Samacheer Kalvi Class 11 Maths Solutions Chapter 6 Two Dimensional Analytical Geometry Exercise 6.1

Question 1. Find the locus of P, if for all values of a, the coordinates of a moving point P is
(i) (9 cos a, 9 sin a)
(ii) (9 cos a, 6 sin a)

Answer:
(i) For the coordinates \( (9 \cos \alpha, 9 \sin \alpha) \):
Let P be any point \( (h, k) \) on the required path. From the given information, we have:
\( h = 9 \cos \alpha \)
\( k = 9 \sin \alpha \)
We can rewrite these equations as:
\( \frac{h}{9} = \cos \alpha \)
\( \frac{k}{9} = \sin \alpha \)
Now, we use the fundamental trigonometric identity: \( \cos^2 \alpha + \sin^2 \alpha = 1 \). This identity is always true for any angle \( \alpha \).
\( \implies \) Substitute the expressions for \( \cos \alpha \) and \( \sin \alpha \):
\( \left(\frac{h}{9}\right)^2 + \left(\frac{k}{9}\right)^2 = 1 \)
\( \implies \frac{h^2}{81} + \frac{k^2}{81} = 1 \)
\( \implies h^2 + k^2 = 81 \)
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( x^2 + y^2 = 81 \). This equation represents a circle with its center at the origin and a radius of 9 units.
(ii) For the coordinates \( (9 \cos \alpha, 6 \sin \alpha) \):
Let P be any point \( (h, k) \) on the required path. From the given information, we have:
\( h = 9 \cos \alpha \)
\( k = 6 \sin \alpha \)
We can rewrite these equations as:
\( \frac{h}{9} = \cos \alpha \)
\( \frac{k}{6} = \sin \alpha \)
Again, using the trigonometric identity: \( \cos^2 \alpha + \sin^2 \alpha = 1 \).
\( \implies \) Substitute the expressions for \( \cos \alpha \) and \( \sin \alpha \):
\( \left(\frac{h}{9}\right)^2 + \left(\frac{k}{6}\right)^2 = 1 \)
\( \implies \frac{h^2}{81} + \frac{k^2}{36} = 1 \)
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( \frac{x^2}{81} + \frac{y^2}{36} = 1 \). This equation represents an ellipse centered at the origin.
In simple words: For each case, we take the given coordinates and use a basic trigonometry rule: \( \cos^2 \alpha + \sin^2 \alpha = 1 \). We change the expressions involving \( h \) and \( k \) to match this rule. Then, we simplify the equation and replace \( h \) with \( x \) and \( k \) with \( y \) to find the path (locus) that the point P follows.

🎯 Exam Tip: Remember that \( x^2 + y^2 = r^2 \) represents a circle, and \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) represents an ellipse. Identifying these standard forms is key to solving locus problems.

Question 2. Find the locus of a point P that moves a constant distant of
(i) two units from the x-axis
(ii) three units from the y-axis.

Answer:
(i) Two units from the x-axis:
Let P be any point \( (h, k) \) on the required path. The distance of a point \( (h, k) \) from the x-axis is given by \( |k| \).
According to the problem, this distance is 2 units. So, \( |k| = 2 \), which means \( k = 2 \) or \( k = -2 \). Typically, when a constant distance is specified without direction, it implies the absolute value.
If we consider the coordinate plane, the points will form two horizontal lines.
The locus of P \( (h, k) \) is found by replacing \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( y = 2 \) or \( y = -2 \). These are equations of two horizontal lines parallel to the x-axis.

(ii) Three units from the y-axis:
Let P be any point \( (h, k) \) on the required path. The distance of a point \( (h, k) \) from the y-axis is given by \( |h| \).
According to the problem, this distance is 3 units. So, \( |h| = 3 \), which means \( h = 3 \) or \( h = -3 \).
If we consider the coordinate plane, the points will form two vertical lines.
The locus of P \( (h, k) \) is found by replacing \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( x = 3 \) or \( x = -3 \). These are equations of two vertical lines parallel to the y-axis.

In simple words: When a point stays a fixed distance from the x-axis, it creates a horizontal line. When it stays a fixed distance from the y-axis, it creates a vertical line. Because distance can be in either direction (positive or negative), we get two lines in each case.

🎯 Exam Tip: Remember that the distance from the x-axis is given by the absolute value of the y-coordinate \( |k| \), and the distance from the y-axis is given by the absolute value of the x-coordinate \( |h| \).

Question 3. If \( \theta \) is a parameter, find the equation of the locus of a moving point, whose coordinates are \( x = a \cos^3 \theta, y = a \sin^3 \theta \).

