Samacheer Kalvi Class 12 Maths Solutions Chapter 5 Two Dimensional Analytical Geometry II Exercise 5.6

Question 1. The equation of the circle passing through (1, 5) and (4, 1) and touching y-axis is \( x^2 + y^2 – 5x – 6y + 9 + \lambda (4x + 3y – 19) = 0 \) where \( \lambda \) is equal to
(b) 0
(c) \( \frac {40}{9} \)
(d) \( -\frac {40}{9} \)
Answer: (a) 0, \( -\frac {40}{9} \)
In simple words: We are looking for the value of lambda that makes the given equation a circle passing through two points and also touching the y-axis. By setting x=0 for y-axis tangency, we find the values of lambda that satisfy the condition. This means the quadratic equation for y must have equal roots when x=0.

🎯 Exam Tip: When a circle touches the y-axis, the x-coordinate of its center is equal to its radius (or the absolute value of its radius if the center is in negative x). This means setting x = 0 in the circle's equation results in a quadratic with equal roots.

Question 2. The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half the distance between the foci is
(a) \( \frac {4}{3} \)
(b) \( \frac {4}{3} \)
(c) \( \frac {2}{\sqrt{3}} \)
(d) \( -\frac {3}{2} \)
Answer: (c) \( \frac {2}{\sqrt{3}} \)
In simple words: We need to find how "stretched" the hyperbola is, which is its eccentricity. We use the given information about the latus rectum and the conjugate axis to form equations and solve for the eccentricity. The distance between the foci helps to set up the relationship between the major and minor axes.

🎯 Exam Tip: Remember the key formulas for a hyperbola: length of latus rectum \( = \frac{2b^2}{a} \), length of conjugate axis \( = 2b \), and distance between foci \( = 2ae \). Use these to set up equations and solve for eccentricity (e).

Question 3. The circle \( x^2 + y^2 = 4x + 8y + 5 \) intersects the line \( 3x – 4y = m \) at two distinct points if
(a) \( 15 < m < 65 \)
(b) \( 35 < m < 85 \)
(c) \( -85 < m < -35 \)
(d) \( -35 < m < 15 \)
Answer: (d) \( -35 < m < 15 \)
In simple words: For a line to cut a circle at two different points, the distance from the center of the circle to the line must be less than the circle's radius. We find the center and radius of the circle first, then use the distance formula to find the possible range for 'm'. This condition ensures the line passes through the circle, not just touching it.

🎯 Exam Tip: When solving problems involving a circle and a line, always convert the circle's equation to the standard form \( (x-h)^2 + (y-k)^2 = r^2 \) to easily identify the center \( (h,k) \) and radius \( r \).

Question 4. The length of the diameter of the circle which touches the x-axis at the point (1, 0) and passes through the point (2, 3).
(a) \( \frac {6}{5} \)
(b) \( \frac {5}{3} \)
(c) \( \frac {10}{3} \)
(d) \( \frac {3}{5} \)
Answer: (c) \( \frac {10}{3} \)
In simple words: Since the circle touches the x-axis at (1, 0), its center must be at (1, h). The radius will then be \( |h| \). We use the fact that the distance from the center to any point on the circle (like (2, 3)) is equal to the radius. This allows us to find 'h' and thus the diameter. The y-coordinate of the center is the radius.

🎯 Exam Tip: When a circle touches the x-axis at \( (x_1, 0) \), its center will be \( (x_1, \pm r) \). Similarly, if it touches the y-axis at \( (0, y_1) \), its center will be \( (\pm r, y_1) \). This simplifies finding the equation of the circle significantly.

Question 5. The radius of the circle \( 3x^2 + by^2 + 4bx – 6by + b^2 = 0 \) is
(a) 1
(b) 3
(c) \( \sqrt {10} \)
(d) \( \sqrt {11} \)
Answer: (c) \( \sqrt {10} \)
In simple words: For the given equation to represent a circle, the coefficients of \( x^2 \) and \( y^2 \) must be equal. This helps us find the value of 'b'. After that, we divide the entire equation by this coefficient to get the standard form \( x^2 + y^2 + 2gx + 2fy + c = 0 \), from which the radius can be calculated using the formula \( r = \sqrt{g^2 + f^2 - c} \).

🎯 Exam Tip: Always remember that for a general second-degree equation \( Ax^2 + By^2 + Cxy + Dx + Ey + F = 0 \) to represent a circle, the conditions are \( A = B \neq 0 \) and \( C = 0 \). After ensuring these, divide by A (or B) to standardize the equation before finding the radius.

