Samacheer Kalvi Class 12 Maths Solutions Chapter 5 Two Dimensional Analytical Geometry II Exercise 5.5

 

Question 1. A bridge has a parabolic arch that is 10 m high in the centre and 30 m wide at the bottom. Find the height of the arch 6 m from the centre, on either sides.
Answer: Let the vertex of the parabolic arch be at the origin \( (0,0) \). Since the arch opens downwards, the equation of the parabola is \( x^2 = -4ay \).
The arch is 30 m wide at the bottom and 10 m high. This means that at a height of 10 m (so \( y = -10 \)), the arch extends \( \pm 15 \) m from the centre. Thus, the point \( (15, -10) \) lies on the parabola.
Substitute this point into the equation:
\( 15^2 = -4a(-10) \)
\( 225 = 40a \)
\( a = \frac{225}{40} = \frac{45}{8} \)
Now, substitute the value of \( a \) back into the parabola equation:
\( x^2 = -4\left(\frac{45}{8}\right)y \)
\( x^2 = -\frac{45}{2}y \)
We need to find the height of the arch 6 m from the centre, on either side. So, we set \( x = 6 \). Let the y-coordinate at this point be \( y_1 \).
\( 6^2 = -\frac{45}{2}y_1 \)
\( 36 = -\frac{45}{2}y_1 \)
\( y_1 = -\frac{36 \times 2}{45} = -\frac{72}{45} = -\frac{8}{5} = -1.6 \) m.
This \( y_1 \) value is the vertical distance from the vertex (top of the arch). Since the total height of the arch is 10 m (from bottom to vertex), the height of the arch 6 m from the centre is \( 10 - |y_1| \).
Height \( = 10 - 1.6 = 8.4 \) m.
In simple words: The bridge forms an upside-down U-shape like a parabola. We first find the mathematical rule for this parabola using its total width and height. Then, we use this rule to calculate how high the arch is at a point 6 meters away from its very middle.

🎯 Exam Tip: When dealing with arches or suspension bridges, drawing a clear diagram and placing the vertex at the origin \( (0,0) \) (or an appropriate point) simplifies the setup of the parabolic equation.

 

Question 2. A tunnel through a mountain for a four lane highway is to have a elliptical opening. The total width of the highway (not the opening) is to be 16 m and the of the road must be sufficient for a truck 4 m high to clear if the highest point of the opening is to be 5 m approximately. How wide must the opening be?
Answer: Let the equation of the ellipse for the tunnel opening be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \).
The highest point of the opening is 5 m, which means the semi-minor axis \( b = 5 \) m. So, the equation becomes \( \frac{x^2}{a^2} + \frac{y^2}{5^2} = 1 \).
The total width of the highway is 16 m. For a truck 4 m high to clear, its path must fit within the tunnel. This means that at a height of 4 m from the center of the road (y-axis), the opening must be at least 16 m wide. Half of this width is 8 m.
So, the point \( (8, 4) \) must lie on the ellipse. Substitute this point into the equation:
\( \frac{8^2}{a^2} + \frac{4^2}{5^2} = 1 \)
\( \frac{64}{a^2} + \frac{16}{25} = 1 \)
\( \frac{64}{a^2} = 1 - \frac{16}{25} \)
\( \frac{64}{a^2} = \frac{25 - 16}{25} \)
\( \frac{64}{a^2} = \frac{9}{25} \)
Now, solve for \( a^2 \):
\( a^2 = \frac{64 \times 25}{9} \)
\( a = \sqrt{\frac{64 \times 25}{9}} = \frac{\sqrt{64} \times \sqrt{25}}{\sqrt{9}} = \frac{8 \times 5}{3} = \frac{40}{3} \) m.
The required width of the opening is \( 2a \).
Width \( = 2 \times \frac{40}{3} = \frac{80}{3} \approx 26.66 \) m.
In simple words: We used the shape of an ellipse to model the tunnel. We knew the tunnel's highest point and how wide it needs to be for a truck to pass. With these numbers, we calculated the total width of the elliptical opening.

🎯 Exam Tip: For ellipse problems, ensure you correctly identify the semi-major axis (a) and semi-minor axis (b) based on whether the ellipse is wider or taller, and how the given dimensions relate to these axes.

