Samacheer Kalvi Class 11 Business Maths Solutions Chapter 3 Analytical Geometry Exercise 3.7

Question 1. If \( m_1 \) and \( m_2 \) are the slopes of the pair of lines given by \( ax^2 + 2hxy + by^2 = 0 \), \( m_1 + m_2 \) is:
(a) \( \frac{2h}{b} \)
(b) \( -\frac{2h}{b} \)
(c) \( \frac{2h}{a} \)
(d) \( -\frac{2h}{a} \)
Answer: (b) \( -\frac{2h}{b} \)
In simple words: When two lines are combined into one equation, the sum of their slopes can be found directly from the numbers in the equation. This is a standard formula that connects the equation to the properties of the lines.

🎯 Exam Tip: Remember this formula as it directly relates the coefficients of the general equation to the properties of the lines. It is a common result derived from polynomial roots.

Question 2. The angle between the pair of straight lines \( x^2 – 7xy + 4y^2 = 0 \) is:
(a) \( \tan ^{-1}\left(\frac{1}{3}\right) \)
(b) \( \tan ^{-1}\left(\frac{1}{2}\right) \)
(c) \( \tan ^{-1}\left(\frac{\sqrt{33}}{5}\right) \)
(d) \( \tan ^{-1}\left(\frac{5}{\sqrt{33}}\right) \)
Answer: (c) \( \tan ^{-1}\left(\frac{\sqrt{33}}{5}\right) \)
In simple words: This question asks for the angle between two lines whose equation is given together. We use a special formula that takes numbers from the equation to find this angle. This helps to understand how the lines are positioned relative to each other.

🎯 Exam Tip: Always identify the values of a, h, and b correctly from the given equation \( ax^2 + 2hxy + by^2 = 0 \) before applying the formula \( \tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right| \) for the angle.

Question 3. If the lines \( 2x – 3y – 5 = 0 \) and \( 3x – 4y – 7 = 0 \) are the diameters of a circle, then its centre is:
(a) (-1, 1)
(b) (1, 1)
(c) (1, -1)
(d) (-1, -1)
Answer: (c) (1, -1)
In simple words: For a circle, all its diameters pass through the center. So, if we have two diameters, their meeting point is the center of the circle. We find this point by solving the two line equations.

🎯 Exam Tip: To find the intersection of two lines, use methods like substitution or elimination to solve the system of equations. The center of a circle is the unique point where all diameters intersect.

Question 4. The x-intercept of the straight line \( 3x + 2y - 1 = 0 \) is
(a) 3
(b) 2
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{2} \)
Answer: (c) \( \frac{1}{3} \)
In simple words: The x-intercept is the point where a line crosses the 'x' line on a graph. At this point, the 'y' value is always zero. We put \( y=0 \) into the equation to find the 'x' value.

🎯 Exam Tip: To find the x-intercept, always set y to 0 in the equation of the line; similarly, for the y-intercept, set x to 0.

Question 5. The slope of the line \( 7x + 5y – 8 = 0 \) is:
(a) \( \frac{7}{5} \)
(b) \( -\frac{7}{5} \)
(c) \( \frac{5}{7} \)
(d) \( -\frac{5}{7} \)
Answer: (b) \( -\frac{7}{5} \)
In simple words: The slope of a line tells us how steep it is. For an equation like \( Ax + By + C = 0 \), we can quickly find its slope by dividing the negative of the 'x' number by the 'y' number. This value helps us understand the direction and steepness of the line.

🎯 Exam Tip: Remember that for an equation in the form \( Ax + By + C = 0 \), the slope can be quickly found using the formula \( -\frac{\text{coefficient of x}}{\text{coefficient of y}} \).

Question 6. The locus of the point P which moves such that P is at equidistance from their coordinate axes is:
(a) \( y = \frac{1}{x} \)
(b) \( y = -x \)
(c) \( y = x \)
(d) \( y = -\frac{1}{x} \)
Answer: (c) \( y = x \)
In simple words: A locus is the path a point makes when it follows a certain rule. Here, the rule is that the point is always the same distance from both the x-axis and the y-axis. This path forms a straight line.

🎯 Exam Tip: When finding a locus, express the given condition using coordinates \( (x, y) \) and then simplify the resulting algebraic equation to find the path.

Question 7. The locus of the point P which moves such that P is always at equidistance from the line \( x + 2y + 7 = 0 \):
(a) \( x + 2y + 2 = 0 \)
(b) \( x – 2y + 1 = 0 \)
(c) \( 2x - y + 2 = 0 \)
(d) \( 3x + y + 1 = 0 \)
Answer: (a) \( x + 2y + 2 = 0 \)
In simple words: If a point always stays the same distance from a straight line, its path will be another straight line that runs next to the first one. This new line is parallel to the original line.

