Find the equation of the circle which touches both axes in the first quadrant and whose radius is \(a\).
\(x^2 + y^2 - 2ax - 2ay + a^2 = 0\)
Show that the point \((x,y)\) given by \(x=\dfrac{2at}{1+t^2}\) and \(y=\dfrac{a(1-t^2)}{1+t^2}\) lies on a circle for all real values of \(t\) such that \(-1\le t\le 1\), where \(a\) is any given real number.
Substitute and compute \(x^2+y^2=\dfrac{4a^2t^2+a^2(1-t^2)^2}{(1+t^2)^2}=a^2\). Hence the point lies on the circle \(x^2+y^2=a^2\).
If a circle passes through the points \((0,0)\), \((a,0)\), \((0,b)\), then find the coordinates of its centre.
\(\left(\dfrac{a}{2},\dfrac{b}{2}\right)\)
Find the equation of the circle which touches the x-axis and whose centre is \((1,2)\).
Centre \((1,2)\) and radius = distance to x-axis = 2. Equation: \((x-1)^2+(y-2)^2=4\), i.e. \(x^2+y^2-2x-4y+1=0\).
If the lines \(3x-4y+4=0\) and \(6x-8y-7=0\) are tangents to a circle, then find the radius of the circle.
The two (parallel) lines can be written with same left-hand side as \(6x-8y+8=0\) and \(6x-8y-7=0\). Distance between them is \(\dfrac{15}{\sqrt{36+64}}=\dfrac{15}{10}=\dfrac{3}{2}\), which is the diameter. Hence radius = \(\dfrac{3}{4}\).
Find the equation of a circle which touches both the axes and the line \(3x-4y+8=0\) and lies in the third quadrant.
Centre is \((-a,-a)\). Distance from centre to line gives \(\dfrac{| -3a+4a+8|}{5}=a\) so \(|a+8|=5a\). Solving gives \(a=2\). Centre \((-2,-2)\), radius 2. Equation: \((x+2)^2+(y+2)^2=4\) or \(x^2+y^2+4x+4y+4=0\).
If one end of a diameter of the circle \(x^2+y^2-4x-6y+11=0\) is \((3,4)\), then find the coordinates of the other end of the diameter.
Center of circle is \((2,3)\). If one end is \((3,4)\), other end is \((2\cdot2-3,2\cdot3-4)=(1,2)\).
Find the equation of the circle having centre \((1,-2)\) and passing through the point of intersection of the lines \(3x+y=14\) and \(2x+5y=18\).
The lines meet at \((4,2)\). Radius = distance from \((1,-2)\) to \((4,2)\) = 5. Equation: \((x-1)^2+(y+2)^2=25\) i.e. \(x^2+y^2-2x+4y-20=0\).
If the line \(y=\sqrt{3}x+k\) touches the circle \(x^2+y^2=16\), then find the value(s) of \(k\).
Distance from origin to line is \(\dfrac{|k|}{\sqrt{1+3}}=\dfrac{|k|}{2}\). Equate to radius 4 gives \(|k|=8\). So \(k=\pm 8\).
Find the equation of a circle concentric with the circle \(x^2+y^2-6x+12y+15=0\) and having double of its area.
Center of given circle is \((3,-6)\) and its radius squared is 30. Double area means new radius squared = 60. Equation: \((x-3)^2+(y+6)^2=60\) or \(x^2+y^2-6x+12y-15=0\).
If the latus rectum of an ellipse is equal to half of the minor axis, then find its eccentricity.
\(\dfrac{\sqrt{3}}{2}\)
Given the ellipse with equation \(9x^2 + 25y^2 = 225\), find the eccentricity and the foci.
Write as \(\dfrac{x^2}{25}+\dfrac{y^2}{9}=1\). So \(a^2=25,\ b^2=9\). Then \(c^2=a^2-b^2=16\) hence \(c=4\) and eccentricity \(e=\dfrac{c}{a}=\dfrac{4}{5}\).
Foci: \((\pm4,0)\).
If the eccentricity of an ellipse is \(\dfrac{5}{8}\) and the distance between its foci is 10, then find the latus rectum of the ellipse.
