of 'e'; if 'e' is between 0 and 1, the conic is an ellipse.

🎯 Exam Tip: The value of eccentricity (e) is key to identifying conic sections: \( e=0 \) for a circle, \( 0 < e < 1 \) for an ellipse, \( e=1 \) for a parabola, and \( e > 1 \) for a hyperbola.