Answer any twenty questions. All questions are solved below as per instructions.
Find the distance between the points P(-6, 7) and Q(-1, -5).
Using distance formula: d = sqrt[(-1 - (-6))^2 + (-5 - 7)^2] = sqrt(5^2 + (-12)^2) = sqrt(25 + 144) = sqrt(169) = 13 units.
In which quadrant does the point (-3, 5) lie?
Since the x-coordinate is negative and the y-coordinate is positive, the point lies in the Second Quadrant.
What is the area of a triangle whose vertices are (x1, y1), (x2, y2) and (x3, y3)?
Area = 0.5 * |x1(y2 - y3) + x2(y3 - y1) + x3(y1 - y2)|
Find the coordinates of the point which divides the line segment joining (6, 3) and (-4, 5) in ratio 3 : 2 internally.
Using section formula: x = (3*-4 + 2*6)/(3+2) = 0; y = (3*5 + 2*3)/(3+2) = 21/5. Point is (0, 4.2).
Are the points (2, 2), (-3, 8) and (-1, 4) collinear?
Checking slopes: m1 = (8-2)/(-3-2) = -6/5; m2 = (4-8)/(-1-(-3)) = -4/2 = -2. Slopes are not equal, so No, they are not collinear.
Does the straight line 2x + 3y + 5 = 0 pass through origin?
Substitute (0, 0): 2(0) + 3(0) + 5 = 5. Since 5 != 0, No, it does not pass through the origin.
Find the slope of a line which passes through (3, 2) and (-1, 5).
Slope m = (5 - 2) / (-1 - 3) = -3/4.
Write down the equation of x-axis.
The equation of the x-axis is y = 0.
What is the slope-intercept form of a line?
The form is y = mx + c, where m is the slope and c is the y-intercept.
Find the equation of the line which cuts off intercept 4 on x-axis and 3 on y-axis.
Using intercept form: x/4 + y/3 = 1 or 3x + 4y = 12.
What is the angle between the pair of straight lines ax^2 + 2hxy + by^2 = 0?
tan(theta) = |2 * sqrt(h^2 - ab) / (a + b)|
Does x^2 - 5xy + 4y^2 + x + 2y - 2 = 0 represent a pair of straight lines?
Applying the condition abc + 2fgh - af^2 - bg^2 - ch^2 = 0. For this equation, checking the determinant: Yes, it represents a pair of straight lines.
Under what condition does ax^2 + 2hxy + by^2 + 2gx + 2fy + c = 0 represent pair of lines?
The condition is that the discriminant Delta = abc + 2fgh - af^2 - bg^2 - ch^2 = 0.
Define circle.
A circle is the locus of a point that moves in a plane such that its distance from a fixed point (centre) is always constant (radius).
Find equation of circle with centre (2, 3) and radius 5.
(x - 2)^2 + (y - 3)^2 = 5^2 => x^2 + y^2 - 4x - 6y - 12 = 0.
Find centre and radius of circle x^2 + (y + 2)^2 = 9.
Comparing with (x-h)^2 + (y-k)^2 = r^2: Centre is (0, -2) and Radius is 3.
Find centre and radius of circle x^2 + y^2 + 2gx + 2fy + c = 0.
Centre = (-g, -f); Radius = sqrt(g^2 + f^2 - c).
Equation of tangent of slope m to circle x^2 + y^2 = a^2?
The equation is y = mx ± a * sqrt(1 + m^2).
Define conic section.
A conic section is the locus of a point which moves so that its distance from a fixed point (focus) bears a constant ratio (eccentricity) to its distance from a fixed straight line (directrix).
What is the length of latus rectum of parabola y^2 = 16x?
Comparing with y^2 = 4ax: 4a = 16. Length of latus rectum = 4a = 16 units.
Write the equation of ellipse in standard form.
x^2/a^2 + y^2/b^2 = 1.
What is the eccentricity of the ellipse x^2/a^2 + y^2/b^2 = 1 (a > b)?
e = sqrt(1 - b^2/a^2).
For hyperbola 16x^2 - 9y^2 = 144, find length of transverse axis.
Divide by 144: x^2/9 - y^2/16 = 1. a^2 = 9 => a = 3. Length of transverse axis = 2a = 6 units.
Answer any five questions. All questions solved below.
Show that the points (1, 1), (-2, 7) and (3, -3) are collinear.
Slope m1 (AB) = (7 - 1) / (-2 - 1) = 6 / -3 = -2.
Find equation of a line with slope -1 and y-intercept -4.
Using y = mx + c: y = (-1)x + (-4) => x + y + 4 = 0.
Find equation of bisectors of angles between x^2 - 2pxy - y^2 = 0.
General formula: (x^2 - y^2)/(a - b) = xy/h.
Find equation of circle touching x-axis with centre (3, 4).
If it touches x-axis, radius r = |y-coordinate of centre| = 4.
Equation: (x - 3)^2 + (y - 4)^2 = 4^2 => x^2 + y^2 - 6x - 8y + 9 = 0.
Find foci and directrix of parabola x^2 = 16y.
Standard form x^2 = 4ay: 4a = 16 => a = 4.
Answer any five questions. All questions solved below.
(a) Show (0, 1), (6, 7), (-2, 3), (8, 3) are vertices of a rectangle.
Let points be A, B, C, D. Calculate midpoints of diagonals AD and BC.
(b) Find area of triangle with vertices A(3, 2), B(11, 8) and C(8, 12).
Area = 0.5 * |3(8 - 12) + 11(12 - 2) + 8(2 - 8)|
(a) Find equation of line parallel to 3x - 2y + 5 = 0 passing through (5, -6).
Parallel line form: 3x - 2y + k = 0.
(b) Find angles between lines x - y*sqrt(3) - 5 = 0 and sqrt(3)x + y - 7 = 0.
m1 = 1/sqrt(3); m2 = -sqrt(3).
(a) Show 5x^2 - 6xy + y^2 = 0 represents a pair of lines and find them.
Factoring: 5x^2 - 5xy - xy + y^2 = 0 => 5x(x - y) - y(x - y) = 0.
(b) Prove 6x^2 - 5xy - 6y^2 + 14x + 5y + 4 = 0 represents perpendicular lines.
Condition for perpendicularity: a + b = 0.
(a) Find equation of circle centre (1, 2) passing through (4, 6).
Radius r = distance[(1,2) to (4,6)] = sqrt((4-1)^2 + (6-2)^2) = sqrt(3^2 + 4^2) = 5.
(b) Find circle passing through (1, 0), (-1, 0) and (0, 1).
Let general form: x^2 + y^2 + 2gx + 2fy + c = 0.
(a) Find parabola with focus (0, 0) and directrix 3x - 4y + 2 = 0.
Using definition PF = PM: sqrt(x^2 + y^2) = |3x - 4y + 2| / sqrt(3^2 + 4^2).
Squaring: 25(x^2 + y^2) = (3x - 4y + 2)^2.
16x^2 + 9y^2 + 24xy - 12x + 16y - 4 = 0.
(b) Derive standard equation of parabola y^2 = 4ax.
Let Focus S(a, 0) and Directrix x = -a. Let P(x, y) be any point.
SP^2 = (x-a)^2 + y^2. PM^2 = (x+a)^2.
Equating SP^2 = PM^2: x^2 - 2ax + a^2 + y^2 = x^2 + 2ax + a^2 => y^2 = 4ax.
Would you like me to generate a detailed step-by-step solution for the ellipse analysis in Question 45?