Here \(2c=10\Rightarrow c=5\). Since \(e=\dfrac{c}{a}=\dfrac{5}{8}\), we get \(a=8\). Then \(b^2=a^2(1-e^2)=64\left(1-\dfrac{25}{64}\right)=39\).
Latus rectum \(=\dfrac{2b^2}{a}=\dfrac{2\cdot39}{8}=\dfrac{39}{4}\).
Find the equation of the ellipse whose eccentricity is \(\dfrac{2}{3}\), latus rectum is 5 and the centre is \((0,0)\).
For ellipse with major axis along x, latus rectum \(=\dfrac{2b^2}{a}=5\) and \(e=\dfrac{2}{3}\Rightarrow b^2=a^2(1-e^2)=a^2\cdot\dfrac{5}{9}\).
So \(\dfrac{2b^2}{a}=\dfrac{2a\cdot5}{9}=\dfrac{10}{9}a=5\Rightarrow a=\dfrac{9}{2}\). Thus \(a^2=\dfrac{81}{4},\ b^2=\dfrac{45}{4}\).
Equation: \(\dfrac{x^2}{81/4}+\dfrac{y^2}{45/4}=1\) or \(\dfrac{4x^2}{81}+\dfrac{4y^2}{45}=1\).
Find the distance between the directrices of the ellipse \(\dfrac{x^2}{36}+\dfrac{y^2}{20}=1\).
Here \(a^2=36\Rightarrow a=6,\ b^2=20\). Then \(e^2=1-\dfrac{b^2}{a^2}=1-\dfrac{20}{36}=\dfrac{16}{36}\Rightarrow e=\dfrac{4}{6}=\dfrac{2}{3}\).
Directrices: \(x=\pm\dfrac{a}{e}=\pm\dfrac{6}{2/3}=\pm9\). Distance between them = \(18\).
Find the coordinates of a point on the parabola \(y^2=8x\) whose focal distance is 4.
For \(y^2=4ax\) we have \(4a=8\Rightarrow a=2\) and focus \((2,0)\). Let point be \((x,y)\) with \(x=\dfrac{y^2}{8}\). Distance to focus:
\(\sqrt{(x-2)^2+y^2}=4\). Solving gives \(y=\pm4,\ x=2\).
Points: \((2,4)\) and \((2,-4)\).
Find the length of the line-segment joining the vertex of the parabola \(y^2=4ax\) and a point on the parabola where the line-segment makes an angle \(\theta\) to the x-axis.
Using parameter \(t\) with slope condition one obtains the distance
\(\dfrac{4a\cos\theta}{\sin^2\theta}\)
If the points \((0,4)\) and \((0,2)\) are respectively the vertex and focus of a parabola, then find the equation of the parabola.
Vertex \((0,4)\), focus \((0,2)\) so \(p= -2\) (parabola opens downward). Equation: \(x^2=4p(y-4)= -8(y-4)\) or \(x^2+8y-32=0\).
If the line \(y=mx+1\) is tangent to the parabola \(y^2=4x\) then find the value of \(m\).
Substitute and apply tangency condition (discriminant = 0) to get \(m=1\).
If the distance between the foci of a hyperbola is 16 and its eccentricity is \(\sqrt{2}\), then obtain the equation of the hyperbola.
Here \(2c=16\Rightarrow c=8\) and \(e=\dfrac{c}{a}=\sqrt{2}\Rightarrow a=\dfrac{c}{e}=\dfrac{8}{\sqrt{2}}=4\sqrt{2}\). Then \(a^2=32,\ b^2=c^2-a^2=64-32=32\).
Equation: \(\dfrac{x^2}{32}-\dfrac{y^2}{32}=1\) or simply \(x^2-y^2=32\).
Find the eccentricity of the hyperbola \(9y^2-4x^2=36\).
Rewrite as \(\dfrac{y^2}{4}-\dfrac{x^2}{9}=1\). So \(a^2=4,\ b^2=9,\ c^2=a^2+b^2=13\). Hence \(e=\dfrac{c}{a}=\dfrac{\sqrt{13}}{2}\).