 

Question 3. At a water fountain, water attains a maximum height of 4 m at horizontal distance of 0.5 m from its origin. If the path of water is a parabola, find the height of water at a horizontal distance of 0.75 m from the point of origin.
Answer: The water path is a parabola opening downwards. The maximum height of 4 m is attained at a horizontal distance of 0.5 m from the origin. This means the vertex of the parabola is at \( (0.5, 4) \).
The general equation of a parabola opening downwards with vertex \( (h, k) \) is \( (x - h)^2 = -4a(y - k) \).
Substitute the vertex \( (0.5, 4) \):
\( (x - 0.5)^2 = -4a(y - 4) \)
The water starts from the origin, so the point \( (0, 0) \) lies on the parabola. Substitute \( (0, 0) \) into the equation:
\( (0 - 0.5)^2 = -4a(0 - 4) \)
\( (-0.5)^2 = -4a(-4) \)
\( 0.25 = 16a \)
\( a = \frac{0.25}{16} = \frac{1/4}{16} = \frac{1}{64} \)
Now, write the complete equation of the parabola:
\( (x - 0.5)^2 = -4\left(\frac{1}{64}\right)(y - 4) \)
\( (x - 0.5)^2 = -\frac{1}{16}(y - 4) \)
We need to find the height of the water at a horizontal distance of 0.75 m from the origin. So, set \( x = 0.75 \) and find \( y_1 \).
\( (0.75 - 0.5)^2 = -\frac{1}{16}(y_1 - 4) \)
\( (0.25)^2 = -\frac{1}{16}(y_1 - 4) \)
\( \left(\frac{1}{4}\right)^2 = -\frac{1}{16}(y_1 - 4) \)
\( \frac{1}{16} = -\frac{1}{16}(y_1 - 4) \)
Divide both sides by \( \frac{1}{16} \):
\( 1 = -(y_1 - 4) \)
\( 1 = -y_1 + 4 \)
\( y_1 = 4 - 1 \)
\( y_1 = 3 \) m.
In simple words: The path of the water is a parabola, which we set up using the highest point it reaches. Then, we used the starting point of the water to find a missing value in our parabola equation. Finally, we used the full equation to find the water's height at a different horizontal distance.

🎯 Exam Tip: When the vertex is not at the origin, remember to use the general form \( (x-h)^2 = 4a(y-k) \) or \( (y-k)^2 = 4a(x-h) \) and substitute the vertex coordinates correctly.

 

Question 4. (a) Position a co-ordinate system with the origin at the vertex and the x-axis on the parabola's axis of symmetry and find an equation of the parabola.
(b) Find the depth of the satellite dish at the vertex.
Answer:
(a) The satellite dish has a parabolic cross-section. We position the coordinate system with the origin at the vertex \( (0,0) \) and the x-axis along the axis of symmetry. Since the dish opens to the right (to receive signals), it is a rightward-opening parabola. The focus is placed 1.2 m from the vertex, so \( a = 1.2 \) m.
The equation of a rightward-opening parabola with vertex at origin is \( y^2 = 4ax \).
Substitute \( a = 1.2 \):
\( y^2 = 4(1.2)x \)
\( y^2 = 4.8x \)
This is the equation of the parabola.
(b) The dish is 5 m wide at the opening. This means the top edge is at \( y = 2.5 \) m (half of 5 m). We need to find the depth of the dish, which is the x-coordinate (\( x_1 \)) when \( y = 2.5 \).
Substitute \( y = 2.5 \) into the equation \( y^2 = 4.8x \):
\( (2.5)^2 = 4.8x_1 \)
\( 6.25 = 4.8x_1 \)
\( x_1 = \frac{6.25}{4.8} \)
\( x_1 \approx 1.302 \) m.
Rounding to one decimal place as in the solution, the depth of the satellite dish at the vertex is approximately 1.3 m.
In simple words: For part (a), we set up the parabola's formula using its starting point and how far its focus (the signal-collecting point) is. For part (b), we used the dish's total width to find out how deep it is from the front to the back.

🎯 Exam Tip: Remember to choose the correct parabolic equation (e.g., \( y^2 = 4ax \) or \( x^2 = 4ay \)) based on the orientation of the dish (right/left opening or up/down opening).

