If a ≠ 0 and the line 2bx + 3cy + 4d = 0 passes through the points of intersection of the parabolas y^{2}+ 4ax = and x^{2}+ 4ay = , then
d^{2} + (2b+3c)^{2} = 0
d^{2} +(3d+2c^{2}) = 0
d^{2} + (2b-3c)^{2} = 0
d^{2} + (3b-2c)^{2} = 0
A.
d^{2} + (2b+3c)^{2} = 0
The equation of parabolas are y^{2} = 4ax and x^{2} = 4ay
On solving these we get x = 0 and x = 4a Also y = 0 and y = 4a
∴ The point of intersection of parabolas are A(0, 0) and B(4a, 4a). Also line 2bx + 3cy + 4d = 0 passes through A and B.
. ∴ d = 0 ............ (i)
or 2b . 4a + 3c . 4a + 4d = 0 2ab + 3ac + d = 0
a(2b + 3c) = 0 (Qd = 0)
⇒ 2b + 3c = 0 ............ (ii) On squaring equation (i) and (ii) and then adding, we get
d^{2} + (2b + 3c)^{2} = 0
If a circle passes through the point (a, b) and cuts the circle x^{2} +y^{2}= 4 orthogonally, then the locus of its centre is
2ax +2by + (a^{2} +b^{2}+4)=0
2ax +2by - (a^{2} +b^{2}+4)=0
2ax -2by - (a^{2} +b^{2}+4)=0
2ax -2by + (a^{2} +b^{2}+4)=0
B.
2ax +2by - (a^{2} +b^{2}+4)=0
Let the equation of circle is
x^{2} + y^{2} + 2gx + 2fy + c = 0
It cut the circle x^{2} + y^{2} = 4 orthogonally
if 2g.0 + 2f.0 = c - 4
⇒ c = 4
∴ Equation of circle is
x^{2} + y^{2} + 2gx + 2fy + 4 = 0
Q It passes through the point (a, b)
∴ a^{2} + b^{2} + 2ag + 2f + 4 = 0
Locus of centre (-g, -f) will be a^{2}+ b^{2}- 2xa - 2yb + 4 = 0
2ax + 2by - (a^{2} + b^{2} + 4) = 0
If the lines 2x + 3y + 1 = 0 and 3x – y – 4 = 0 lie along diameters of a circle of circumference 10π, then the equation of the circle is
x^{2} + y^{2}- 2x +2y -23 = 0
x^{2} - y^{2}- 2x -2y -23 = 0
x^{2} - y^{2}- 2x -2y +23 = 0
x^{2} + y^{2}+ 2x +2y -23 = 0
A.
x^{2} + y^{2}- 2x +2y -23 = 0
The lines 2x + 3y + 1 = 0 and 3x - y - 4 = 0 are diameters of circle.
On solving these equations we get x = 1, y = -1
Therefore the centre of circle = (1, -1) and circumference = 10 π
2πr = 10π
⇒ r = 5
∴ Equation of circle (x - x_{1} )^{2} + (y - y_{1} )^{2} = r_{2}
(x - 1)^{2} + (y + 1)^{2} = 52
x^{2} + 1 - 2x + y^{2} + 2y + 1 = 25
x^{2} + y^{2} - 2x + 2y - 23 = 0
Area of the greatest rectangle that can be inscribed in the ellipse
2ab
ab
a/b
A.
2ab
The eccentricity of an ellipse, with its centre at the origin, is 1 /2 . If one of the directrices is x = 4, then the equation of the ellipse is
3x^{2} +4y^{2} = 1
3x^{2}+ 4y^{2} = 12
4x^{2} +3y^{2} = 12
4x^{2}+ 3y^{2} = 1
B.
3x^{2}+ 4y^{2} = 12
Since the directrix is x = 4 then ellipse is parallel to X-axis.
⇒ a/e = 4
⇒ a = 4e = 4 x (1/2)
⇒ a = 2
Also we know that
b^{2} =a2 (1-e2)
b^{2} = 4(1-1/4) = 4 x 3/4
b^{2} = 3
therefore equation of ellipse is
an ellipse
a circle
a straight line
a parabola
C.
a straight line
As given ⇒ distance of z from origin and point (0,1/3) is same hence z lies on the bisector of the line joining points (0, 0) and (0, 1/3). Hence z lies on a straight line.
Let P be the point (1, 0) and Q a point on the locus y^{2} = 8x. The locus of mid point of PQ is
y^{2} – 4x + 2 = 0
y^{2} + 4x + 2 = 0
x^{2} + 4y + 2 =
x^{2} – 4y + 2 = 0
A.
y^{2} – 4x + 2 = 0
P = (1, 0) Q = (h, k) such that
k^{2} = 8h
Let (α, β) be the midpoint of PQ
If |z^{2}-1|=|z|^{2}+1, then z lies on
the real axis
an ellipse
a circle
the imaginary axis
B.
an ellipse
Given that
|z^{2}- 1| = |z|^{2}+ 2
|z^{2} + (-1)| = |z^{2}| + |-1|
It shows that the origin, -1 and z^{2} lies on a line and z^{2} and -1 lies on one side of the origin, therefore
z2 is a negative number. Hence z will be purely imaginary. So we can say that z lies on y-axis.
The locus of a point P (α, β) moving under the condition that the line y = αx + β is a tangent to the hyperbola
an ellipse
a circle
a parabola
a hyperbola
D.
a hyperbola
Tangent to the hyperbola
Given that y = αx + β is the tangent of hyperbola
⇒ m = α and a^{2} m^{2} – b^{2} = β^{2}
∴ a^{2} α^{2} – b^{2} = β^{2}
Locus is a^{2} x^{2} – y^{2} = b^{2} which is hyperbola.
A variable circle passes through the fixed point A (p, q) and touches x-axis. The locus of the other end of the diameter through A is
(x-p)^{2} = 4qy
(x-q)^{2} = 4py
(y-p)^{2} = 4qx
(y-p)^{2} = 4px
A.
(x-p)^{2} = 4qy
In a circle, AB is a diameter where the co-ordinate of A is (p, q) and let the co-ordinate of B is (x_{1} , y_{1} ).
Equation of circle in diameter form is (x - p)(x - x_{1} ) + (y - q)(y - y_{1} ) = 0
x^{2} - (p + x_{1} )x + px_{1} + y^{2} - (y_{1} + q)y + qy_{1} = 0
x^{2} - (p + x_{1} )x + y^{2} - (y_{1} + q)y + px_{1} + qy_{1} = 0
Since this circle touches X-axis
∴ y = 0
⇒ x^{2} - (p + x1 )x + px_{1} + qy_{1} = 0 Also the discriminant of above equation will be equal to zero because circle touches X-axis.
∴ (p + x_{1} )^{2} = 4(px_{1} + qy_{1}) p^{2} + x^{2}_{1} + 2px_{1}
= 4px_{1} + 4qy_{1} x^{2}_{1} - 2px_{1} + p^{2} = 4qy1
Therefore the locus of point B is, (x - p)^{2} = 4qy