🎯 Exam Tip: The locus of points equidistant from a given line is a line parallel to the given line. The constant term will change, but the coefficients of x and y will remain the same.

Question 8. If \( kx^2 + 3xy - 2y^2 = 0 \) represent a pair of lines which are perpendicular then k is equal to:
(a) \( \frac{1}{2} \)
(b) \( -\frac{1}{2} \)
(c) 2
(d) -2
Answer: (c) 2
In simple words: When two lines are perpendicular, their combined equation has a special property: the sum of the number in front of \( x^2 \) and the number in front of \( y^2 \) is always zero. This rule helps us find the missing number 'k'.

🎯 Exam Tip: For a homogeneous equation of a pair of straight lines \( ax^2 + 2hxy + by^2 = 0 \), remember the condition for perpendicularity is \( a + b = 0 \).

Question 9. (1, -2) is the centre of the circle \( x^2 + y^2 + ax + by - 4 = 0 \), then its radius:
(a) 3
(b) 2
(c) 4
(d) 1
Answer: (a) 3
In simple words: The center and radius are key parts of a circle's definition. We can use the given center of the circle and the equation to find the radius. The formula for the radius uses the 'g', 'f', and 'c' values from the circle's general equation.

🎯 Exam Tip: When the equation of a circle is \( x^2 + y^2 + 2gx + 2fy + c = 0 \), its center is \( (-g, -f) \) and its radius is \( \sqrt{g^2 + f^2 - c} \). Carefully compare coefficients.

Question 10. The length of the tangent from (4, 5) to the circle \( x^2 + y^2 = 16 \) is:
(a) 4
(b) 5
(c) 16
(d) 25
Answer: (b) 5
In simple words: A tangent is a line that just touches a circle at one point. We can find the length of this line from a given point outside the circle to where it touches the circle using a specific formula. This formula connects the point's coordinates to the circle's equation.

🎯 Exam Tip: The length of the tangent from an external point \( (x_1, y_1) \) to the circle \( x^2 + y^2 = r^2 \) is given by \( \sqrt{x_1^2 + y_1^2 - r^2} \). Just substitute the coordinates and the radius squared.

Question 11. The focus of the parabola \( x^2 = 16y \) is:
(c) (0, 4)
(d) (0, -4)
Answer: (c) (0, 4)
In simple words: A parabola has a special point inside it called the focus. Its position depends on the parabola's equation. By comparing the given equation with the standard form, we can find the exact coordinates of this important point.

🎯 Exam Tip: To find the focus of a parabola, first identify its standard form (e.g., \( x^2 = 4ay \) or \( y^2 = 4ax \)). Then, compare coefficients to find the value of 'a' and determine the focus coordinates accordingly.

Question 12. Length of the latus rectum of the parabola \( y^2 = -25x \):
(a) 25
(b) -5
(c) 5
(d) -25
Answer: (a) 25
In simple words: The latus rectum is a special line segment inside a parabola that passes through its focus and is perpendicular to its axis. Its length is a key property of the parabola. We find this length by comparing the given equation with the standard form and identifying the '4a' part.

🎯 Exam Tip: The length of the latus rectum of a parabola (e.g., from \( y^2 = 4ax \) or \( x^2 = 4ay \)) is always \( |4a| \). Remember that a length must always be positive.

Question 13. The centre of the circle \( x^2 + y^2 – 2x + 2y – 9 = 0 \) is:
(a) (1, 1)
(b) (-1, 1)
(c) (-1, -1)
(d) (1, -1)
Answer: (d) (1, -1)
In simple words: Every circle has a center point. For a circle's equation written in the general form, we can find its center by looking at the numbers in front of 'x' and 'y'. This helps us understand the circle's position on a graph.

🎯 Exam Tip: To find the center \( (-g, -f) \) from the general equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \), divide the coefficient of x by -2 to get g, and the coefficient of y by -2 to get f.

Question 14. The equation of the circle with centre on the x axis and passing through the origin is:
(a) \( x^2 – 2ax + y^2 = 0 \)
(b) \( y^2 – 2ay + x^2 = 0 \)
(c) \( x^2 + y^2 = a^2 \)
(d) \( x^2 – 2ay + y^2 = 0 \)
Answer: (a) \( x^2 – 2ax + y^2 = 0 \)
In simple words: If a circle's center is on the x-axis, its y-coordinate is zero. If it also passes through the origin (0,0), then the distance from its center to the origin is its radius. We use these facts to write the circle's equation.