 

Question 5. The parabolic cable of a 60 m portion of the roadbed of a suspension bridge are positioned as shown below. Vertical cables are to be spaced every 6 m along this portion of the roadbed. Calculate; the lengths of first two of these vertical cables: from the vertex.
Answer: Let the vertex of the parabolic cable be at the origin \( (0,0) \). The cable opens upwards, so the equation is \( x^2 = 4ay \).
The roadbed is 60 m long, and the cable supports at the ends are 13 m high. This means the points \( (\pm 30, 13) \) lie on the parabola.
Substitute the point \( (30, 13) \) into the equation:
\( 30^2 = 4a(13) \)
\( 900 = 52a \)
\( a = \frac{900}{52} = \frac{225}{13} \)
So, the equation of the parabola is \( x^2 = 4\left(\frac{225}{13}\right)y \implies x^2 = \frac{900}{13}y \).
The vertical cables are spaced every 6 m from the vertex. The roadbed itself is 3 m below the vertex (from the diagram).
(i) Length of the first vertical cable (6 m from the vertex):
Set \( x = 6 \). Let the height of the cable from the vertex be \( y_1 \).
\( 6^2 = \frac{900}{13}y_1 \)
\( 36 = \frac{900}{13}y_1 \)
\( y_1 = \frac{36 \times 13}{900} = \frac{468}{900} = 0.52 \) m.
The length of the cable from the roadbed is this height plus the 3 m distance between the roadbed and the vertex. So, the length is \( 3 + 0.52 = 3.52 \) m.
(ii) Length of the second vertical cable (12 m from the vertex):
Set \( x = 12 \). Let the height of the cable from the vertex be \( y_2 \).
\( 12^2 = \frac{900}{13}y_2 \)
\( 144 = \frac{900}{13}y_2 \)
\( y_2 = \frac{144 \times 13}{900} = \frac{1872}{900} = 2.08 \) m.
The length of this cable from the roadbed is \( 3 + 2.08 = 5.08 \) m.
In simple words: We used the shape of the bridge's main cable, which is a parabola, and its highest points to write down its mathematical equation. Then, we calculated the lengths of the smaller vertical cables that support the road by finding their height at specific distances from the center.

🎯 Exam Tip: Remember to account for any offset of the roadbed from the vertex when calculating the total length of the vertical cables, adding or subtracting as necessary.

 

Question 6. Cross-section of a Nuclear cooling tower is in the shape of a hyperbola with equation \( \frac {x^2}{30^2} – \frac {y^2}{44^2} = 1 \). The tower is 150 m tall and the distance from the top of the tower to the centre of the hyperbola is half the distance from the base of the tower to the centre of the hyperbola. Find the diameter of the top and base of the tower.
Answer: The equation of the hyperbola is \( \frac{x^2}{30^2} - \frac{y^2}{44^2} = 1 \). This means \( a^2 = 30^2 \) and \( b^2 = 44^2 \).
The total height of the tower (CD) is 150 m. Let O be the centre of the hyperbola.
Given that the distance from the top of the tower to the centre (OC) is half the distance from the base to the centre (OD). So, \( OC = \frac{1}{2}OD \).
Also, \( OC + OD = CD = 150 \).
Let \( OC = h \). Then \( OD = 2h \).
\( h + 2h = 150 \implies 3h = 150 \implies h = 50 \) m.
So, the top of the tower is at \( y = 50 \) m (from the centre of the hyperbola), and the base is at \( y = -100 \) m.

**Diameter of the top of the tower:**
Let the radius of the top of the tower be \( x_1 \). The point corresponding to the top is \( (x_1, 50) \). Substitute this into the hyperbola equation:
\( \frac{x_1^2}{30^2} - \frac{50^2}{44^2} = 1 \)
\( \frac{x_1^2}{900} - \frac{2500}{1936} = 1 \)
\( \frac{x_1^2}{900} = 1 + \frac{2500}{1936} = \frac{1936 + 2500}{1936} = \frac{4436}{1936} \)
\( x_1^2 = 900 \times \frac{4436}{1936} \)
\( x_1 = \sqrt{900 \times \frac{4436}{1936}} = 30 \times \sqrt{\frac{4436}{1936}} \)
\( x_1 = 30 \times \frac{66.6033}{44} \approx 30 \times 1.5137 \approx 45.41 \) m.
The radius of the top is \( x_1 \approx 45.41 \) m. So, the diameter of the top is \( 2x_1 = 2 \times 45.41 = 90.82 \) m.