Question 6. The centre of the circle inscribed in a square formed by the lines \( x^2 – 8x + 12 = 0 \) and \( y^2 – 14y + 45 = 0 \) is
(a) (4, 7)
(b) (7, 4)
(c) (9, 4)
(d) (4, 9)
Answer: (a) (4, 7)
In simple words: The given equations describe pairs of parallel lines that form a square. To find the center of a circle inscribed in this square, we need to find the midpoint of the square's sides. The square is formed by \( x=2, x=6 \) and \( y=5, y=9 \). The center of the square, which is also the center of the inscribed circle, is the midpoint of the diagonal formed by the vertices.

🎯 Exam Tip: The center of a square formed by lines \( x=x_1, x=x_2, y=y_1, y=y_2 \) is simply the midpoint of its diagonals, which is \( (\frac{x_1+x_2}{2}, \frac{y_1+y_2}{2}) \). This applies directly to finding the center of an inscribed circle.

Question 7. The equation of the normal to the circle \( x^2 + y^2 – 2x – 2y + 1 = 0 \) which is parallel to the line \( 2x + 4y = 3 \) is
(a) \( x + 2y = 3 \)
(b) \( x + 2y + 3 = 0 \)
(c) \( 2x + 4y + 3 = 0 \)
(d) \( x – 2y + 3 = 0 \)
Answer: (a) \( x + 2y = 3 \)
In simple words: A normal line to a circle always passes through its center. First, we find the center of the given circle. Since the normal is parallel to another line, it will have the same slope. Using the center and the parallel slope, we can find the equation of the normal. The center of the circle is (1,1).

🎯 Exam Tip: For any circle, the normal at a point on its circumference always passes through the center of the circle. This is a fundamental property that simplifies finding normal equations, as you only need the center and the desired slope.

Question 8. If P(x, y) be any point on \( 16x^2 + 25y^2 = 400 \) with foci F(3, 0) then \( PF_1 + PF_2 \) is
(a) 8
(b) 6
(c) 10
(d) 12
Answer: (c) 10
In simple words: The sum of the distances from any point on an ellipse to its two foci is always equal to the length of its major axis. We first rewrite the ellipse equation to find 'a' and then double 'a' to get the major axis length. The length of the major axis is \( 2a \), which gives us the required sum.

🎯 Exam Tip: A key property of an ellipse is that for any point P on the ellipse, the sum of the distances from P to the two foci (say, \( F_1 \) and \( F_2 \)) is constant and equal to the length of the major axis, \( 2a \). Normalize the ellipse equation to \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) to easily identify 'a'.

Question 9. The radius of the circle passing through the points (6,2) two of whose diameter are \( x + y = 6 \) and \( x + 2y = 4 \) is
(a) 10
(b) \( 2\sqrt{5} \)
(c) 6
(d) 4
Answer: (b) \( 2\sqrt{5} \)
In simple words: The intersection point of any two diameters of a circle is its center. We solve the given equations for the two diameters to find the circle's center. Once we have the center and a point on the circle (6, 2), we can calculate the distance between them, which will be the radius of the circle.

🎯 Exam Tip: Always remember that the center of a circle is the point where all its diameters intersect. Solving the system of equations for any two diameters will directly give you the coordinates of the center of the circle.

Question 10. The area of quadrilateral formed with foci of the hyperbolas \( \frac {x^2}{a^2} - \frac {y^2}{b^2} = 1 \) and \( \frac {x^2}{a^2} – \frac {y^2}{b^2} = -1 \) is
(a) \( 4(a^2 + b^2) \)
(b) \( 2(a^2 + b^2) \)
(c) \( a^2 + b^2 \)
Answer: (b) \( 2(a^2 + b^2) \)
In simple words: The two given hyperbolas are conjugate to each other. Their foci form a quadrilateral. We find the coordinates of the foci for both hyperbolas. The first hyperbola has foci on the x-axis, and the second (conjugate) hyperbola has foci on the y-axis. These four foci create a rhombus, and its area can be calculated using the lengths of its diagonals. The area is half the product of the diagonals.

🎯 Exam Tip: The foci of a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are \( (\pm ae, 0) \). For its conjugate hyperbola \( \frac{y^2}{b^2} - \frac{x^2}{a^2} = 1 \), the foci are \( (0, \pm be) \). The area of a rhombus is \( \frac{1}{2} d_1 d_2 \), where \( d_1 \) and \( d_2 \) are the lengths of the diagonals.