🎯 Exam Tip: For a circle with its center on the x-axis, its center can be written as \( (h, 0) \). If it passes through the origin \( (0, 0) \), then its radius \( r = \sqrt{(h-0)^2 + (0-0)^2} = |h| \).

Question 15. If the centre of the circle is (-a, -b) and radius is \( \sqrt{a^{2}-b^{2}} \) then the equation of circle is:
(a) \( x^2 + y^2 + 2ax + 2by + 2b^2 = 0 \)
(b) \( x^2 + y^2 + 2ax + 2by – 2b^2 = 0 \)
(c) \( x^2 + y^2 – 2ax – 2by – 2b^2 = 0 \)
(d) \( x^2 + y^2 – 2ax – 2by + 2b^2 = 0 \)
Answer: (a) \( x^2 + y^2 + 2ax + 2by + 2b^2 = 0 \)
In simple words: The equation of a circle is found using its center and radius. We use a standard formula where we plug in the given center coordinates and the radius squared. Then, we simplify this expression to get the final circle equation.

🎯 Exam Tip: Always use the standard equation of a circle \( (x - h)^2 + (y - k)^2 = r^2 \). Be very careful with signs when substituting \( h = -a \), \( k = -b \) and \( r^2 = a^2 - b^2 \) and expanding the equation.

Question 16. Combined equation of co-ordinate axes is:
(a) \( x^2 – y^2 = 0 \)
(b) \( x^2 + y^2 = 0 \)
(c) \( xy = c \)
(d) \( xy = 0 \)
Answer: (d) \( xy = 0 \)
In simple words: The x-axis is a line where all 'y' values are zero, and the y-axis is a line where all 'x' values are zero. When we combine their individual equations by multiplying them, we get one equation that describes both axes together.

🎯 Exam Tip: To find the combined equation of two lines, set each line's equation equal to zero and then multiply these two expressions together. For the axes, this means \( x \cdot y = 0 \).

Question 17. \( ax^2 + 4xy + 2y^2 = 0 \) represents a pair of parallel lines then 'a' is:
(a) 2
(b) -2
(c) 4
(d) -4
Answer: (a) 2
In simple words: When a single equation describes two parallel lines, there's a special rule involving the numbers in front of \( x^2 \), \( xy \), and \( y^2 \). We use this rule to find the unknown number 'a'. This property helps classify the type of lines represented.

🎯 Exam Tip: For a pair of straight lines given by \( ax^2 + 2hxy + by^2 = 0 \), the condition for them to be parallel is \( h^2 = ab \). Remember to correctly identify '2h' from the \( xy \) term.

Question 18. In the equation of the circle \( x^2 + y^2 = 16 \) then y intercept is (are):
(a) 4
(b) 16
(c) \( \pm 4 \)
(d) \( \pm 16 \)
Answer: (c) \( \pm 4 \)
In simple words: The y-intercept is where a shape crosses the 'y' line on a graph. For a circle, this happens when the 'x' value is zero. We put \( x=0 \) into the circle's equation to find the 'y' values where it touches the y-axis.

🎯 Exam Tip: To find y-intercepts, substitute \( x = 0 \) into the equation of the curve and solve for y. For a circle, there can be two y-intercepts (positive and negative values).

Question 19. If the perimeter of the circle is \( 8\pi \) units and centre is (2, 2) then the equation of the circle is:
(a) \( (x – 2)^2 + (y – 2)^2 = 4 \)
(b) \( (x – 2)^2 + (y – 2)^2 = 16 \)
(c) \( (x - 4)^2 + (y - 4)^2 = 16 \)
(d) \( x^2 + y^2 = 4 \)
Answer: (b) \( (x – 2)^2 + (y – 2)^2 = 16 \)
In simple words: The perimeter of a circle is its circumference, which helps us find its radius. Once we know the radius and the center point, we can easily write the circle's equation using the standard formula. The radius is squared in the final equation.

🎯 Exam Tip: The circumference of a circle is \( 2\pi r \). Use this to first calculate the radius \( r \). Then, substitute the center coordinates \( (h, k) \) and the calculated radius into the standard circle equation: \( (x-h)^2 + (y-k)^2 = r^2 \).

Question 20. The equation of the circle with centre (3, -4) and touches the x-axis is:
(a) \( (x − 3)^2 + (y − 4)^2 = 4 \)
(b) \( (x – 3)^2 + (y + 4)^2 = 16 \)
(c) \( (x – 3)^2 + (y + 4)^2 = 4 \)
(d) \( x^2 + y^2 = 16 \)
Answer: (b) \( (x – 3)^2 + (y + 4)^2 = 16 \)
In simple words: When a circle touches the x-axis, it means its radius is equal to the up-and-down distance of its center from the x-axis. This distance is simply the absolute value of the y-coordinate of the center. Knowing the center and radius helps write the circle's equation.