**Diameter of the base of the tower:**
Let the radius of the base of the tower be \( x_2 \). The point corresponding to the base is \( (x_2, -100) \). Substitute this into the hyperbola equation:
\( \frac{x_2^2}{30^2} - \frac{(-100)^2}{44^2} = 1 \)
\( \frac{x_2^2}{900} - \frac{10000}{1936} = 1 \)
\( \frac{x_2^2}{900} = 1 + \frac{10000}{1936} = \frac{1936 + 10000}{1936} = \frac{11936}{1936} \)
\( x_2^2 = 900 \times \frac{11936}{1936} \)
\( x_2 = \sqrt{900 \times \frac{11936}{1936}} = 30 \times \sqrt{\frac{11936}{1936}} \)
\( x_2 = 30 \times \frac{109.2519}{44} \approx 30 \times 2.4829 \approx 74.49 \) m.
The radius of the base is \( x_2 \approx 74.49 \) m. So, the diameter of the base is \( 2x_2 = 2 \times 74.49 = 148.98 \) m.
In simple words: First, we figured out the exact heights of the tower's top and base relative to its center, using the given total height and the distance relationship. Then, using the hyperbola's formula, we calculated the radius at both the top and the base, which allowed us to find their diameters.

🎯 Exam Tip: When dealing with geometric shapes that are part of a larger structure, carefully determine the coordinates of key points relative to the chosen origin. This helps in correctly substituting values into the conic section equations.

 

Question 7. A rod of length 1.2 m moves with its ends always touching the co-ordinate axes. The locus of a point P on the rod, which is 0.3 m from the end in contact with x axis is an ellipse. Find the eccentricity.
Answer: Let the rod be \( BD = 1.2 \) m. Let its end B be on the x-axis and end D be on the y-axis. Let \( P(x,y) \) be a point on the rod.
The point P is 0.3 m from the end in contact with the x-axis (let's call this end A in the context of the solution, or B in the diagram). So, \( PA = 0.3 \) m. This means the other segment \( PC = BD - PA = 1.2 - 0.3 = 0.9 \) m.
For a rod of length \( L \) with a point P dividing it into segments of length \( m \) and \( n \) (where \( m \) is the distance from the x-axis contact point, and \( n \) is from the y-axis contact point), the locus of P is an ellipse with equation \( \frac{x^2}{n^2} + \frac{y^2}{m^2} = 1 \).
In this case, \( n = 0.9 \) m and \( m = 0.3 \) m.
So, the equation of the ellipse is \( \frac{x^2}{(0.9)^2} + \frac{y^2}{(0.3)^2} = 1 \).
Here, the semi-major axis squared is \( A^2 = (0.9)^2 = 0.81 \) and the semi-minor axis squared is \( B^2 = (0.3)^2 = 0.09 \). (Since \( 0.9 > 0.3 \), \( A \) is the semi-major axis and \( B \) is the semi-minor axis.)
The eccentricity \( e \) of an ellipse is given by the formula \( e = \sqrt{1 - \frac{B^2}{A^2}} \).
\( e = \sqrt{1 - \frac{(0.3)^2}{(0.9)^2}} \)
\( e = \sqrt{1 - \frac{0.09}{0.81}} \)
\( e = \sqrt{1 - \frac{1}{9}} \)
\( e = \sqrt{\frac{9-1}{9}} \)
\( e = \sqrt{\frac{8}{9}} \)
\( e = \frac{\sqrt{8}}{\sqrt{9}} = \frac{2\sqrt{2}}{3} \).
In simple words: Imagine a ladder sliding down a wall. A specific point on this ladder traces out an elliptical path. We used the lengths of the parts of the rod to find the mathematical formula for this ellipse. Then, we calculated its 'eccentricity', which tells us how "stretched out" or "circular" the ellipse is.

🎯 Exam Tip: For the sliding ladder problem, remember the general form of the ellipse created by a point P(x,y) on a ladder of length \( L \), where P divides the ladder into segments of length \( n \) (from y-axis) and \( m \) (from x-axis): \( \frac{x^2}{n^2} + \frac{y^2}{m^2} = 1 \).