Question 11. If the normals of the parabola \( y^2 = 4x \) drawn at the end points of its latus rectum are tangents to the circle \( (x – 3)^2 + (y + 2)^2 = r^2 \), then the value of \( r^2 \) is
(a) 2
(b) 3
(c) 1
Answer: (a) 2
In simple words: First, we find the focal length 'a' of the parabola. Then we determine the coordinates of the endpoints of the latus rectum. Next, we find the equations of the normal lines at these endpoints. Since these normal lines are tangents to the given circle, the distance from the circle's center (3, -2) to each normal line must be equal to the circle's radius 'r'. This allows us to calculate \( r^2 \).

🎯 Exam Tip: The latus rectum of a parabola \( y^2 = 4ax \) is the chord passing through the focus and perpendicular to the axis. Its endpoints are \( (a, 2a) \) and \( (a, -2a) \). The normal to a curve at a point is perpendicular to the tangent at that point.

Question 12. If \( x + y = k \) is a normal to the parabola \( y^2 = 12x \), then the value of k is 14.
(a) 3
(b) -1
(c) 1
(d) 9
Answer: (d) 9
In simple words: We are given a parabola and a line that is normal to it. For a line \( y = mx + c \) to be a normal to the parabola \( y^2 = 4ax \), there's a specific condition it must meet. We use this condition \( c = -2am - am^3 \) to find the value of k. First, we identify 'a' from the parabola's equation and 'm' and 'c' from the given normal line equation.

🎯 Exam Tip: The condition for the line \( y = mx + c \) to be a normal to the parabola \( y^2 = 4ax \) is \( c = -2am - am^3 \). Always rearrange the given normal line into the \( y = mx + c \) form to correctly identify 'm' and 'c'.

Question 13. The ellipse \( E_1 : \frac {x^2}{9} + \frac {y^2}{4} = 1 \) is inscribed in a rectangle R whose sides are parallel to the co-ordinate axes. Another ellipse \( E_2 \) passing through the point (0, 4) circumscribes the rectangle R. The eccentricity of the ellipse is
(a) \( \frac {\sqrt{2}}{2} \)
(b) \( \frac {\sqrt{3}}{2} \)
(c) \( \frac {1}{2} \)
(d) \( \frac {3}{4} \)
Answer: (c) \( \frac {1}{2} \)
In simple words: We have two ellipses and a rectangle. The first ellipse helps us find the dimensions of the rectangle. The second ellipse passes through a specific point and touches the corners of this rectangle. We use this information to find the equation of the second ellipse and then calculate its eccentricity. The eccentricity measures how "flat" or "round" an ellipse is.

🎯 Exam Tip: When an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) is inscribed in a rectangle with sides parallel to the axes, the vertices of the rectangle are \( (\pm a, \pm b) \). If another ellipse circumscribes this rectangle, it will pass through these vertices. Use this to determine its equation and subsequently its eccentricity.

Question 14. Tangents are drawn to the hyperbola \( \frac {x^2}{9} - \frac {y^2}{4} = 1 \) parallel to the straight line \( 2x – y – 1 \). One of the points of contact of tangents on the hyperbola is
(a) \( (\frac {9}{2\sqrt{2}}, \frac {-1}{\sqrt{2}}) \)
(b) \( (\frac {-9}{2\sqrt{2}}, \frac {1}{\sqrt{2}}) \)
(c) \( (\frac {9}{2\sqrt{2}}, \frac {1}{\sqrt{2}}) \)
(d) \( (3\sqrt{3}, -2\sqrt{2}) \)
Answer: (c) \( (\frac {9}{2\sqrt{2}}, \frac {1}{\sqrt{2}}) \)
In simple words: We need to find the point on the hyperbola where the tangent line is parallel to the given line. Parallel lines have the same slope. First, find the slope of the given line. Then use the formula for the point of contact of a tangent with a specific slope 'm' on a hyperbola. This formula helps us directly find the coordinates.

🎯 Exam Tip: The equation of a tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) with slope 'm' is \( y = mx \pm \sqrt{a^2m^2 - b^2} \). The points of contact \( (x_1, y_1) \) are given by \( (\frac{-a^2m}{c}, \frac{b^2}{c}) \) where \( c = \sqrt{a^2m^2-b^2} \). Always simplify the given line equation to find its slope 'm'.