🎯 Exam Tip: If a circle with center \( (h,k) \) touches the x-axis, its radius is \( |k| \). If it touches the y-axis, its radius is \( |h| \).

Question 21. If the circle touches the x-axis, y-axis, and the line x = 6 then the length of the diameter of the circle is:
(a) 6
(b) 3
(c) 12
(d) 4
Answer: (a) 6
In simple words: A circle that touches both the x and y axes must have its center at \( (\pm r, \pm r) \). If it also touches the line \( x=6 \), this means its radius must be 6. The diameter is simply twice the radius.

🎯 Exam Tip: When a circle touches both coordinate axes, its center is \( (r, r) \) or similar combinations depending on the quadrant. If it also touches a vertical line \( x=k \), then its radius must be \( |k| \).

Question 22. The eccentricity of the parabola is:
(a) 3
(b) 2
(c) 0
(d) 1
Answer: (d) 1
In simple words: Eccentricity is a number that tells us how "oval" or "stretched" a conic shape is. For a parabola, this number is always exactly 1, which means it has a specific, open curve shape.

🎯 Exam Tip: Remember the eccentricity values for the different conic sections: circle (e=0), ellipse (0 < e < 1), parabola (e=1), and hyperbola (e > 1).

Question 23. The double ordinate passing through the focus is:
(a) focal chord
(b) latus rectum
(c) directrix
(d) axis
Answer: (b) latus rectum
In simple words: In a parabola, a "focal chord" is any straight line segment that goes through the focus. If this chord is also straight up and down (perpendicular to the axis), it has a special name: the latus rectum.

🎯 Exam Tip: Differentiate between a focal chord (any chord passing through the focus) and the latus rectum (a specific focal chord perpendicular to the axis of the parabola).

Question 24. The distance between directrix and focus of a parabola \( y^2 = 4ax \) is
(a) a
(b) 2a
(c) 4a
(d) 3a
Answer: (b) 2a
In simple words: A parabola is defined by a fixed point (focus) and a fixed line (directrix). The distance between these two defining elements is an important property of the parabola. For standard parabolas, this distance is always twice the value of 'a'.

🎯 Exam Tip: For the standard parabola \( y^2 = 4ax \), the focus is at \( (a, 0) \) and the directrix is the line \( x = -a \). The distance between them is \( a - (-a) = 2a \).

Question 25. The equation of directrix of the parabola \( y^2 = -x \) is:
(a) \( 4x + 1 = 0 \)
(b) \( 4x - 1 = 0 \)
(c) \( x - 1 = 0 \)
(d) \( x + 4 = 0 \)
Answer: (b) \( 4x - 1 = 0 \)
In simple words: The directrix is a straight line outside a parabola that helps define its shape. For a parabola that opens sideways, the directrix is a vertical line. We find its equation by comparing the given parabola equation with a standard form and finding the 'a' value.

🎯 Exam Tip: For a parabola of the form \( y^2 = -4ax \), the directrix is \( x = a \). First, identify the value of 'a' from the given equation, then substitute it into the directrix formula.

Samacheer Kalvi 11th Business Maths Guide Chapter 3 Analytical Geometry Ex 3.7

Question 1. If \( m_1 \) and \( m_2 \) are the slopes of the pair of lines given by \( ax^2 + 2hxy + by^2 = 0 \), \( m_1 + m_2 \) is:
(a) \( \frac{2h}{b} \)
(b) \( -\frac{2h}{b} \)
(c) \( \frac{2h}{a} \)
(d) \( -\frac{2h}{a} \)
Answer: (b) \( -\frac{2h}{b} \)
In simple words: When you have a combined equation for two lines, the sum of their slopes \( (m_1 + m_2) \) is found by dividing minus two times the 'h' term by the 'b' term. This formula helps you understand the relationship between the slopes and the coefficients in the given equation.

🎯 Exam Tip: Remember the formulas for the sum and product of slopes for a pair of straight lines given by a general quadratic equation; they are frequently tested.

Question 2. The angle between the pair of straight lines \( x^2 – 7xy + 4y^2 = 0 \) is:
(a) \( \tan^{-1}\left(\frac{1}{3}\right) \)
(b) \( \tan^{-1}\left(\frac{1}{2}\right) \)
(c) \( \tan^{-1}\left(\frac{\sqrt{33}}{5}\right) \)
(d) \( \tan^{-1}\left(\frac{5}{\sqrt{33}}\right) \)
Answer: (c) \( \tan^{-1}\left(\frac{\sqrt{33}}{5}\right) \)
In simple words: To find the angle between these two lines, we use a special formula involving the 'a', 'b', and 'h' values from the equation. After putting in the numbers and doing the math, we get \( \tan^{-1} \) of \( \frac{\sqrt{33}}{5} \). This calculation gives us the measure of the angle between the two straight lines represented by the given combined equation.

