Samacheer Kalvi Class 11 Business Maths Solutions Chapter 3 Analytical Geometry Exercise 3.6

Tamilnadu Samacheer Kalvi 11th Business Maths Solutions Chapter 3 Analytical Geometry Ex 3.6

Samacheer Kalvi 11th Business Maths Analytical Geometry Ex 3.6 Text Book Back Questions and Answers

 

Question 1. Find the equation of the parabola whose focus is the point \( F(-1, -2) \) and the directrix is the line \( 4x – 3y + 2 = 0 \).
Answer: Given the focus \( F(-1, -2) \) and the directrix \( l: 4x – 3y + 2 = 0 \).
Let \( P(x, y) \) be any point on the parabola. By definition, any point on a parabola is equidistant from its focus and its directrix.
So, \( FP = PM \)
Now, we can square both sides to remove the square root:
\( \implies FP^2 = PM^2 \)
\( \implies (x + 1)^2 + (y + 2)^2 = \left(\frac{4x - 3y + 2}{\sqrt{4^2 + (-3)^2}}\right)^2 \)
\( \implies x^2 + 2x + 1 + y^2 + 4y + 4 = \frac{(4x - 3y + 2)^2}{16 + 9} \)
\( \implies x^2 + y^2 + 2x + 4y + 5 = \frac{16x^2 + 9y^2 + 4 - 24xy + 16x - 12y}{25} \)
Next, multiply both sides by 25 to clear the fraction:
\( \implies 25(x^2 + y^2 + 2x + 4y + 5) = 16x^2 + 9y^2 - 24xy + 16x - 12y + 4 \)
\( \implies 25x^2 + 25y^2 + 50x + 100y + 125 = 16x^2 + 9y^2 - 24xy + 16x - 12y + 4 \)
Rearrange all terms to one side to form the general equation of the parabola:
\( \implies (25 - 16)x^2 + (25 - 9)y^2 + 24xy + (50 - 16)x + (100 + 12)y + 125 - 4 = 0 \)
\( \implies 9x^2 + 16y^2 + 24xy + 34x + 112y + 121 = 0 \)
This is the general equation of the parabola. This equation shows how the coordinates of any point on the parabola relate to each other.
In simple words: We found the equation of the parabola by using the rule that any point on it is the same distance from a special point (the focus) and a special line (the directrix). We used a formula for distance and some algebra to get the final equation.

🎯 Exam Tip: Remember the fundamental definition of a parabola: the locus of points equidistant from a fixed point (focus) and a fixed line (directrix). This definition is key to solving such problems.

 

Question 2. The parabola \( y^2 = kx \) passes through the point \( (4, -2) \). Find its latus rectum and focus.
Answer: The equation of the parabola is given as \( y^2 = kx \).
It passes through the point \( (4, -2) \). We can substitute these coordinates into the equation to find the value of k.
\( \implies (-2)^2 = k(4) \)
\( \implies 4 = 4k \)
Divide both sides by 4 to solve for k:
\( \implies k = 1 \)
So, the equation of the parabola is \( y^2 = x \).
To find the latus rectum and focus, we compare this to the standard form \( y^2 = 4ax \).
\( \implies y^2 = 4\left(\frac{1}{4}\right)x \)
From this, we can see that \( a = \frac{1}{4} \).
The length of the latus rectum is \( 4a \).
\( \implies \) Length of latus rectum \( = 4 \times \frac{1}{4} = 1 \).
The equation of the latus rectum (LR) for \( y^2 = 4ax \) is \( x = a \).
\( \implies x = \frac{1}{4} \)
Rearrange this to a standard form:
\( \implies 4x = 1 \)
\( \implies 4x - 1 = 0 \)
The focus for a parabola of the form \( y^2 = 4ax \) is \( (a, 0) \).
\( \implies \) Focus \( = \left(\frac{1}{4}, 0\right) \).
The latus rectum is a line segment that passes through the focus and is perpendicular to the axis of symmetry of the parabola. Its length helps in understanding the width of the parabola at its focus.
In simple words: We used the given point to find the missing number 'k' in the parabola's equation. Then, by comparing it to a standard shape, we found 'a'. This 'a' value helped us find the focus (a special point) and the length of the latus rectum (a measure of the parabola's width).