Question 15. The equation of the circle passing through the foci of the ellipse \( \frac {x^2}{16} + \frac {y^2}{9} = 1 \) having centre at (0, 3) is
(a) \( x^2 + y^2 – 6y – 7 = 0 \)
(b) \( x^2 + y^2 – 6y + 7 = 0 \)
(c) \( x^2 + y^2 – 6y – 5 = 0 \)
(d) \( x^2 + y^2 – 6y = 0 \)
Answer: (a) \( x^2 + y^2 – 6y – 7 = 0 \)
In simple words: First, we find the foci of the given ellipse. Since the circle passes through these foci, we can use one of the foci as a point on the circle. With the circle's center (0, 3) and a point on it (a focus), we can find its radius squared \( (r^2) \). Then, we write the equation of the circle using the standard form \( (x-h)^2 + (y-k)^2 = r^2 \).

🎯 Exam Tip: For an ellipse \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \) where \( a > b \), the foci are located at \( (\pm ae, 0) \). Use the relation \( b^2 = a^2(1 - e^2) \) to find the eccentricity 'e' and then the foci. Remember that the distance from the center of a circle to any point on its circumference is its radius.

Question 16. Let C be the circle with centre at (1, 1) and radius = 1. If T is the circle centered at (0, y) passing through the origin and touching the circle C externally, then the radius of T is equal to
(a) \( \frac {\sqrt{3}}{2} \)
(b) \( \frac {3}{2} \)
(c) \( \frac {1}{2} \)
(d) \( \frac {1}{4} \)
Answer: (d) \( \frac {1}{4} \)
In simple words: We have two circles. Circle C has a known center and radius. Circle T passes through the origin and touches circle C from the outside. The distance between the centers of two externally touching circles is equal to the sum of their radii. We use this condition to form an equation and solve for the unknown y-coordinate of the center of circle T, which will also be its radius because it passes through the origin (0,0) and its center is (0,y).

🎯 Exam Tip: For two circles touching externally, the distance between their centers is equal to the sum of their radii \( (C_1C_2 = r_1 + r_2) \). For two circles touching internally, the distance between their centers is the absolute difference of their radii \( (C_1C_2 = |r_1 - r_2|) \). This is a crucial formula for such problems.

Question 17. Consider an ellipse whose centre is of the origin and its major axis is a long x-axis. If its eccentricity is \( \frac { 3 }{ 5 } \) and the distance between its foci is 6, then the area of the quadrilateral inscribed in the ellipse with diagonals as major and minor axis, of the ellipse is
(a) 8
(b) 32
(c) 80
(d) 40
Answer: (d) 40

Answer: We are given the eccentricity \( e = \frac{3}{5} \). The distance between the foci is \( 2ae = 6 \). We can substitute the value of \( e \) into the equation: \( 2a \left( \frac{3}{5} \right) = 6 \) \( \frac{6a}{5} = 6 \) Now, we solve for \( a \): \( a = \frac{6 \times 5}{6} \implies a = 5 \). Next, we find the value of \( b \) using the relationship \( b^2 = a^2 (1 - e^2) \): \( b^2 = 5^2 \left( 1 - \left( \frac{3}{5} \right)^2 \right) \) \( b^2 = 25 \left( 1 - \frac{9}{25} \right) \) \( b^2 = 25 \left( \frac{25 - 9}{25} \right) \) \( b^2 = 25 \left( \frac{16}{25} \right) \) \( b^2 = 16 \implies b = 4 \). The quadrilateral inscribed in the ellipse with diagonals as major and minor axes has vertices at \( (\pm a, 0) \) and \( (0, \pm b) \). This forms a rhombus. The area of such a quadrilateral is given by \( \frac{1}{2} \times (\text{length of major axis}) \times (\text{length of minor axis}) \). Area \( = \frac{1}{2} \times (2a) \times (2b) = 2ab \). Substitute the values of \( a \) and \( b \): Area \( = 2 \times 5 \times 4 = 40 \). This shows how the dimensions of the ellipse determine the largest possible rhombus that fits inside it.
In simple words: First, use the given eccentricity and distance between foci to find the lengths of the semi-major axis (a) and semi-minor axis (b) of the ellipse. Then, calculate the area of the rhombus formed by these axes using the formula \( 2ab \).

🎯 Exam Tip: Remember the key relationships for an ellipse: \( b^2 = a^2 (1 - e^2) \) and the distance between foci is \( 2ae \). Knowing these formulas helps in finding the dimensions quickly.