🎯 Exam Tip: When finding the angle between two lines represented by a combined equation \( ax^2 + 2hxy + by^2 = 0 \), always identify 'a', 'h', and 'b' correctly from the equation, and use the formula \( \tan \theta = \left| \frac{2\sqrt{h^2 - ab}}{a+b} \right| \).

Question 3. If the lines \( 2x – 3y – 5 = 0 \) and \( 3x – 4y – 7 = 0 \) are the diameters of a circle, then its centre is:
(a) (-1, 1)
(b) (1, 1)
(c) (1, -1)
(d) (-1, -1)
Answer: (c) (1, -1)
In simple words: The center of a circle is where all its diameters cross. So, to find the center, we need to solve the two equations of the diameter lines at the same time. The point where they both meet is the center of the circle.

🎯 Exam Tip: The intersection point of any two diameters of a circle is always its center. Therefore, to find the center, simply solve the equations of the given diameters simultaneously.

Question 4. The x-intercept of the straight line \( 3x + 2y – 1 = 0 \) is
(a) 3
(b) 2
(c) \( \frac{1}{3} \)
(d) \( \frac{1}{2} \)
Answer: (c) \( \frac{1}{3} \)
In simple words: To find where a line crosses the x-axis, you just need to set the 'y' value to zero in the line's equation. Then, solve for 'x'. This 'x' value is the point where the line touches or crosses the x-axis.

🎯 Exam Tip: The x-intercept is the point where the line crosses the x-axis, meaning the y-coordinate is always 0. Similarly, for the y-intercept, the x-coordinate is 0.

Question 5. The slope of the line \( 7x + 5y – 8 = 0 \) is:
(a) \( \frac{7}{5} \)
(b) \( -\frac{7}{5} \)
(c) \( \frac{5}{7} \)
(d) \( -\frac{5}{7} \)
Answer: (b) \( -\frac{7}{5} \)
In simple words: For a straight line equation like \( Ax + By + C = 0 \), the slope is always \( -\frac{A}{B} \). Here, 'A' is 7 and 'B' is 5, so the slope is \( -\frac{7}{5} \). The slope tells us how steep the line is.

🎯 Exam Tip: When a line is in the form \( Ax + By + C = 0 \), its slope is given by the formula \( -\frac{\text{coefficient of x}}{\text{coefficient of y}} \). Be careful with the signs.

Question 6. The locus of the point P which moves such that P is at equidistance from their coordinate axes is:
(a) \( y = \frac{1}{x} \)
(b) \( y = -x \)
(c) \( y = x \)
(d) \( y = -\frac{1}{x} \)
Answer: (c) \( y = x \)
In simple words: If a point is equally far from the x-axis and the y-axis, it means its x-coordinate and y-coordinate are the same in absolute value. This forms a line that passes through the origin at a 45-degree angle. This is often represented by the equation \( y = x \) or \( y = -x \).

🎯 Exam Tip: A point equidistant from the coordinate axes lies on the lines \( y = x \) or \( y = -x \). If it's simply "equidistant," both possibilities should be considered, but \( y = x \) is typically the standard representation when referring to positive distances.

Question 7. The locus of the point P which moves such that P is always at equidistance from the line \( x + 2y + 7 = 0 \):
(a) \( x + 2y + 2 = 0 \)
(b) \( x – 2y + 1 = 0 \)
(c) \( 2x - y + 2 = 0 \)
(d) \( 3x + y + 1 = 0 \)
Answer: (a) \( x + 2y + 2 = 0 \)
In simple words: If a point is always the same distance from a given line, the path it traces is another line that runs parallel to the first one. So, to find the locus, we look for an equation that has the same slope as the original line but with a different constant term.

🎯 Exam Tip: The locus of a point equidistant from a straight line is a parallel line. The general equation of a line parallel to \( Ax + By + C = 0 \) is \( Ax + By + K = 0 \).

Question 8. If \( kx^2 + 3xy - 2y^2 = 0 \) represent a pair of lines which are perpendicular then k is equal to:
(a) \( \frac{1}{2} \)
(b) \( -\frac{1}{2} \)
(c) 2
(d) -2
Answer: (c) 2
In simple words: For a pair of lines represented by \( ax^2 + 2hxy + by^2 = 0 \), if the lines are perpendicular, it means that the sum of the 'a' and 'b' coefficients must be zero. By applying this rule, we can easily find the value of 'k'.