Find the equation of the hyperbola with eccentricity \(\dfrac{3}{2}\) and foci at \((\pm2,0)\).
Given \(c=2\) and \(e=\dfrac{3}{2}\Rightarrow a=\dfrac{c}{e}=\dfrac{2}{3/2}=\dfrac{4}{3}\), so \(a^2=\dfrac{16}{9},\ b^2=c^2-a^2=4-\dfrac{16}{9}=\dfrac{20}{9}\).
Equation: \(\dfrac{x^2}{16/9}-\dfrac{y^2}{20/9}=1\) or equivalently \(\dfrac{9x^2}{16}-\dfrac{9y^2}{20}=1\).
If the lines \(2x - 3y = 5\) and \(3x - 4y = 7\) are the diameters of a circle of area 154 square units, then obtain the equation of the circle.
\(x^2 + y^2 - 2x + 2y = 47\)
Find the equation of the circle which passes through the points \((2,3)\) and \((4,5)\) and the centre lies on the straight line \(y - 4x + 3 = 0\).
\(x^2 + y^2 - 4x - 10y + 25 = 0\)
Find the equation of a circle whose centre is \((3,-1)\) and which cuts off a chord of length 6 units on the line \(2x - 5y + 18 = 0\).
\((x - 3)^2 + (y + 1)^2 = 38\)
Find the equation of a circle of radius 5 which is touching another circle \(x^2 + y^2 - 2x - 4y - 20 = 0\) at \((5,5)\).
\(x^2 + y^2 - 18x - 16y + 120 = 0\)
Find the equation of a circle passing through the point \((7,3)\) having radius 3 units and whose centre lies on the line \(y = x - 1\).
\(x^2 + y^2 - 8x - 6y + 16 = 0\)
Find the equation of each of the following parabolas:
(a) Directrix \(x = 0\), focus at \((6,0)\)
(b) Vertex at \((0,4)\), focus at \((0,2)\)
(c) Focus at \((-1,-2)\), directrix \(x - 2y + 3 = 0\)
(a) \(y^2 = 12x - 36\)
(b) \(x^2 = 32 - 8y\)
(c) \(4x^2 + 4xy + y^2 + 4x + 32y + 16 = 0\)
Find the equation of the set of all points the sum of whose distances from the points \((3,0)\) and \((9,0)\) is 12.
\(3x^2 + 4y^2 - 36x = 0\)
Find the equation of the set of all points whose distance from \((0,4)\) are \(\tfrac{2}{3}\) of their distance from the line \(y = 9\).
\(9x^2 + 5y^2 = 180\)
Show that the set of all points such that the difference of their distances from \((4,0)\) and \((-4,0)\) is always equal to 2 represent a hyperbola.
\(15x^2 - y^2 = 15\)
Find the equation of the hyperbola with
(a) Vertices \((\pm5,0)\), foci \((\pm7,0)\)
(b) Vertices \((0,\pm7)\), eccentricity \(\dfrac{4}{3}\)
(c) Foci \((0,\pm\sqrt{10})\), passing through \((2,3)\)
(a) \(15x^2 - y^2 = 15\)
(b) \(9x^2 - 7y^2 + 343 = 0\)
(c) \(y^2 - x^2 = 5\)
The line \(x + 3y = 0\) is a diameter of the circle \(x^2 + y^2 + 6x + 2y = 0\).
False
The shortest distance from the point \((2, -7)\) to the circle \(x^2 + y^2 - 14x - 10y - 151 = 0\) is equal to 5.
False
If the line \(lx + my = 1\) is a tangent to the circle \(x^2 + y^2 = a^2\), then the point \((l,m)\) lies on a circle.
True
The point \((1,2)\) lies inside the circle \(x^2 + y^2 - 2x + 6y + 1 = 0\).
False
The line \(lx + my + n = 0\) will touch the parabola \(y^2 = 4ax\) if \(ln = am^2\).
True
If P is a point on the ellipse \(\dfrac{x^2}{16} + \dfrac{y^2}{25} = 1\) whose foci are S and S′, then \(PS + PS′ = 8\).
False
The line \(2x + 3y = 12\) touches the ellipse \(\dfrac{x^2}{9} + \dfrac{y^2}{4} = 2\) at the point \((3,2)\).