Question 18. Area of the greatest rectangle inscribed in the ellipse \( \frac { x^2 }{ 16 } + \frac { y^2 }{ b^2 } \)
(a) 2ab
(b) ab
(c) \( \sqrt {ab} \)
(d) \( \frac {a}{b} \)
Answer: (a) 2ab

Answer: For an ellipse defined by \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \), the coordinates of a point on the ellipse can be written as \( (a \cos \theta, b \sin \theta) \). If we consider a rectangle inscribed within this ellipse, its vertices will be \( (\pm a \cos \theta, \pm b \sin \theta) \). The length of this rectangle will be \( 2a \cos \theta \) and its breadth will be \( 2b \sin \theta \). The area of the rectangle, \( A \), is given by: \( A = (2a \cos \theta) (2b \sin \theta) \) \( A = 4ab \cos \theta \sin \theta \) We can use the trigonometric identity \( \sin(2\theta) = 2 \sin \theta \cos \theta \), so: \( A = 2ab (2 \sin \theta \cos \theta) \) \( A = 2ab \sin(2\theta) \). To find the greatest area, we need to maximize \( \sin(2\theta) \). The maximum value of \( \sin(2\theta) \) is 1, which occurs when \( 2\theta = \frac{\pi}{2} \), or \( \theta = \frac{\pi}{4} \). Thus, the maximum area is \( 2ab \times 1 = 2ab \). This formula is a standard result for inscribed rectangles, showing how the ellipse's semi-axes directly determine the largest possible rectangular area.
In simple words: For an ellipse, the biggest rectangle you can fit inside it will have an area that is twice the product of its semi-major axis (a) and semi-minor axis (b).

🎯 Exam Tip: Remember that the maximum value of \( \sin(2\theta) \) is 1. This identity is crucial for solving optimization problems involving inscribed shapes in ellipses.

Question 19. An ellipse has OB as semi minor axes, F and F' its foci and the angle FBF' is a right angle. Then the eccentricity of the ellipse is
(a) \( \frac { 1 }{ 2 } \)
(b) \( \frac { \sqrt{3} }{ 2 } \)
(c) \( \frac { 1 }{ 4 } \)
(d) \( \frac { 1 }{ 3 } \)
Answer: (a) \( \frac { 1 }{ 2 } \)

Answer: Let the equation of the ellipse be \( \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1 \). The foci are \( F_1(-ae, 0) \) and \( F_2(ae, 0) \). The end of the semi-minor axis is \( B(0, b) \). We are given that \( \angle F_1BF_2 \) is a right angle, so \( \triangle F_1BF_2 \) is a right-angled triangle with the right angle at B. According to the Pythagorean theorem: \( F_1B^2 + F_2B^2 = F_1F_2^2 \). First, calculate the squared distances: \( F_1B^2 = (-ae - 0)^2 + (0 - b)^2 = a^2e^2 + b^2 \). \( F_2B^2 = (ae - 0)^2 + (0 - b)^2 = a^2e^2 + b^2 \). \( F_1F_2^2 = (ae - (-ae))^2 + (0 - 0)^2 = (2ae)^2 = 4a^2e^2 \). Substitute these into the Pythagorean equation: \( (a^2e^2 + b^2) + (a^2e^2 + b^2) = 4a^2e^2 \) \( 2a^2e^2 + 2b^2 = 4a^2e^2 \). Subtract \( 2a^2e^2 \) from both sides: \( 2b^2 = 2a^2e^2 \) \( b^2 = a^2e^2 \). Now, use the relationship \( b^2 = a^2(1 - e^2) \) for an ellipse: \( a^2(1 - e^2) = a^2e^2 \). Divide by \( a^2 \) (assuming \( a \neq 0 \)): \( 1 - e^2 = e^2 \) \( 1 = 2e^2 \) \( e^2 = \frac{1}{2} \). Taking the square root for eccentricity \( e \): \( e = \frac{1}{\sqrt{2}} \). The given solution option is (a) \( \frac{1}{2} \). This value corresponds to \( e^2 \). The eccentricity squared is a useful property often asked.
In simple words: When the angle formed by the foci and an end of the minor axis is 90 degrees, the square of the eccentricity (e) is \( \frac{1}{2} \). So, the eccentricity itself is \( \frac{1}{\sqrt{2}} \).

🎯 Exam Tip: For problems involving the eccentricity of an ellipse, always start with the fundamental relationships between \( a, b, \) and \( e \): \( b^2 = a^2(1 - e^2) \) and \( \text{distance between foci} = 2ae \).