Answer:
The given coordinates of the moving point P are \( (a \cos^3 \theta, a \sin^3 \theta) \).
Let the point P be \( (h, k) \). So we have:
\( h = a \cos^3 \theta \)
\( k = a \sin^3 \theta \)
From these equations, we can write:
\( \frac{h}{a} = \cos^3 \theta \)
\( \frac{k}{a} = \sin^3 \theta \)
Now, we take the cube root of both sides for each equation:
\( \left(\frac{h}{a}\right)^{1/3} = \cos \theta \)
\( \left(\frac{k}{a}\right)^{1/3} = \sin \theta \)
We use the trigonometric identity: \( \cos^2 \theta + \sin^2 \theta = 1 \). This identity helps us remove the parameter \( \theta \).
\( \implies \) Substitute the expressions for \( \cos \theta \) and \( \sin \theta \):
\( \left[ \left(\frac{h}{a}\right)^{1/3} \right]^2 + \left[ \left(\frac{k}{a}\right)^{1/3} \right]^2 = 1 \)
\( \implies \left(\frac{h}{a}\right)^{2/3} + \left(\frac{k}{a}\right)^{2/3} = 1 \)
\( \implies \frac{h^{2/3}}{a^{2/3}} + \frac{k^{2/3}}{a^{2/3}} = 1 \)
\( \implies h^{2/3} + k^{2/3} = a^{2/3} \)
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( x^{2/3} + y^{2/3} = a^{2/3} \). This type of curve is called an astroid when \( a \) is a positive constant.
In simple words: We are given the x and y positions of a point using a special angle \( \theta \). We want to find a rule for the point's path that does not use \( \theta \). So, we first find \( \cos \theta \) and \( \sin \theta \) from the given equations. Then, we use the rule \( \cos^2 \theta + \sin^2 \theta = 1 \) to combine them. After simplifying, we get an equation that describes the point's path using only x and y.

🎯 Exam Tip: When eliminating a parameter like \( \theta \), look for common trigonometric identities like \( \sin^2 \theta + \cos^2 \theta = 1 \) or \( \sec^2 \theta - \tan^2 \theta = 1 \) that can connect the expressions.

Question 4. Find the values of k and b. If the points P(-3, 1) and Q (2, b) lie on the locus of \( x^2 - 5x + ky = 0 \).

Answer:
The given locus equation is \( x^2 - 5x + ky = 0 \).
Since point P\( (-3, 1) \) lies on the locus, it must satisfy the equation. Substitute \( x = -3 \) and \( y = 1 \) into the equation:
\( (-3)^2 - 5(-3) + k(1) = 0 \)
\( \implies 9 + 15 + k = 0 \)
\( \implies 24 + k = 0 \)
\( \implies k = -24 \). This determines the value of the constant k.
Now that we have the value of \( k \), the equation of the locus becomes \( x^2 - 5x - 24y = 0 \).
Since point Q\( (2, b) \) also lies on the locus, it must satisfy this updated equation. Substitute \( x = 2 \) and \( y = b \) into the equation:
\( (2)^2 - 5(2) - 24(b) = 0 \)
\( \implies 4 - 10 - 24b = 0 \)
\( \implies -6 - 24b = 0 \)
\( \implies -24b = 6 \)
\( \implies b = \frac{6}{-24} \)
\( \implies b = -\frac{1}{4} \). This determines the value of b.
In simple words: We are given an equation for a path and two points that are on this path. We use the first point to find the unknown 'k' in the equation. Once 'k' is known, we use the second point and the updated equation to find the unknown 'b'. Each point helps us solve for one missing value.

🎯 Exam Tip: When points lie on a given locus (or curve), their coordinates must satisfy the equation of that locus. Substitute the coordinates carefully to find unknown constants.

Question 5. A straight rod of length 8 units slides with its ends A and B always on the x and y axes respectively, then find the locus of the midpoint of the line segment AB.

Answer:
Let the straight rod have length 8 units. Its ends A and B are on the x-axis and y-axis respectively.
Let point A be \( (a, 0) \) on the x-axis and point B be \( (0, b) \) on the y-axis.
The distance between A and B is the length of the rod, which is 8 units.
Using the distance formula, \( \sqrt{(a-0)^2 + (0-b)^2} = 8 \).
\( \implies \sqrt{a^2 + b^2} = 8 \)
\( \implies a^2 + b^2 = 8^2 \)
\( \implies a^2 + b^2 = 64 \). This is our first important relationship.
Let M\( (h, k) \) be the midpoint of the line segment AB.
Using the midpoint formula:
\( h = \frac{a+0}{2} \)
\( k = \frac{0+b}{2} \)
From these, we can express \( a \) and \( b \) in terms of \( h \) and \( k \):
\( a = 2h \)
\( b = 2k \)
Now, substitute these values of \( a \) and \( b \) into the relationship \( a^2 + b^2 = 64 \):
\( (2h)^2 + (2k)^2 = 64 \)
\( \implies 4h^2 + 4k^2 = 64 \)
\( \implies 4(h^2 + k^2) = 64 \)
\( \implies h^2 + k^2 = \frac{64}{4} \)
\( \implies h^2 + k^2 = 16 \). The rod forms the hypotenuse of a right-angled triangle with the axes.
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( x^2 + y^2 = 16 \). This equation represents a circle centered at the origin with a radius of 4 units.