🎯 Exam Tip: Always remember the standard forms of parabolas and their corresponding features like vertex, focus, directrix, and latus rectum. Matching the given equation to a standard form is the first crucial step.

 

Question 3. Find the vertex, focus, axis, directrix, and the length of the latus rectum of the parabola \( y^2 – 8y – 8x + 24 = 0 \).
Answer: We are given the equation of the parabola: \( y^2 – 8y – 8x + 24 = 0 \).
To find its features, we need to convert it into a standard form, usually by completing the square for the y-terms.
First, move the x-term and constant to the right side:
\( \implies y^2 – 8y = 8x - 24 \)
Complete the square for \( y^2 – 8y \) by adding \( \left(\frac{-8}{2}\right)^2 = (-4)^2 = 16 \) to both sides:
\( \implies y^2 – 8y + 16 = 8x - 24 + 16 \)
\( \implies (y - 4)^2 = 8x - 8 \)
Factor out 8 from the right side:
\( \implies (y - 4)^2 = 8(x - 1) \)
Now, compare this with the standard form \( Y^2 = 4aX \), where \( Y = y - 4 \) and \( X = x - 1 \).
From \( (y - 4)^2 = 4(2)(x - 1) \), we can identify \( a = 2 \).
This means the parabola opens towards the positive x-axis.
Here are the features of the parabola:

Completing the square helps to transform the general equation into a standard form, making it easy to identify the key features of the parabola like its vertex and axis of symmetry.
In simple words: We changed the given parabola equation into a simpler form by completing the square. From this new form, we could easily find its main parts: the vertex (turning point), the focus (a special point), the axis (the line it's symmetrical around), the directrix (a special line), and the length of its latus rectum (how wide it is at the focus).

🎯 Exam Tip: When given a general quadratic equation for a parabola, always complete the square for the squared variable to transform it into one of the standard forms \((y-k)^2 = 4a(x-h)\) or \((x-h)^2 = 4a(y-k)\). This is essential for easily identifying all the properties.

 

Question 4. Find the co-ordinates of the focus, vertex, equation of the directrix, axis and the length of latus rectum of the parabola (a) \( y^2 = 20x \), (b) \( x^2 = 8y \), (c) \( x^2 = -16y \)
Answer: We will find the features for each parabola by comparing them to their standard forms.
(a) For the parabola \( y^2 = 20x \):
Compare with \( y^2 = 4ax \).
\( 4a = 20 \implies a = 5 \).
This parabola opens to the right.

(b) For the parabola \( x^2 = 8y \):
Compare with \( x^2 = 4ay \).
\( 4a = 8 \implies a = 2 \).
This parabola opens upwards.
(c) For the parabola \( x^2 = -16y \):
Compare with \( x^2 = -4ay \).
\( 4a = 16 \implies a = 4 \).
This parabola opens downwards.
Understanding the sign of 'a' is important as it determines the direction in which the parabola opens, whether right, left, up, or down.
In simple words: For each parabola, we matched its equation to a standard type. Then, we found the value of 'a', which is a key number for parabolas. Using 'a', we identified the vertex (its turning point), focus (a special point), axis (its line of symmetry), directrix (a special line), and the length of its latus rectum (how wide it is at the focus).

🎯 Exam Tip: Memorize the four basic standard forms of parabolas (\(y^2=4ax\), \(y^2=-4ax\), \(x^2=4ay\), \(x^2=-4ay\)) and their corresponding properties (vertex, focus, directrix, axis, latus rectum). This makes it easy to quickly identify features for questions like this.