 

Question 8. Assume that water issuing from the end of a horizontal pipe, 7.5 m. above the ground, describes a parabolic path. The vertex of the parabolic path is at The end of the pipe. At position 2.5 in below the line of the pipe, the flow of water has curved outward 3 m beyond the vertical, line through the end of the pipe. How far beyond this vertical line will the water strike the ground?
Answer: Let's place the origin \( (0,0) \) at the end of the pipe (which is the vertex of the parabolic path). Since the water flows horizontally and then curves downwards, the parabola opens downwards. The equation of such a parabola is \( x^2 = -4ay \).
We are given a point on the water's path: at a position 2.5 m below the pipe line, the water has curved outward 3 m. So, this point is \( (3, -2.5) \).
Substitute this point into the parabola equation:
\( 3^2 = -4a(-2.5) \)
\( 9 = 10a \)
\( a = \frac{9}{10} \)
Now, the equation of the parabola is \( x^2 = -4\left(\frac{9}{10}\right)y \implies x^2 = -\frac{18}{5}y \).
The pipe is 7.5 m above the ground. The water strikes the ground when \( y = -7.5 \) m (relative to the pipe's height). We need to find the horizontal distance \( x_1 \) at which this happens.
Substitute \( y = -7.5 \) into the equation:
\( x_1^2 = -\frac{18}{5}(-7.5) \)
\( x_1^2 = \frac{18 \times 7.5}{5} \)
\( x_1^2 = 18 \times 1.5 \)
\( x_1^2 = 27 \)
\( x_1 = \sqrt{27} = \sqrt{9 \times 3} = 3\sqrt{3} \) m.
Thus, the water will strike the ground \( 3\sqrt{3} \) m beyond the vertical line through the end of the pipe.
In simple words: The path of water from the pipe is a parabola. We used one known point on this path to find the specific formula for this parabola. Then, knowing how high the pipe is from the ground, we used the formula to calculate how far the water travels horizontally before hitting the ground.

🎯 Exam Tip: When setting up coordinates for projectile motion problems, choosing the launch point (or vertex) as the origin can often simplify the parabolic equation and subsequent calculations.

 

Question 9. On lighting a rocket cracker it gets projected in a parabolic path and reaches a maximum height of 4 in when it is 6 m away from the point of projection. Finally, it reaches the ground 12 in away from the starting-point. Find the angle of! projection.
Answer: Let's set up the coordinate system with the vertex of the parabola at the origin \( (0,0) \) and the parabola opening downwards, so its equation is \( x^2 = -4ay \). The solution implies this setup by using \( x^2 = -4ay \) initially and then considering points.
The problem states a maximum height of 4 m is reached. The text also mentions "6 m away from the point of projection" and "12 m away from the starting-point". The solution uses `B(6,-4)` and then calculates the slope at `(-6,-4)`. This implies that `(-6,-4)` is the point of projection, and `(6,-4)` is a symmetric point on the ground. The maximum height relative to the projection point `(-6,-4)` would be 4m if the vertex is at `(0,0)`. In this setup, the problem's "maximum height of 4m" (likely a typo, should be meters) refers to the change in y from `y=-4` to `y=0`.
Using the point \( B(6, -4) \) (or any point on the parabola, such as `(-6,-4)`):
\( 6^2 = -4a(-4) \)
\( 36 = 16a \)
\( a = \frac{36}{16} = \frac{9}{4} \)
So, the equation of the parabolic path is \( x^2 = -4\left(\frac{9}{4}\right)y \implies x^2 = -9y \).
To find the angle of projection, we need to find the slope of the tangent to the parabola at the point of projection. The solution calculates this at \( (-6, -4) \).
Differentiate the equation \( x^2 = -9y \) with respect to \( x \):
\( 2x = -9 \frac{dy}{dx} \)
\( \frac{dy}{dx} = -\frac{2x}{9} \)
Now, evaluate the slope \( \frac{dy}{dx} \) at the point of projection \( (-6, -4) \):
\( \frac{dy}{dx} \Big|_{x=-6} = -\frac{2(-6)}{9} = \frac{12}{9} = \frac{4}{3} \)
The angle of projection \( \theta \) is given by \( \tan \theta = \frac{dy}{dx} \).
\( \tan \theta = \frac{4}{3} \)
\( \theta = \tan^{-1}\left(\frac{4}{3}\right) \).
In simple words: We used the given points to define the parabola's shape. Then, we found the slope of the path exactly at the point where the rocket starts flying. This slope tells us the angle at which the rocket was launched.