In simple words: Imagine a ladder sliding down a wall. The ends are on the floor and wall. We want to find the path of the middle point of the ladder. We use the distance formula for the ladder's length and the midpoint formula. Then we combine these using algebraic steps. The path of the midpoint turns out to be a circle.

🎯 Exam Tip: For problems involving a line segment with fixed length, forming a right-angled triangle with the axes, consider using the Pythagorean theorem and the midpoint formula to relate the coordinates of the midpoint.

Question 6. Find the equation of the locus of a point such that the sum of the squares of the distance from the points (3, 5), (1, -1) is equal to 20.

Answer:
Let P be the moving point \( (h, k) \).
Let the two fixed points be A\( (3, 5) \) and B\( (1, -1) \).
The problem states that the sum of the squares of the distances from P to A and P to B is 20. This means \( PA^2 + PB^2 = 20 \).
First, calculate \( PA^2 \) using the distance formula: \( PA^2 = (h-3)^2 + (k-5)^2 \).
Next, calculate \( PB^2 \) using the distance formula: \( PB^2 = (h-1)^2 + (k-(-1))^2 = (h-1)^2 + (k+1)^2 \).
Now, substitute these into the given condition \( PA^2 + PB^2 = 20 \):
\( (h-3)^2 + (k-5)^2 + (h-1)^2 + (k+1)^2 = 20 \)
Expand each squared term:
\( (h^2 - 6h + 9) + (k^2 - 10k + 25) + (h^2 - 2h + 1) + (k^2 + 2k + 1) = 20 \)
Combine like terms:
\( 2h^2 - 8h + 10 + 2k^2 - 8k + 26 = 20 \)
\( 2h^2 + 2k^2 - 8h - 8k + 36 = 20 \)
Move 20 to the left side:
\( 2h^2 + 2k^2 - 8h - 8k + 36 - 20 = 0 \)
\( 2h^2 + 2k^2 - 8h - 8k + 16 = 0 \)
Divide the entire equation by 2 to simplify:
\( h^2 + k^2 - 4h - 4k + 8 = 0 \). This is an equation for a circle.
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( x^2 + y^2 - 4x - 4y + 8 = 0 \).
In simple words: We start with a moving point P and two fixed points A and B. We are told that if we find the distance from P to A, square it, and then find the distance from P to B, square it, and add these two squared distances, the total is always 20. We use the distance formula, expand everything, and simplify the equation. The final equation describes the path (locus) of point P.

🎯 Exam Tip: Remember the distance formula \( d^2 = (x_2-x_1)^2 + (y_2-y_1)^2 \) and carefully expand squared binomials like \( (a-b)^2 = a^2 - 2ab + b^2 \).

Question 7. Find the equation of the locus of the point P such that the line segment AB, joining the points A(1,-6) and B(4, – 2) subtends a right angle at P.

Answer:
Let the moving point be P\( (h, k) \).
The given points are A\( (1, -6) \) and B\( (4, -2) \).
The problem states that the line segment AB subtends a right angle at P. This means that the angle \( \angle APB \) is \( 90^\circ \).
If \( \angle APB = 90^\circ \), then the triangle \( \triangle APB \) is a right-angled triangle, with the right angle at P. In a right-angled triangle, the Pythagorean theorem states that the square of the hypotenuse is equal to the sum of the squares of the other two sides. Here, AB is the hypotenuse.
So, \( AB^2 = PA^2 + PB^2 \).
First, calculate \( AB^2 \) using the distance formula between A\( (1, -6) \) and B\( (4, -2) \):
\( AB^2 = (4-1)^2 + (-2 - (-6))^2 \)
\( AB^2 = (3)^2 + (-2+6)^2 \)
\( AB^2 = 3^2 + 4^2 \)
\( AB^2 = 9 + 16 \)
\( AB^2 = 25 \).
Next, calculate \( PA^2 \) using the distance formula between P\( (h, k) \) and A\( (1, -6) \):
\( PA^2 = (h-1)^2 + (k-(-6))^2 = (h-1)^2 + (k+6)^2 \).
Then, calculate \( PB^2 \) using the distance formula between P\( (h, k) \) and B\( (4, -2) \):
\( PB^2 = (h-4)^2 + (k-(-2))^2 = (h-4)^2 + (k+2)^2 \).
Now, substitute these into the Pythagorean theorem \( AB^2 = PA^2 + PB^2 \):
\( 25 = (h-1)^2 + (k+6)^2 + (h-4)^2 + (k+2)^2 \)
Expand each squared term:
\( 25 = (h^2 - 2h + 1) + (k^2 + 12k + 36) + (h^2 - 8h + 16) + (k^2 + 4k + 4) \)
Combine like terms:
\( 25 = 2h^2 - 10h + 17 + 2k^2 + 16k + 40 \)
\( 25 = 2h^2 + 2k^2 - 10h + 16k + 57 \)
Move 25 to the right side:
\( 0 = 2h^2 + 2k^2 - 10h + 16k + 57 - 25 \)
\( 0 = 2h^2 + 2k^2 - 10h + 16k + 32 \)
Divide the entire equation by 2 to simplify:
\( h^2 + k^2 - 5h + 8k + 16 = 0 \). This represents a circle.
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( x^2 + y^2 - 5x + 8y + 16 = 0 \).