🎯 Exam Tip: For a pair of straight lines given by \( ax^2 + 2hxy + by^2 = 0 \), the condition for them to be perpendicular is \( a + b = 0 \). The 'a' and 'b' values are the coefficients of \( x^2 \) and \( y^2 \) respectively.

Question 9. (1, -2) is the centre of the circle \( x^2 + y^2 + ax + by – 4 = 0 \), then its radius:
(a) 3
(b) 2
(c) 4
(d) 1
Answer: (a) 3
In simple words: The center of a circle and its general equation are connected. If the center is \( (-g, -f) \) and the constant term is 'c', the radius can be found using the formula \( \sqrt{g^2 + f^2 - c} \). We just plug in the numbers to find the radius.

🎯 Exam Tip: For a general circle equation \( x^2 + y^2 + 2gx + 2fy + c = 0 \), the center is \( (-g, -f) \) and the radius is \( \sqrt{g^2 + f^2 - c} \). Make sure to correctly identify 'g', 'f', and 'c' from the equation.

Question 10. The length of the tangent from (4, 5) to the circle \( x^2 + y^2 = 16 \) is:
(a) 4
(b) 5
(c) 16
(d) 25
Answer: (b) 5
In simple words: To find the length of a tangent line from a point to a circle, we can use a formula that involves the coordinates of the point and the circle's equation. For a circle \( x^2 + y^2 = r^2 \), and a point \( (x_1, y_1) \), the length of the tangent is \( \sqrt{x_1^2 + y_1^2 - r^2} \).

🎯 Exam Tip: The length of the tangent from an external point \( (x_1, y_1) \) to the circle \( x^2 + y^2 + 2gx + 2fy + c = 0 \) is \( \sqrt{S_1} \), where \( S_1 = x_1^2 + y_1^2 + 2gx_1 + 2fy_1 + c \). For a circle centered at the origin \( x^2 + y^2 = r^2 \), it simplifies to \( \sqrt{x_1^2 + y_1^2 - r^2} \).

Question 11. The focus of the parabola \( x^2 = 16y \) is:
(a) (-4, 0)
(b) (-4, 0)
(c) (0, 4)
(d) (0, -4)
Answer: (c) (0, 4)
In simple words: For a parabola that opens upwards, like \( x^2 = 4ay \), the focus is a special point located at \( (0, a) \). First, we find 'a' by comparing the given equation with the standard form, then we can easily find the focus point.

🎯 Exam Tip: For parabolas of the form \( x^2 = 4ay \), the focus is at \( (0, a) \). For \( y^2 = 4ax \), the focus is at \( (a, 0) \). Always identify the correct standard form to find 'a' accurately.

Question 12. Length of the latus rectum of the parabola \( y^2 = -25x \):
(a) 25
(b) -5
(c) 5
(d) -25
Answer: (a) 25
In simple words: The latus rectum is a special line segment in a parabola that passes through the focus and is parallel to the directrix. Its length is always \( |4a| \). Even if the 'a' value is negative in the equation, the length must be a positive number.

🎯 Exam Tip: The length of the latus rectum for any parabola (e.g., \( y^2 = 4ax \), \( x^2 = 4ay \), etc.) is always \( |4a| \). Remember that length is a positive quantity, so use the absolute value.

Question 13. The centre of the circle \( x^2 + y^2 – 2x + 2y – 9 = 0 \) is:
(a) (1, 1)
(b) (-1, 1)
(c) (-1, 1)
(d) (1, -1)
Answer: (d) (1, -1)
In simple words: The general equation of a circle is \( x^2 + y^2 + 2gx + 2fy + c = 0 \). The center of this circle is found at the coordinates \( (-g, -f) \). By comparing our equation with the general one, we can easily find the values of g and f, and then determine the center.

🎯 Exam Tip: To find the center \( (-g, -f) \) from \( x^2 + y^2 + 2gx + 2fy + c = 0 \), simply divide the coefficients of x and y by -2. For example, if the x-term is \( -2x \), then \( 2g = -2 \implies g = -1 \), so \( -g = 1 \).

Question 14. The equation of the circle with centre on the x axis and passing through the origin is:
(a) \( x^2 – 2ax + y^2 = 0 \)
(b) \( y^2 – 2ay + x^2 = 0 \)
(c) \( x^2 + y^2 = a^2 \)
(d) \( x^2 – 2ay + y^2 = 0 \)
Answer: (a) \( x^2 – 2ax + y^2 = 0 \)
In simple words: If a circle's center is on the x-axis, its y-coordinate will be zero. If it also passes through the starting point (origin), its radius will be the distance from the center to the origin. We can then use the standard circle equation \( (x-h)^2 + (y-k)^2 = r^2 \) to form the specific equation.

