Transform to axes inclined at  to the original axes the equation

The perpendicular bisector of the line segment joining P (1, 4) and Q (k, 3) has a Y–intercept of – 4. Then the possible value of k is

a) – 4

b) 1

c) 2

d) – 2

A parabola has the origin as its focus and the line x = 2 as the directrix. Then the vertex of the parabola is

a) (2, 0)

b) (0, 2)

c) (1, 0)

d) (0, 1)

The point diametrically opposite to the point P (1, 0) on the circle x² + y² + 2x + 4y – 3 = 0 is

a) (3, 4)

b) (3, − 4)

c) (− 3, 4)

d) (−3, −4)

Consider a family of circles passing through the point (−1, 1) and are tangent to the X–axis. If (h, k) are co-ordinates of the centre of circles then the set of values of k is given by the interval

a)

b)

c)

d)

For the hyperbola  which of the following remains constant when α varies

a) Eccentricity

b) Directrix

c) Abscissa of vertices

d) Abscissa of focus

Let A (h, k), B (1, 1) and C (2, 1) be the vertices of a right angled triangle with AC as its diagonal. If the area of the triangle is 1, the set of values of k are given by

a) {1, 3}

b) {0, 2}

c) {−1, 3}

d) {−3, −2}

Let P = (−1, 0), Q = (0, 0) and  be three points then the equation of the bisector of the angle PQR is

a)

b)

c)

d)

If one of the lines of my² + (1 – m²) xy – mx² = 0 is a bisector of the angles between the lines xy = 0 then m is

a)

b)  2

c)

d) 2

In an ellipse the distance between the foci is 6 and the minor axis is 8. Then the eccentricity is

a)

b)

c)

d)

A straight line through the point A (3, 4) is such that its intercepts between the axis is bisected at A. Its equation is

a) 3x – 4y + 7 = 0

b) 4x + 3y = 24

c) 3x + 4y = 25

d) x + y = 7

If (a, a²) lies inside the angle formed by the lines and y = 3x, x > 0 the a ε

a) (3, ∞)

b)

c)

d)

If two lines 3x – 4y – 7 = 0 and 2x – 3y – 5 = 0 are two diameters of a circle of area 49π square units, the equation of the circle is

a) x² + y² + 2x – 2y – 62 = 0

b) x² + y² – 2x + 2y – 62 = 0

c) x² + y² – 2x + 2y – 47 = 0

d) x² + y² + 2x – 2y – 47 = 0

Let C be a circle with centre (0, 0) and radius 3 units. The equation of the locus of the middle point of the chords of the circle that subtend an angle of  at its centre is

a) x² + y² = 1

b)

c)

d)

Let P be the point (1, 0) and Q be any point on y² = 8x. Then the locus of the middle point of PQ is

a) x² – 4y + 2 = 0

b) x² + 4y + 2 = 0

c) y² + 4x + 2 = 0

d) y² – 4x + 2 = 0

The area of the greatest rectangle that can be inscribed in the ellipse  is

a) square units

b)  square units

c) ab square units

d) 2ab square units

The line parallel to the X–axis and passing through the intersection of the lines
ax + 2by + 3b = 0 and
bx – 2ay – 3a = 0 where (a, b) ≠ (0, 0) is

a) Above the X–axis at a distance of  from it

b) Above the X–axis at a distance of  from it

c) Below the X–axis at a distance of  from it

d) Below the X–axis at a distance of  from it

If the non–zero numbers a, b, c are in Harmonic Progression then the straight line  always passes through a fixed point. This fixed point is

a)

b) 2)

c)

d)

If the vertex of a triangle is (1, 1) and the mid points of the two sides through the vertex are (−1, 2) and (3, 2) then the centroid of the triangle is

a)

b)

c)

d)

A circle touches the X–axis and touches the circle with centre (0, 3) and radius 2. The locus of the centre of the circle is

a) A parabola

b) A hyperbola

c) A circle

d) An ellipse

A circle passes through the point (a, b) and cuts the circle x² + y² = p² orthogonally then the equation of the locus of its centre is

a) 2ax + 2by – (a² + b² + p²) = 0

b) x² + y² – 2ax – 3by + (a² – b² – p²) = 0

c) 2ax + 2by – (a² – b² + p²) = 0

d) x² + y² – 3ax – 4by + (a² + b² – p²) = 0

The locus of a point P (α, β) moving under the condition that y = αx + β is a tangent to the hyperbola  is

a) A hyperbola

b) A parabola

c) A circle

d) An ellipse

If the pair of lines ax² + 2 (a + b) xy + by² = 0 lie along the diameter of a circle and divide the circle into four sectors such that the area of one of the sector is three times that of another sector then

a) 3a² + 2ab + 3b² = 0

b) 3a² + 10ab + 3b² = 0

c) 3a² − 2ab + 3b² = 0

d) 3a² − 10ab + 3b² = 0