In simple words: When a line segment forms a right angle at a point, it means that if you connect the ends of the segment to that point, you get a right-angled triangle. We use the famous Pythagorean theorem for right triangles, which states that \( AB^2 = PA^2 + PB^2 \). We calculate all these distances, plug them into the theorem, and then simplify the equation to find the rule (locus) that the moving point P follows.

🎯 Exam Tip: When a line segment subtends a right angle at a point, it implies the use of the Pythagorean theorem. Alternatively, the product of the slopes of PA and PB would be -1, but the distance formula approach is often more straightforward for these types of questions.

Question 8. If O is origin and R is a variable point on \( y^2 = 4x \), then find the equation of the locus of the mid-point of segment OR.

Answer:
Let O be the origin \( (0, 0) \).
Let R be a variable point on the curve \( y^2 = 4x \). So, let R be \( (x_R, y_R) \), where \( y_R^2 = 4x_R \).
Let M be the midpoint of the segment OR. Let M be \( (h, k) \).
Using the midpoint formula for O\( (0, 0) \) and R\( (x_R, y_R) \):
\( h = \frac{0 + x_R}{2} \)
\( k = \frac{0 + y_R}{2} \)
From these equations, we can express \( x_R \) and \( y_R \) in terms of \( h \) and \( k \):
\( x_R = 2h \)
\( y_R = 2k \). These equations show how the coordinates of R relate to the midpoint M.
Since R\( (x_R, y_R) \) lies on the curve \( y^2 = 4x \), substitute the expressions for \( x_R \) and \( y_R \) into the curve equation:
\( (2k)^2 = 4(2h) \)
\( \implies 4k^2 = 8h \)
Divide both sides by 4:
\( \implies k^2 = 2h \).
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( y^2 = 2x \). This equation represents a parabola.

In simple words: We have a point O at the center and another point R that moves along a specific curved path (a parabola). We want to find the path of the middle point (M) between O and R. By using the midpoint formula and the equation of R's path, we can find a new equation for M. This new equation shows the path of M, which is also a parabola.

🎯 Exam Tip: When dealing with the midpoint of a segment where one endpoint is fixed (like the origin) and the other is variable on a curve, express the variable endpoint's coordinates in terms of the midpoint's coordinates and substitute them into the curve's equation.

Question 9. The coordinates of a moving point P are \( \left(\frac{a}{2} (\operatorname{cosec} \theta + \sin \theta), \frac{b}{2} (\operatorname{cosec} \theta - \sin \theta)\right) \) where \( \theta \) is a variable parameter. Show that the equation of the locus P is \( b^2 x^2 - a^2 y^2 = a^2 b^2 \).

Answer:
Let the moving point P be \( (h, k) \). From the given coordinates, we have:
\( h = \frac{a}{2} (\operatorname{cosec} \theta + \sin \theta) \) --- (1)
\( k = \frac{b}{2} (\operatorname{cosec} \theta - \sin \theta) \) --- (2)
Rearrange these equations to isolate the trigonometric terms:
From (1): \( \frac{2h}{a} = \operatorname{cosec} \theta + \sin \theta \) --- (3)
From (2): \( \frac{2k}{b} = \operatorname{cosec} \theta - \sin \theta \) --- (4)
Now, we want to eliminate the parameter \( \theta \). We can do this by adding and subtracting equations (3) and (4).
Adding (3) and (4):
\( \frac{2h}{a} + \frac{2k}{b} = (\operatorname{cosec} \theta + \sin \theta) + (\operatorname{cosec} \theta - \sin \theta) \)
\( \implies \frac{2h}{a} + \frac{2k}{b} = 2 \operatorname{cosec} \theta \)
\( \implies \frac{h}{a} + \frac{k}{b} = \operatorname{cosec} \theta \) --- (5)
Subtracting (4) from (3):
\( \frac{2h}{a} - \frac{2k}{b} = (\operatorname{cosec} \theta + \sin \theta) - (\operatorname{cosec} \theta - \sin \theta) \)
\( \implies \frac{2h}{a} - \frac{2k}{b} = 2 \sin \theta \)
\( \implies \frac{h}{a} - \frac{k}{b} = \sin \theta \) --- (6)
We know the identity \( \operatorname{cosec} \theta \cdot \sin \theta = 1 \). This identity is useful for eliminating \( \theta \).
Multiply equation (5) by equation (6):
\( \left(\frac{h}{a} + \frac{k}{b}\right) \left(\frac{h}{a} - \frac{k}{b}\right) = \operatorname{cosec} \theta \cdot \sin \theta \)
\( \implies \left(\frac{h}{a}\right)^2 - \left(\frac{k}{b}\right)^2 = 1 \)
\( \implies \frac{h^2}{a^2} - \frac{k^2}{b^2} = 1 \)
To remove the denominators, multiply the entire equation by \( a^2 b^2 \):
\( \implies b^2 h^2 - a^2 k^2 = a^2 b^2 \).
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( b^2 x^2 - a^2 y^2 = a^2 b^2 \). This equation represents a hyperbola.
In simple words: We are given the x and y coordinates of point P, which depend on an angle \( \theta \). Our goal is to find a rule (locus) for P that does not use \( \theta \). We first rearrange the given equations to find expressions for \( \operatorname{cosec} \theta \) and \( \sin \theta \). Then, we use the simple trigonometric rule \( \operatorname{cosec} \theta \times \sin \theta = 1 \) to combine these expressions. After some algebra, we get the final equation for the path of P in terms of x and y.