🎯 Exam Tip: When the center of a circle is on the x-axis, its coordinates are \( (h, 0) \). If it passes through the origin \( (0, 0) \), then its radius 'r' is simply \( |h| \), leading to the equation \( (x-h)^2 + y^2 = h^2 \).

Question 15. If the centre of the circle is \( (-a, -b) \) and radius is \( \sqrt{a^2-b^2} \) then the equation of circle is:
(a) \( x^2 + y^2 + 2ax + 2by + 2b^2 = 0 \)
(b) \( x^2 + y^2 + 2ax + 2by – 2b^2 = 0 \)
(c) \( x^2 + y^2 – 2ax – 2by – 2b^2 = 0 \)
(d) \( x^2 + y^2 – 2ax – 2by + 2b^2 = 0 \)
Answer: (a) \( x^2 + y^2 + 2ax + 2by + 2b^2 = 0 \)
In simple words: We use the standard formula for a circle's equation: \( (x - h)^2 + (y - k)^2 = r^2 \), where \( (h, k) \) is the center and 'r' is the radius. We just substitute the given center \( (-a, -b) \) for \( (h, k) \) and \( \sqrt{a^2-b^2} \) for 'r' and then simplify the equation to find the final form.

🎯 Exam Tip: The standard equation of a circle is \( (x-h)^2 + (y-k)^2 = r^2 \). Expand this equation carefully, paying close attention to signs, especially when the center coordinates are negative.

Question 16. Combined equation of co-ordinate axes is:
(a) \( x^2 – y^2 = 0 \)
(b) \( x^2 + y^2 \)
(c) \( xy = c \)
(d) \( xy = 0 \)
Answer: (d) \( xy = 0 \)
In simple words: The x-axis has the equation \( y = 0 \), and the y-axis has the equation \( x = 0 \). When we combine these two equations to represent both axes together, we multiply them. So, \( x \) multiplied by \( y \) equals zero, which represents both axes at once.

🎯 Exam Tip: The combined equation of two lines \( L_1 = 0 \) and \( L_2 = 0 \) is simply \( L_1 \cdot L_2 = 0 \). For the coordinate axes, \( x=0 \) and \( y=0 \), so their combined equation is \( xy=0 \).

Question 17. \( ax^2 + 4xy + 2y^2 = 0 \) represents a pair of parallel lines then 'a' is:
(a) 2
(b) -2
(c) 4
(d) -4
Answer: (a) 2
In simple words: For a general equation representing a pair of lines \( Ax^2 + 2Hxy + By^2 = 0 \), if these lines are parallel, a specific condition must be met: \( H^2 - AB = 0 \). We use this rule to find the unknown value 'a'.

🎯 Exam Tip: For a pair of straight lines given by \( Ax^2 + 2Hxy + By^2 = 0 \), the condition for them to be parallel is \( H^2 - AB = 0 \). Always identify the correct values for A, H, and B from the given equation.

Question 18. In the equation of the circle \( x^2 + y^2 = 16 \) then y intercept is (are):
(a) 4
(b) 16
(c) \( \pm 4 \)
(d) \( \pm 16 \)
Answer: (c) \( \pm 4 \)
In simple words: The y-intercept is the point where the circle crosses the y-axis. To find it, we set the x-coordinate to zero in the circle's equation and then solve for 'y'. This will give us the y-values where the circle intersects the y-axis.

🎯 Exam Tip: To find the y-intercept(s) of any curve, substitute \( x = 0 \) into its equation and solve for y. Similarly, for x-intercept(s), substitute \( y = 0 \) and solve for x.

Question 19. If the perimeter of the circle is \( 8\pi \) units and centre is (2, 2) then the equation of the circle is:
(a) \( (x – 2)^2 + (y − 2)^2 = 4 \)
(b) \( (x – 2)^2 + (y − 2)^2 = 16 \)
(c) \( (x - 4)^2 + (y − 4)^2 = 16 \)
(d) \( x^2 + y^2 = 4 \)
Answer: (c) \( (x – 2)^2 + (y − 2)^2 = 16 \)
In simple words: The perimeter of a circle is its circumference, which is found using the formula \( 2\pi r \). From the given perimeter, we can find the radius 'r'. Once we have the radius and the given center \( (h, k) \), we use the standard circle equation \( (x-h)^2 + (y-k)^2 = r^2 \) to write the full equation.