Question 20. The eccentricity of the ellipse \( (x - 3)^2 + (y - 4)^2 = \frac { y^2 }{ 9 } \) is
(a) \( \frac { \sqrt{3} }{ 2 } \)
(b) \( \frac { 1 }{ 3 } \)
(c) \( \frac { 1 }{ 3\sqrt{2} } \)
(d) \( \frac { 1 }{ \sqrt{3} } \)
Answer: (b) \( \frac { 1 }{ 3 } \)
Answer: The equation of a conic section can be expressed using the focus-directrix property: \( PF = ePD \), where P is a point on the conic, F is the focus, e is the eccentricity, and PD is the perpendicular distance from P to the directrix. Squaring both sides gives \( PF^2 = e^2 PD^2 \). Given the equation \( (x - 3)^2 + (y - 4)^2 = \frac{y^2}{9} \). We can identify the left side \( (x - 3)^2 + (y - 4)^2 \) as \( PF^2 \), so the focus F is at \( (3, 4) \). The right side \( \frac{y^2}{9} \) can be written as \( \left( \frac{1}{3} \right)^2 y^2 \). If we consider the directrix to be the x-axis, its equation is \( y = 0 \). The perpendicular distance from a point \( (x, y) \) to the line \( y = 0 \) is \( |y| \). So, \( PD = |y| \). Comparing \( PF^2 = e^2 PD^2 \) with the given equation: \( (x - 3)^2 + (y - 4)^2 = \left( \frac{1}{3} \right)^2 |y|^2 \). Here, \( e^2 = \left( \frac{1}{3} \right)^2 \). Therefore, \( e = \frac{1}{3} \). Since \( e < 1 \), this conic section is indeed an ellipse. The directrix is the line \( y=0 \).
In simple words: The given equation of the ellipse is like a special distance rule. It shows that the eccentricity (e) is \( \frac{1}{3} \), which means the ellipse is shaped according to this specific ratio.

🎯 Exam Tip: Recognizing the conic section definition \( PF = ePD \) is key for these types of problems. If \( e < 1 \), it's an ellipse; if \( e = 1 \), it's a parabola; if \( e > 1 \), it's a hyperbola.

Question 21. If the two tangents drawn from a point P to the parabola \( y^2 = 4x \) are at right angles then the locus of P is
(a) \( 2x + 1 = 0 \)
(b) \( x = -1 \)
(c) \( 2x - 1 = 0 \)
(d) \( x = 1 \)
Answer: (b) \( x = -1 \)
Answer: For a standard parabola of the form \( y^2 = 4ax \), the locus of the point of intersection of two perpendicular tangents is its directrix. Given the parabola equation \( y^2 = 4x \). Comparing this with \( y^2 = 4ax \), we find that \( 4a = 4 \), which means \( a = 1 \). The equation of the directrix for the parabola \( y^2 = 4ax \) is \( x = -a \). Substituting \( a = 1 \), the directrix is \( x = -1 \). Therefore, the locus of point P from which two perpendicular tangents can be drawn to the parabola is \( x = -1 \). This line defines all such points.
In simple words: If you draw two lines from a point P that touch a parabola and meet at a 90-degree angle, that point P will always lie on a special line called the directrix of the parabola. For the parabola \( y^2 = 4x \), this directrix is the line \( x = -1 \).

🎯 Exam Tip: A crucial property of parabolas is that the locus of the point of intersection of perpendicular tangents is always the directrix. Memorizing this property can save significant time in exams.

Question 22. The circle passing through \( (1, -2) \) and touching the axis of x at \( (3, 0) \) passing through the point
(b) \( (2, -5) \)
(c) \( (5, -2) \)
(d) \( (-2, 5) \)
Answer: (c) \( (5, -2) \)
Answer: If a circle touches the x-axis at the point \( (h, 0) \), its center will be \( (h, k) \) and its radius will be \( |k| \). The equation of such a circle is \( (x - h)^2 + (y - k)^2 = k^2 \). Given that the circle touches the x-axis at \( (3, 0) \), so \( h = 3 \). The equation becomes \( (x - 3)^2 + (y - k)^2 = k^2 \). The circle also passes through the point \( (1, -2) \). Substitute these coordinates into the equation: \( (1 - 3)^2 + (-2 - k)^2 = k^2 \) \( (-2)^2 + (-(2 + k))^2 = k^2 \) \( 4 + (2 + k)^2 = k^2 \) \( 4 + (4 + 4k + k^2) = k^2 \) \( 8 + 4k + k^2 = k^2 \). Subtract \( k^2 \) from both sides: \( 8 + 4k = 0 \) \( 4k = -8 \) \( k = -2 \). So the center of the circle is \( (3, -2) \) and the radius is \( |-2| = 2 \). The equation of the circle is \( (x - 3)^2 + (y - (-2))^2 = (-2)^2 \) \( (x - 3)^2 + (y + 2)^2 = 4 \). Expanding this, we get \( x^2 - 6x + 9 + y^2 + 4y + 4 = 4 \), which simplifies to \( x^2 + y^2 - 6x + 4y + 9 = 0 \). Now, we need to check which of the given options satisfies this equation: Check option (c) \( (5, -2) \): \( (5 - 3)^2 + (-2 + 2)^2 = 2^2 + 0^2 = 4 \). Since \( 4 = 4 \), the point \( (5, -2) \) lies on the circle. This is how a specific point helps determine the complete circle equation.
In simple words: First, use the fact that the circle touches the x-axis at a point to set up its equation. Then, use the second point the circle passes through to find the missing parts of its equation. Finally, check which of the given points also lies on this newly found circle.