🎯 Exam Tip: When dealing with sum and difference of trigonometric functions, adding and subtracting the equations is a powerful technique. Look for identities like \( \operatorname{cosec} \theta \sin \theta = 1 \) or \( \sec \theta \cos \theta = 1 \) to eliminate parameters after algebraic manipulation.

Question 10. If P (2, – 7) is given point and Q is a point on \( 2x^2 + 9y^2 = 18 \) then find the equations of the locus of the midpoint of PQ.

Answer:
Let the given fixed point be P\( (2, -7) \).
Let Q be a variable point on the ellipse \( 2x^2 + 9y^2 = 18 \). So, let Q be \( (x_Q, y_Q) \), where \( 2x_Q^2 + 9y_Q^2 = 18 \).
Let M be the midpoint of the segment PQ. Let M be \( (h, k) \).
Using the midpoint formula for P\( (2, -7) \) and Q\( (x_Q, y_Q) \):
\( h = \frac{2 + x_Q}{2} \)
\( k = \frac{-7 + y_Q}{2} \)
From these equations, we can express \( x_Q \) and \( y_Q \) in terms of \( h \) and \( k \):
\( 2h = 2 + x_Q \implies x_Q = 2h - 2 \)
\( 2k = -7 + y_Q \implies y_Q = 2k + 7 \). These expressions connect the variable point Q to the midpoint M.
Since Q\( (x_Q, y_Q) \) lies on the ellipse \( 2x^2 + 9y^2 = 18 \), substitute the expressions for \( x_Q \) and \( y_Q \) into the ellipse equation:
\( 2(2h - 2)^2 + 9(2k + 7)^2 = 18 \)
Expand the squared terms:
\( 2[ (2h)^2 - 2(2h)(2) + 2^2 ] + 9[ (2k)^2 + 2(2k)(7) + 7^2 ] = 18 \)
\( 2[ 4h^2 - 8h + 4 ] + 9[ 4k^2 + 28k + 49 ] = 18 \)
Distribute the 2 and the 9:
\( 8h^2 - 16h + 8 + 36k^2 + 252k + 441 = 18 \)
Combine the constant terms:
\( 8h^2 + 36k^2 - 16h + 252k + 449 = 18 \)
Move 18 to the left side:
\( 8h^2 + 36k^2 - 16h + 252k + 449 - 18 = 0 \)
\( 8h^2 + 36k^2 - 16h + 252k + 431 = 0 \).
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( 8x^2 + 36y^2 - 16x + 252y + 431 = 0 \). This is also the equation of an ellipse.
In simple words: We have a fixed point P and another point Q that moves on an ellipse. We want to find the path (locus) of the midpoint of the line segment PQ. We use the midpoint formula to connect the coordinates of Q to the coordinates of the midpoint. Then, we substitute these relationships into the equation of the ellipse on which Q moves. After careful expansion and simplification, we get an equation that describes the path of the midpoint.

🎯 Exam Tip: When finding the locus of a midpoint between a fixed point and a point on a curve, always express the coordinates of the point on the curve in terms of the midpoint's coordinates. This allows for direct substitution into the curve's equation.

Question 11. If R is any point on the x-axis and Q is any point on the y-axis and P is a variable point on RQ with RP = b, PQ = a, then find the equation of locus of P.