🎯 Exam Tip: The angle of projection is found by calculating the derivative \( \frac{dy}{dx} \) at the point of projection, which gives the tangent's slope \( \tan\theta \).

 

Question 10. Points A and B are 10 km apart and it is determined from the sound of an explosion heard at those points at different times that the location of the explosion is 6 km closer to A than 5. Show that the location of the explosion is a particular curve and find an equation of it.
Answer: Let points A and B be located on the x-axis. Since they are 10 km apart, we can set their coordinates as \( A(-5, 0) \) and \( B(5, 0) \).
Let the location of the explosion be \( C(x, y) \).
The sound of the explosion is heard at different times, and it is 6 km closer to A than to B. This means the difference in distances from the explosion to A and B is constant. Specifically, the distance from C to B minus the distance from C to A is 6 km: \( BC - AC = 6 \).
The definition of a hyperbola is the locus of all points for which the absolute difference of the distances to two fixed points (foci) is constant.
Here, the foci are A and B, and the constant difference is 6 km. So, \( 2a = 6 \implies a = 3 \).
The distance between the foci is \( 2c = 10 \implies c = 5 \).
For a hyperbola, the relationship between \( a \), \( b \), and \( c \) is \( c^2 = a^2 + b^2 \).
\( 5^2 = 3^2 + b^2 \)
\( 25 = 9 + b^2 \)
\( b^2 = 25 - 9 = 16 \)
The standard equation for a hyperbola with foci on the x-axis centered at the origin is \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \).
Substitute \( a^2 = 9 \) and \( b^2 = 16 \):
\( \frac{x^2}{9} - \frac{y^2}{16} = 1 \)

Alternatively, using the distance formula:
\( AC = \sqrt{(x - (-5))^2 + (y - 0)^2} = \sqrt{(x+5)^2 + y^2} \)
\( BC = \sqrt{(x - 5)^2 + (y - 0)^2} = \sqrt{(x-5)^2 + y^2} \)
Given \( BC - AC = 6 \):
\( \sqrt{(x-5)^2 + y^2} - \sqrt{(x+5)^2 + y^2} = 6 \)
Rearrange the terms:
\( \sqrt{(x-5)^2 + y^2} = 6 + \sqrt{(x+5)^2 + y^2} \)
Square both sides:
\( (x-5)^2 + y^2 = 36 + (x+5)^2 + y^2 + 12\sqrt{(x+5)^2 + y^2} \)
\( x^2 - 10x + 25 + y^2 = 36 + x^2 + 10x + 25 + y^2 + 12\sqrt{(x+5)^2 + y^2} \)
Simplify the equation by canceling \( x^2 \), \( y^2 \), and 25 from both sides:
\( -10x = 36 + 10x + 12\sqrt{(x+5)^2 + y^2} \)
\( -20x - 36 = 12\sqrt{(x+5)^2 + y^2} \)
Divide by 4:
\( -5x - 9 = 3\sqrt{(x+5)^2 + y^2} \)
Square both sides again:
\( (-5x - 9)^2 = (3\sqrt{(x+5)^2 + y^2})^2 \)
\( (5x + 9)^2 = 9((x+5)^2 + y^2) \)
\( 25x^2 + 90x + 81 = 9(x^2 + 10x + 25 + y^2) \)
\( 25x^2 + 90x + 81 = 9x^2 + 90x + 225 + 9y^2 \)
Rearrange the terms to form the hyperbola equation:
\( 25x^2 - 9x^2 - 9y^2 + 90x - 90x + 81 - 225 = 0 \)
\( 16x^2 - 9y^2 - 144 = 0 \)
Divide by 144:
\( \frac{16x^2}{144} - \frac{9y^2}{144} = \frac{144}{144} \)
\( \frac{x^2}{9} - \frac{y^2}{16} = 1 \)
This is the required equation of the hyperbola, showing that the location of the explosion lies on a hyperbola.
In simple words: We used the given information about two points and the constant difference in distance to the explosion to figure out its path. This kind of path is called a hyperbola. We then derived the mathematical formula for this specific hyperbola.

🎯 Exam Tip: Recognize that problems involving a constant difference in distances from two fixed points are related to hyperbolas. Start by defining the coordinates of the foci and using the distance formula, or directly apply the standard hyperbola equation.

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