 

Question 5. The average variable cost of the monthly output of x tonnes of a firm producing a valuable metal is \( y = \frac{1}{5} x^2 – 6x + 100 \). Show that the average variable cost curve is a parabola. Also, find the output and the average cost at the vertex of the parabola.
Answer: The average variable cost (AVC) is given by the equation: \( y = \frac{1}{5} x^2 – 6x + 100 \).
To show that this curve is a parabola, we can rewrite it in a standard parabolic form. First, multiply the entire equation by 5 to clear the fraction:
\( \implies 5y = x^2 – 30x + 500 \)
Now, we want to complete the square for the x-terms. Move the \( x^2 \) and \( -30x \) terms to one side and the \( 5y \) and \( 500 \) to the other side conceptually, then add \( \left(\frac{-30}{2}\right)^2 = (-15)^2 = 225 \) to both sides related to the x-terms:
\( \implies x^2 - 30x + 225 = 5y - 500 + 225 \)
\( \implies (x - 15)^2 = 5y - 275 \)
Factor out 5 from the right side:
\( \implies (x - 15)^2 = 5(y - 55) \)
This equation is in the standard form \( X^2 = 4AY \), where \( X = x - 15 \), \( Y = y - 55 \), and \( 4A = 5 \implies A = \frac{5}{4} \).
Since the equation is in the form \( X^2 = 4AY \), it represents a parabola. In economics, cost curves are often parabolic because they reflect initial economies of scale followed by diminishing returns.
To find the output and average cost at the vertex, we set \( X=0 \) and \( Y=0 \).
\( \implies x - 15 = 0 \implies x = 15 \)
\( \implies y - 55 = 0 \implies y = 55 \)
Therefore, at the vertex of the parabola, the output is 15 tonnes, and the average variable cost is Rs. 55.
In simple words: We changed the given cost equation into a standard parabola shape, showing it is indeed a parabola. Then, we found the lowest point of this parabola, called the vertex. At this point, the factory makes 15 tonnes of metal, and the average cost for each tonne is Rs. 55. This vertex tells us the most efficient production level.

🎯 Exam Tip: To show an equation represents a parabola, transform it into one of the standard forms \((y-k)^2 = 4a(x-h)\) or \((x-h)^2 = 4a(y-k)\) by completing the square. The vertex of the parabola gives the minimum or maximum value of the cost or profit function.

 

Question 6. The profit y accumulated in thousand in x months is given by \( y = -x^2 + 10x - 15 \). Find the best time to end the project.
Answer: The profit function is given by \( y = -x^2 + 10x - 15 \).
This is a quadratic equation, and its graph is a parabola. Since the coefficient of \( x^2 \) is negative (it's -1), the parabola opens downwards, which means it will have a maximum point.
To find the maximum profit and the time when it occurs, we need to find the vertex of this parabola. We can do this by completing the square:
First, factor out -1 from the \( x^2 \) and \( x \) terms:
\( \implies y = -(x^2 - 10x + 15) \)
Now, complete the square inside the parenthesis for \( x^2 - 10x \). We need to add and subtract \( \left(\frac{-10}{2}\right)^2 = (-5)^2 = 25 \):
\( \implies y = -(x^2 - 10x + 25 - 25 + 15) \)
\( \implies y = -[(x - 5)^2 - 10] \)
Now, distribute the negative sign back:
\( \implies y = -(x - 5)^2 + 10 \)
Rearrange this to the standard vertex form \( (x-h)^2 = -4a(y-k) \), which is \( (x - 5)^2 = -(y - 10) \).
From this, we can see the vertex is at \( (h, k) = (5, 10) \).
The vertex of a downward-opening parabola represents its maximum point. This maximum profit occurs when \( x - 5 = 0 \), which means \( x = 5 \) months.
At \( x = 5 \) months, the profit is \( y = 10 \) (thousand).
After this point, the profit will gradually decrease. Thus, the best time to end the project is after 5 months to ensure maximum profit is achieved.
In simple words: We have a profit equation that makes a downward-curving shape. The highest point of this curve shows when profit is best. We found this peak point by rearranging the equation. It tells us that after 5 months, the profit will be at its highest, so that's the best time to finish the project.

🎯 Exam Tip: For problems involving maximum or minimum profit/cost, convert the quadratic function into vertex form \( y = a(x-h)^2 + k \). The vertex \((h, k)\) will give the x-value (time/quantity) at which the maximum/minimum occurs and the corresponding y-value (profit/cost).