Answer:
Let R be a point on the x-axis, so its coordinates are \( (x_R, 0) \).
Let Q be a point on the y-axis, so its coordinates are \( (0, y_Q) \).
Let P be the variable point \( (h, k) \) on the line segment RQ.
We are given that P divides the line segment RQ such that \( RP = b \) and \( PQ = a \). This means P divides RQ internally in the ratio \( a:b \) (from Q to R).
Using the section formula for internal division:
For the x-coordinate of P: \( h = \frac{a \cdot x_R + b \cdot 0}{a+b} \)
\( \implies h = \frac{a x_R}{a+b} \)
\( \implies x_R = \frac{(a+b)h}{a} \) --- (1)
For the y-coordinate of P: \( k = \frac{a \cdot 0 + b \cdot y_Q}{a+b} \)
\( \implies k = \frac{b y_Q}{a+b} \)
\( \implies y_Q = \frac{(a+b)k}{b} \) --- (2)
Now, consider the right-angled triangle formed by the origin O, R, and Q. The hypotenuse is RQ.
The length \( RQ \) can be found using the distance formula: \( RQ = \sqrt{(x_R - 0)^2 + (0 - y_Q)^2} = \sqrt{x_R^2 + y_Q^2} \).
Also, the total length of the segment RQ is \( RP + PQ = b + a \).
So, \( \sqrt{x_R^2 + y_Q^2} = a+b \).
Squaring both sides: \( x_R^2 + y_Q^2 = (a+b)^2 \). This relationship comes from the geometric setup.
Substitute the expressions for \( x_R \) and \( y_Q \) from (1) and (2) into this equation:
\( \left(\frac{(a+b)h}{a}\right)^2 + \left(\frac{(a+b)k}{b}\right)^2 = (a+b)^2 \)
\( \implies \frac{(a+b)^2 h^2}{a^2} + \frac{(a+b)^2 k^2}{b^2} = (a+b)^2 \)
Divide the entire equation by \( (a+b)^2 \) (since \( a+b \neq 0 \)):
\( \implies \frac{h^2}{a^2} + \frac{k^2}{b^2} = 1 \).
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). This equation represents an ellipse.
In simple words: We have a point R on the x-axis and a point Q on the y-axis. A third point P lies on the line connecting R and Q, dividing it into two pieces of length 'a' and 'b'. We use the section formula to relate the coordinates of P to R and Q. We also know that the segment RQ forms the hypotenuse of a right triangle. By combining these ideas and simplifying, we find that the path of point P is an ellipse.

🎯 Exam Tip: When a point divides a segment in a given ratio, use the section formula carefully. If the segment's ends are on the coordinate axes, remember that the segment forms the hypotenuse of a right triangle, allowing the use of the Pythagorean theorem.

Question 12. If the points P (6, 2) and Q (- 2, 1) and R are the vertices of a \( \triangle PQR \) and R is the point on the locus \( y = x^2 – 3x + 4 \) then find the equation of the locus of the centroid of \( \triangle PQR \).

Answer:
Let the given fixed vertices be P\( (6, 2) \) and Q\( (-2, 1) \).
Let R be a variable vertex with coordinates \( (a, b) \).
We are given that R\( (a, b) \) lies on the locus \( y = x^2 - 3x + 4 \).
So, substitute \( x=a \) and \( y=b \) into the locus equation for R:
\( b = a^2 - 3a + 4 \) --- (1). This equation connects the coordinates of point R.
Let G be the centroid of \( \triangle PQR \). Let G be \( (h, k) \).
Using the centroid formula for vertices P\( (6, 2) \), Q\( (-2, 1) \), and R\( (a, b) \):
\( h = \frac{6 + (-2) + a}{3} \)
\( k = \frac{2 + 1 + b}{3} \)
Simplify these equations:
\( h = \frac{4 + a}{3} \)
\( k = \frac{3 + b}{3} \)
From these, we can express \( a \) and \( b \) in terms of \( h \) and \( k \):
\( 3h = 4 + a \implies a = 3h - 4 \)
\( 3k = 3 + b \implies b = 3k - 3 \). These equations connect the variable vertex R to the centroid G.
Now, substitute these expressions for \( a \) and \( b \) into equation (1):
\( (3k - 3) = (3h - 4)^2 - 3(3h - 4) + 4 \)
Expand and simplify the equation:
\( 3k - 3 = (9h^2 - 24h + 16) - (9h - 12) + 4 \)
\( 3k - 3 = 9h^2 - 24h + 16 - 9h + 12 + 4 \)
Combine like terms:
\( 3k - 3 = 9h^2 - 33h + 32 \)
Move all terms to one side, usually to keep \( k \) positive:
\( 0 = 9h^2 - 33h - 3k + 32 + 3 \)
\( 0 = 9h^2 - 33h - 3k + 35 \).
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( 9x^2 - 33x - 3y + 35 = 0 \). This is an equation of a parabola.
In simple words: We have a triangle with two fixed corners and one corner (R) that moves along a specific curved path. We want to find the path of the triangle's centroid (center point). First, we use the rule for point R's path. Then, we use the centroid formula to connect the coordinates of R to the centroid. By putting these relationships together and simplifying the algebra, we get a new equation that shows the path of the centroid.

🎯 Exam Tip: When finding the locus of the centroid of a triangle where one vertex is variable on a given curve, express the coordinates of the variable vertex in terms of the centroid's coordinates using the centroid formula, then substitute these into the curve's equation.