True
The locus of the point of intersection of lines \(\sqrt{3}x - y - 4\sqrt{3}k = 0\) and \(\sqrt{3}kx + ky - 4\sqrt{3} = 0\) for different values of k is a hyperbola whose eccentricity is 2.
True
The equation of the circle having centre at (3, −4) and touching the line \(5x + 12y - 12 = 0\) is ____.
\((x-3)^2 + (y+4)^2 = \left(\dfrac{45}{13}\right)^2\)
The equation of the circle circumscribing the triangle whose sides are the lines \(y = x + 2\), \(3y = 4x\), \(2y = 3x\) is ____.
\(x^2 + y^2 - 46x + 22y = 0\)
An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the length of the string and distance between the pins are ____.
Length of string = \(6 + 2\sqrt{5}\), distance between pins = \(2\sqrt{5}\).
The equation of the ellipse having foci \((0,1)\), \((0,-1)\) and minor axis of length 1 is ____.
\(4x^2 + \dfrac{4}{5}y^2 = 1\)
The equation of the parabola having focus at \((-1,-2)\) and the directrix \(x - 2y + 3 = 0\) is ____.
\(4x^2 + 4xy + y^2 + 4x + 32y + 16 = 0\)
The equation of the hyperbola with vertices at \((0, \pm 6)\) and eccentricity \(\dfrac{5}{3}\) is ____ and its foci are ____.
Equation: \(\dfrac{y^2}{36} - \dfrac{x^2}{64} = 1\). Foci: \((0, \pm 10)\).
The area of the circle centred at (1, 2) and passing through (4, 6) is
5π
10π
25π
none of these
Equation of a circle which passes through (3, 6) and touches the axes is
x² + y² + 6x + 6y + 3 = 0
x² + y² − 6x − 6y − 9 = 0
x² + y² − 6x − 6y = 0
none of these
Equation of the circle with centre on the y-axis and passing through the origin and the point (2, 3) is
x² + y² + 13y = 0
3x² + 3y² + 13x + 3 = 0
6x² + 6y² − 13x = 0
x² + y² + 13x + 3 = 0
The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
x² + y² = 9a²
x² + y² = 16a²
x² + y² = 4a²
x² + y² = a²
If the focus of a parabola is (0, −3) and its directrix is y = 3, then its equation is
x² = −12y
x² = 12y
y² = −12x
y² = 12x
If the parabola y² = 4ax passes through the point (3, 2), then the length of its latus rectum is
2/3
4/3
1/3
4
If the vertex of the parabola is the point (−3, 0) and the directrix is the line x + 5 = 0, then its equation is
y² = 8(x + 3)
x² = 8(y + 3)
y² = −8(x + 3)
y² = 8(x + 5)
The equation of the ellipse whose focus is (1, −1), the directrix the line x − y − 3 = 0 and eccentricity \(\dfrac{1}{2}\) is
7x² + 2xy + 7y² − 10x + 10y + 7 = 0
7x² + 2xy + 7y² + 7 = 0
7x² + 2xy + 7y² + 10x − 10y − 7 = 0
none
The length of the latus rectum of the ellipse 3x² + y² = 12 is
4
3
8
4/√3
If e is the eccentricity of the ellipse \(\dfrac{x²}{a²} + \dfrac{y²}{b²} = 1\) (a < b), then
b² = a²(1 − e²)
a² = b²(1 − e²)
a² = b²(e² − 1)
b² = a²(e² − 1)
The eccentricity of the hyperbola whose latus rectum is 8 and conjugate axis is equal to half of the distance between the foci is
4/3
4/√3
2/√3
none of these
The distance between the foci of a hyperbola is 16 and its eccentricity is \(\sqrt{2}\). Its equation is
x² − y² = 32
x²/4 − y²/9 = 1
2x − 3y² = 7
none of these
Equation of the hyperbola with eccentricity \(\dfrac{3}{2}\) and foci at (±2, 0) is
x²/4 − y²/5 = 4/9
x²/9 − y²/9 = 4/9
x²/4 − y²/9 = 1
none of these