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 \(\dfrac{9x^2}{16}-\dfrac{9y^2}{20}=1\).