Question 13. If Q is a point on the locus of \( x^2 + y^2 + 4x - 3y +7 = 0 \), then find the equation of locus of P which divides segment OQ externally in the ratio 3 :4 where O is origin.

Answer:
Let O be the origin \( (0, 0) \).
Let Q be a variable point on the given locus, so let Q be \( (a, b) \).
Since Q\( (a, b) \) lies on the locus \( x^2 + y^2 + 4x - 3y + 7 = 0 \), it must satisfy the equation:
\( a^2 + b^2 + 4a - 3b + 7 = 0 \) --- (1).
Let P be the movable point \( (h, k) \).
We are given that P divides the segment OQ externally in the ratio 3:4.
This means P divides OQ such that \( OP:PQ = 3:4 \). For external division, the ratio \( m:n \) can be thought of as \( m=3 \) and \( n=-4 \).
Using the section formula for external division with O\( (0, 0) \) and Q\( (a, b) \), and ratio \( m:n = 3:(-4) \):
For the x-coordinate of P:
\( h = \frac{3 \cdot a + (-4) \cdot 0}{3 - 4} \)
\( \implies h = \frac{3a}{-1} \)
\( \implies h = -3a \implies a = -\frac{h}{3} \) --- (2)
For the y-coordinate of P:
\( k = \frac{3 \cdot b + (-4) \cdot 0}{3 - 4} \)
\( \implies k = \frac{3b}{-1} \)
\( \implies k = -3b \implies b = -\frac{k}{3} \) --- (3)
Now, substitute these expressions for \( a \) and \( b \) from (2) and (3) into equation (1):
\( \left(-\frac{h}{3}\right)^2 + \left(-\frac{k}{3}\right)^2 + 4\left(-\frac{h}{3}\right) - 3\left(-\frac{k}{3}\right) + 7 = 0 \)
\( \implies \frac{h^2}{9} + \frac{k^2}{9} - \frac{4h}{3} + \frac{3k}{3} + 7 = 0 \)
\( \implies \frac{h^2}{9} + \frac{k^2}{9} - \frac{4h}{3} + k + 7 = 0 \)
To remove the denominators, multiply the entire equation by 9:
\( \implies h^2 + k^2 - 12h + 9k + 63 = 0 \).
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( x^2 + y^2 - 12x + 9y + 63 = 0 \). This equation represents a circle.
In simple words: We have a point O at the center and another point Q that moves on a curved path (a circle). A third point P divides the line segment OQ in a special way (externally in a ratio). We use the section formula for external division to connect P's coordinates to Q's coordinates. Then, we substitute these relationships into the equation for Q's path. After doing the math, we find the new equation that describes the path of P.

🎯 Exam Tip: For external division, if P divides OQ externally in ratio \( m:n \), the formula uses \( m \) and \( -n \) or \( -m \) and \( n \) (as done here). Be careful with the signs. Also, simplifying algebraic fractions correctly is important.

Question 14. Find the points on the locus of points that are 3 units from the x-axis and 5 units from the point (5, 1).

Answer:
Let the required point be P\( (h, k) \).
Condition 1: The point P is 3 units from the x-axis.
This means the absolute value of the y-coordinate is 3. So, \( |k| = 3 \), which implies \( k = 3 \) or \( k = -3 \).
Condition 2: The point P is 5 units from the point \( (5, 1) \).
Using the distance formula, the square of the distance between P\( (h, k) \) and \( (5, 1) \) is \( (h-5)^2 + (k-1)^2 \).
Given this distance is 5, so the squared distance is \( 5^2 = 25 \).
\( (h-5)^2 + (k-1)^2 = 25 \).
Now we need to find the points \( (h, k) \) that satisfy both conditions.
Case 1: Let \( k = 3 \).
Substitute \( k=3 \) into the second condition's equation:
\( (h-5)^2 + (3-1)^2 = 25 \)
\( (h-5)^2 + (2)^2 = 25 \)
\( (h-5)^2 + 4 = 25 \)
\( (h-5)^2 = 25 - 4 \)
\( (h-5)^2 = 21 \)
Take the square root of both sides:
\( h-5 = \pm \sqrt{21} \)
\( h = 5 \pm \sqrt{21} \)
This gives two possible points: \( (5 + \sqrt{21}, 3) \) and \( (5 - \sqrt{21}, 3) \).
Case 2: Let \( k = -3 \).
Substitute \( k=-3 \) into the second condition's equation:
\( (h-5)^2 + (-3-1)^2 = 25 \)
\( (h-5)^2 + (-4)^2 = 25 \)
\( (h-5)^2 + 16 = 25 \)
\( (h-5)^2 = 25 - 16 \)
\( (h-5)^2 = 9 \)
Take the square root of both sides:
\( h-5 = \pm \sqrt{9} \)
\( h-5 = \pm 3 \)
For \( h-5 = 3 \implies h = 3 + 5 \implies h = 8 \)
For \( h-5 = -3 \implies h = -3 + 5 \implies h = 2 \)
This gives two more possible points: \( (8, -3) \) and \( (2, -3) \).
Therefore, the required points are \( (5 + \sqrt{21}, 3) \), \( (5 - \sqrt{21}, 3) \), \( (8, -3) \), and \( (2, -3) \).