🎯 Exam Tip: Remember the relationship between circumference \( C = 2\pi r \) and radius 'r'. With the center \( (h,k) \) and radius 'r', the circle's equation is always \( (x-h)^2 + (y-k)^2 = r^2 \).

Question 20. The equation of the circle with centre (3, -4) and touches the x-axis is:
(a) \( (x − 3)^2 + (y − 4)^2 = 4 \)
(b) \( (x – 3)^2 + (y + 4)^2 = 16 \)
(c) \( (x − 3)^2 + (y − 4)^2 = 16 \)
(d) \( x^2 + y^2 = 16 \)
Answer: (b) \( (x – 3)^2 + (y + 4)^2 = 16 \)
In simple words: If a circle touches the x-axis, its radius is equal to the absolute value of the y-coordinate of its center. Here, the y-coordinate is -4, so the radius is 4. Then we use the center coordinates \( (h, k) \) and the radius 'r' in the general circle equation.

🎯 Exam Tip: If a circle with center \( (h, k) \) touches the x-axis, its radius is \( |k| \). If it touches the y-axis, its radius is \( |h| \). This fact simplifies finding the radius significantly.

Question 21. If the circle touches the x-axis, y-axis, and the line x = 6 then the length of the diameter of the circle is:
(a) 6
(b) 3
(c) 12<
(d) 4
Answer: (a) 6
In simple words: When a circle touches both the x and y axes, its center must be at \( (r, r) \) or \( (-r, r) \), where 'r' is the radius. If it also touches the line \( x = 6 \), then the distance from the center to this line must also be 'r'. This helps us find the radius, and then we can double it to get the diameter. The only possibility for the center given \( x=6 \) is \( (r, r) \) where \( 6-r = r \).

🎯 Exam Tip: If a circle touches both coordinate axes, its center is \( (\pm r, \pm r) \). If it also touches a vertical line \( x=k \), then \( |k-h| = r \). Similarly, if it touches a horizontal line \( y=k' \), then \( |k'-k| = r \). Combine these conditions to solve for 'r'.

Question 22. The eccentricity of the parabola is:
(a) 3
(b) 2
(c) 0
(d) 1
Answer: (d) 1
In simple words: Eccentricity is a measure that tells us how much a conic section (like a parabola, ellipse, or hyperbola) curves. For a parabola, this value is always exactly 1. It means the distance from any point on the parabola to the focus is equal to its distance to the directrix.

🎯 Exam Tip: Remember the eccentricities for different conic sections: for a parabola, eccentricity \( e=1 \); for an ellipse, \( 0 < e < 1 \); for a hyperbola, \( e > 1 \); and for a circle, \( e = 0 \).

Question 23. The double ordinate passing through the focus is:
(a) focal chord
(b) latus rectum
(c) directrix
(d) axis
Answer: (b) latus rectum
In simple words: In a parabola, a "double ordinate" is a line segment that is perpendicular to the axis of symmetry and connects two points on the parabola. If this specific line also passes through the focus of the parabola, it is given the special name "latus rectum".

🎯 Exam Tip: Understand the key terms related to conic sections: directrix, focus, vertex, axis, and latus rectum. The latus rectum is specifically the chord through the focus perpendicular to the axis.

Question 24. The distance between directrix and focus of a parabola \( y^2 = 4ax \) is
(a) a
(b) 2a
(c) 4a
(d) 3a
Answer: (b) 2a
In simple words: For a standard parabola like \( y^2 = 4ax \), the focus is at \( (a, 0) \) and the directrix is the line \( x = -a \). The distance between these two is found by subtracting the x-coordinate of the directrix from the x-coordinate of the focus, which gives \( a - (-a) = 2a \).

🎯 Exam Tip: For a parabola, the distance from the focus to the vertex is 'a', and the distance from the vertex to the directrix is also 'a'. Thus, the total distance from the focus to the directrix is \( 2a \).

Question 25. The equation of directrix of the parabola \( y^2 = -x \) is:
(a) \( 4x + 1 = 0 \)
(b) \( 4x – 1 = 0 \)
(c) \( x – 1 = 0 \)
(d) \( x + 4 = 0 \)
Answer: (b) \( 4x – 1 = 0 \)
In simple words: First, we compare the given equation with the standard form of a parabola that opens leftward, which is \( y^2 = -4ax \). From this comparison, we can find the value of 'a'. Once 'a' is known, the equation for the directrix of such a parabola is \( x = a \). We then rearrange this to the desired form.

🎯 Exam Tip: Identify the orientation of the parabola first. For \( y^2 = -4ax \) (opens left), the directrix is \( x = a \). For \( y^2 = 4ax \) (opens right), the directrix is \( x = -a \). Be careful with the sign of 'a' based on the direction of opening.