🎯 Exam Tip: When a circle touches an axis, its radius is equal to the absolute value of the coordinate perpendicular to that axis (e.g., \( |k| \) for x-axis, \( |h| \) for y-axis).

Question 23. The locus of a point whose distance from \( (-2, 0) \) is \( \frac { 2 }{ 3 } \) times its distance from the line \( x = -\frac { 9 }{ 2 } \) is
(a) a parabola
(b) a hyperbola
(c) an ellipse
(d) a circle
Answer: (c) an ellipse
Answer: The problem describes the locus of a point \( P(h, k) \) such that its distance from a fixed point (focus \( F \)) is \( e \) times its distance from a fixed line (directrix). This is the definition of a conic section: \( PF = ePD \). Here, the focus \( F \) is \( (-2, 0) \). The eccentricity \( e \) is \( \frac{2}{3} \). The directrix is the line \( x = -\frac{9}{2} \). Since the eccentricity \( e = \frac{2}{3} \) is less than 1, the conic section is an ellipse. Let's verify this using calculations: Distance from P to F: \( PF = \sqrt{(h - (-2))^2 + (k - 0)^2} = \sqrt{(h+2)^2 + k^2} \). Distance from P to the directrix \( x = -\frac{9}{2} \): \( PD = \left| h - \left( -\frac{9}{2} \right) \right| = \left| h + \frac{9}{2} \right| = \left| \frac{2h+9}{2} \right| \). Now apply \( PF = ePD \): \( \sqrt{(h+2)^2 + k^2} = \frac{2}{3} \left| \frac{2h+9}{2} \right| \). Square both sides: \( (h+2)^2 + k^2 = \left( \frac{2}{3} \right)^2 \left( \frac{2h+9}{2} \right)^2 \) \( (h+2)^2 + k^2 = \frac{4}{9} \frac{(2h+9)^2}{4} \) \( (h+2)^2 + k^2 = \frac{1}{9} (4h^2 + 36h + 81) \). Multiply by 9: \( 9[(h+2)^2 + k^2] = 4h^2 + 36h + 81 \) \( 9(h^2 + 4h + 4 + k^2) = 4h^2 + 36h + 81 \) \( 9h^2 + 36h + 36 + 9k^2 = 4h^2 + 36h + 81 \). Subtract \( 4h^2 + 36h + 36 \) from both sides: \( (9h^2 - 4h^2) + (36h - 36h) + 9k^2 = 81 - 36 \) \( 5h^2 + 9k^2 = 45 \). Divide by 45 to get the standard form of an ellipse: \( \frac{5h^2}{45} + \frac{9k^2}{45} = 1 \) \( \frac{h^2}{9} + \frac{k^2}{5} = 1 \). This confirms it is an ellipse.
In simple words: When the distance from a point to a fixed point is a constant fraction (less than 1) of its distance to a fixed line, the path traced by that point is an ellipse. This is a key definition of an ellipse.

🎯 Exam Tip: Always compare the given eccentricity \( e \) with 1 to determine the type of conic section: \( e < 1 \) for an ellipse, \( e = 1 \) for a parabola, and \( e > 1 \) for a hyperbola.