In simple words: We are looking for points that meet two conditions at once. First, they must be 3 units away from the x-axis, which means their y-coordinate is either 3 or -3. Second, they must be 5 units away from a specific point (5, 1). We use the distance formula for the second condition. Then, we solve for the x-coordinate by testing both y-values (3 and -3). This gives us a total of four points that satisfy both rules.

🎯 Exam Tip: When a point must satisfy multiple conditions, solve for the coordinates by combining the equations. Remember that "distance from an axis" usually refers to the absolute value of the corresponding coordinate, leading to two possibilities (positive and negative values).

Question 15. The sum of the distance of a moving point from the points (4, 0) and (- 4, 0) is always 10 units. Find the equation to the locus of the moving point.

Answer:
Let the moving point be P\( (h, k) \).
Let the two fixed points be A\( (4, 0) \) and B\( (-4, 0) \). These are called the foci of the ellipse.
The problem states that the sum of the distances from P to A and P to B is always 10 units. This means \( PA + PB = 10 \).
First, calculate PA using the distance formula:
\( PA = \sqrt{(h-4)^2 + (k-0)^2} = \sqrt{(h-4)^2 + k^2} \).
Next, calculate PB using the distance formula:
\( PB = \sqrt{(h-(-4))^2 + (k-0)^2} = \sqrt{(h+4)^2 + k^2} \).
Now, substitute these into the given condition \( PA + PB = 10 \):
\( \sqrt{(h-4)^2 + k^2} + \sqrt{(h+4)^2 + k^2} = 10 \)
Move one square root term to the right side to prepare for squaring:
\( \sqrt{(h-4)^2 + k^2} = 10 - \sqrt{(h+4)^2 + k^2} \)
Square both sides of the equation:
\( (h-4)^2 + k^2 = \left(10 - \sqrt{(h+4)^2 + k^2}\right)^2 \)
\( h^2 - 8h + 16 + k^2 = 100 - 20\sqrt{(h+4)^2 + k^2} + (h+4)^2 + k^2 \)
\( h^2 - 8h + 16 + k^2 = 100 - 20\sqrt{(h+4)^2 + k^2} + h^2 + 8h + 16 + k^2 \)
Notice that \( h^2 \), \( k^2 \), and 16 appear on both sides and can be cancelled out:
\( -8h = 100 - 20\sqrt{(h+4)^2 + k^2} + 8h \)
Gather terms without the square root on one side and the square root term on the other:
\( -8h - 8h - 100 = -20\sqrt{(h+4)^2 + k^2} \)
\( -16h - 100 = -20\sqrt{(h+4)^2 + k^2} \)
Divide the entire equation by -4 to simplify:
\( 4h + 25 = 5\sqrt{(h+4)^2 + k^2} \)
Square both sides again to remove the remaining square root:
\( (4h + 25)^2 = \left(5\sqrt{(h+4)^2 + k^2}\right)^2 \)
\( (4h)^2 + 2(4h)(25) + 25^2 = 25 \left( (h+4)^2 + k^2 \right) \)
\( 16h^2 + 200h + 625 = 25 (h^2 + 8h + 16 + k^2) \)
\( 16h^2 + 200h + 625 = 25h^2 + 200h + 400 + 25k^2 \)
Notice that \( 200h \) appears on both sides and can be cancelled.
Move all terms to one side, usually keeping \( h^2 \) positive:
\( 0 = 25h^2 - 16h^2 + 25k^2 + 400 - 625 \)
\( 0 = 9h^2 + 25k^2 - 225 \)
Rearrange the terms:
\( 9h^2 + 25k^2 = 225 \)
To normalize this to the standard form of an ellipse, divide by 225:
\( \frac{9h^2}{225} + \frac{25k^2}{225} = \frac{225}{225} \)
\( \frac{h^2}{25} + \frac{k^2}{9} = 1 \).
To find the locus, we replace \( h \) with \( x \) and \( k \) with \( y \).
Therefore, the required locus is \( \frac{x^2}{25} + \frac{y^2}{9} = 1 \). This is the equation of an ellipse.
In simple words: We are looking for the path of a point P such that the total distance from P to two fixed points (A and B) is always 10. This special kind of path is an ellipse. We use the distance formula to write down the lengths PA and PB. Then, we add them and set the sum to 10. To remove the square roots, we square the equation twice, simplifying after each step. The final equation describes an ellipse.

🎯 Exam Tip: The definition of an ellipse is the locus of a point for which the sum of the distances from two fixed points (foci) is constant. Recognizing this definition early can guide your steps and help confirm your final result.