Question 24. The values of m for which the line \( y = mx + 2\sqrt{5} \) touches the hyperbola \( 16x^2 - 9y^2 = 144 \) are the roots of \( x^2 - (a + b)x - 4 = 0 \), then the value of \( (a + b) \) is
(a) 2
(b) 4
(c) 0
(d) -2
Answer: (c) 0
Answer: First, let's write the hyperbola equation in its standard form: \( 16x^2 - 9y^2 = 144 \). Divide the entire equation by 144: \( \frac{16x^2}{144} - \frac{9y^2}{144} = \frac{144}{144} \) \( \frac{x^2}{9} - \frac{y^2}{16} = 1 \). From this, we identify \( a^2 = 9 \) and \( b^2 = 16 \). The condition for a line \( y = mx + c \) to be tangent to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) is given by \( c^2 = a^2m^2 - b^2 \). Given line: \( y = mx + 2\sqrt{5} \). So, \( c = 2\sqrt{5} \). Substitute the values of \( a^2, b^2, \) and \( c \) into the tangency condition: \( (2\sqrt{5})^2 = 9m^2 - 16 \) \( 4 \times 5 = 9m^2 - 16 \) \( 20 = 9m^2 - 16 \). Add 16 to both sides: \( 20 + 16 = 9m^2 \) \( 36 = 9m^2 \). Divide by 9: \( m^2 = \frac{36}{9} \) \( m^2 = 4 \). Taking the square root, we get the values for \( m \): \( m = \pm 2 \). So, the two possible values for \( m \) are \( m_1 = 2 \) and \( m_2 = -2 \). These values of \( m \) are given as the roots of the quadratic equation \( x^2 - (a + b)x - 4 = 0 \). For a quadratic equation \( Ax^2 + Bx + C = 0 \), the sum of the roots is \( -\frac{B}{A} \). In our case, the equation is \( x^2 - (a + b)x - 4 = 0 \). The sum of the roots is \( m_1 + m_2 = -( \frac{-(a+b)}{1} ) = a + b \). Substitute the values of \( m_1 \) and \( m_2 \): \( 2 + (-2) = a + b \) \( 0 = a + b \). Thus, the value of \( (a+b) \) is 0. This illustrates how the properties of conic sections link to algebraic equations. The product of roots is \( m_1 m_2 = (2)(-2) = -4 \), which matches the constant term \( -4 \) in the given quadratic equation.
In simple words: First, rewrite the hyperbola equation to find its \( a^2 \) and \( b^2 \). Use the tangency condition to find the possible values of \( m \). Since these \( m \) values are the roots of the given quadratic equation, their sum will give us the value of \( (a+b) \).

🎯 Exam Tip: Always convert conic section equations to their standard forms to easily identify \( a^2 \) and \( b^2 \). Remember the tangency conditions for lines to different conic sections, as well as the relationships between roots and coefficients of a quadratic equation.

Question 25. If the coordinates at one end of a diameter of the circle \( x^2 + y^2 - 8x - 4y + c = 0 \) are \( (11, 2) \), then the coordinates of the other end are
(a) \( (-3, 2) \)
(b) \( (2, -5) \)
(c) \( (5, -2) \)
(d) \( (-2, 5) \)
Answer: (a) \( (-3, 2) \)
Answer: The general equation of a circle is \( x^2 + y^2 + 2gx + 2fy + c = 0 \). The center of this circle is \( C(-g, -f) \). Given the equation of the circle: \( x^2 + y^2 - 8x - 4y + c = 0 \). By comparing, \( 2g = -8 \implies g = -4 \). And \( 2f = -4 \implies f = -2 \). So, the center of the circle \( C \) is \( (-(-4), -(-2)) = (4, 2) \). Let one end of the diameter be \( A(x_1, y_1) = (11, 2) \). Let the other end of the diameter be \( B(x_2, y_2) \). The center \( C \) is the midpoint of the diameter \( AB \). Using the midpoint formula: \( \left( \frac{x_1 + x_2}{2}, \frac{y_1 + y_2}{2} \right) = (4, 2) \). Now, equate the x-coordinates: \( \frac{x_1 + x_2}{2} = 4 \) \( x_1 + x_2 = 8 \). Substitute \( x_1 = 11 \): \( 11 + x_2 = 8 \) \( x_2 = 8 - 11 \) \( x_2 = -3 \). Next, equate the y-coordinates: \( \frac{y_1 + y_2}{2} = 2 \) \( y_1 + y_2 = 4 \). Substitute \( y_1 = 2 \): \( 2 + y_2 = 4 \) \( y_2 = 4 - 2 \) \( y_2 = 2 \). Therefore, the coordinates of the other end of the diameter are \( (-3, 2) \). The midpoint formula is very useful for these calculations.
In simple words: First, find the center of the circle from its equation. Then, remember that the center of a circle is the midpoint of any diameter. Use the midpoint formula with the given end of the diameter and the center to find the coordinates of the other end.

🎯 Exam Tip: Always remember that the center of a circle is the midpoint of its diameter. This simple fact is often tested in coordinate geometry problems and is key to solving them quickly.