The parabola \(C\) has equation \(y^2=8x\). Write down the coordinates of the focus and the equation of the directrix.
Standard form \(y^2=4ax\) with \(4a=8\Rightarrow a=2\). Focus \((2,0)\), directrix \(x=-2\).
Conic Sections
12 exam-style questions on parabolas, rectangular hyperbolas, ellipses, tangents, normals and loci. Tap Show answer for full working.
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Find the equation of the tangent to the parabola \(y^2=12x\) at the point \((3,6)\).
\(4a=12\Rightarrow a=3\). Point \((at^2,2at)=(3,6)\Rightarrow t=1\). Tangent: \(ty=x+at^2\Rightarrow y=x+3\), or \(x-y+3=0\).
The rectangular hyperbola \(H\) has equation \(xy=16\). Find the coordinates of the points where \(H\) meets the line \(y=2x\).
\(x\cdot2x=16\Rightarrow 2x^2=16\Rightarrow x^2=8\Rightarrow x=\pm2\sqrt{2}\). Points \((2\sqrt{2},4\sqrt{2})\) and \((-2\sqrt{2},-4\sqrt{2})\).
A point \(P(at^2,2at)\) lies on the parabola \(y^2=4ax\). Show that the equation of the normal at \(P\) is \(y+tx=2at+at^3\).
Differentiate: \(2y\dfrac{\mathrm{d}y}{\mathrm{d}x}=4a\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}x}=\dfrac{2a}{y}=\dfrac{1}{t}\) at \(P\). Gradient of normal \(=-t\).
Normal: \(y-2at=-t(x-at^2)\Rightarrow y-2at=-tx+at^3\Rightarrow y+tx=2at+at^3\).
Find the equations of the tangents from the point \((1,4)\) to the parabola \(y^2=4x\).
\(a=1\). Tangent with parameter \(t\): \(ty=x+t^2\). Passes through \((1,4)\): \(4t=1+t^2\Rightarrow t^2-4t+1=0\Rightarrow t=2\pm\sqrt{3}\).
Tangents: \((2+\sqrt{3})y=x+(2+\sqrt{3})^2\) and \((2-\sqrt{3})y=x+(2-\sqrt{3})^2\).
The point \(P\bigl(ct,\dfrac{c}{t}\bigr)\) lies on the rectangular hyperbola \(xy=c^2\). Find the equation of the tangent at \(P\).
Differentiate \(xy=c^2\): \(x\dfrac{\mathrm{d}y}{\mathrm{d}x}+y=0\Rightarrow\dfrac{\mathrm{d}y}{\mathrm{d}x}=-\dfrac{y}{x}=-\dfrac{1}{t^2}\) at \(P\).
Tangent: \(y-\dfrac{c}{t}=-\dfrac{1}{t^2}\bigl(x-ct\bigr)\). Multiply by \(t^2\): \(t^2y-ct=-x+ct\Rightarrow x+t^2y=2ct\).
Show that the chord joining the points with parameters \(t_1\) and \(t_2\) on the parabola \(y^2=4ax\) has equation \(y(t_1+t_2)=2x+2at_1t_2\). Deduce the condition for the chord to pass through the focus.
Points \((at_1^2,2at_1)\), \((at_2^2,2at_2)\). Gradient \(\dfrac{2a(t_2-t_1)}{a(t_2^2-t_1^2)}=\dfrac{2}{t_1+t_2}\).
Equation using first point: \(y-2at_1=\dfrac{2}{t_1+t_2}(x-at_1^2)\). Multiply by \(t_1+t_2\): \(y(t_1+t_2)-2at_1(t_1+t_2)=2x-2at_1^2\Rightarrow y(t_1+t_2)=2x+2at_1t_2\).
Passes through focus \((a,0)\): \(0=2a+2at_1t_2\Rightarrow t_1t_2=-1\).
Find the equation of the normal to the rectangular hyperbola \(xy=9\) at the point \(\bigl(6,\dfrac{3}{2}\bigr)\).
\(c^2=9\), point corresponds to \(t=2\) (since \(ct=6\Rightarrow t=2\), \(c/t=3/2\)). Gradient of tangent \(=-\dfrac{1}{t^2}=-\dfrac{1}{4}\). Gradient of normal \(=4\).
Normal: \(y-\dfrac{3}{2}=4(x-6)\Rightarrow y-\dfrac{3}{2}=4x-24\Rightarrow y=4x-\dfrac{45}{2}\), or \(8x-2y-45=0\).
The point \(P(at^2,2at)\) lies on the parabola \(y^2=4ax\). The normal at \(P\) meets the parabola again at \(Q\). Show that the coordinates of \(Q\) are \(\bigl(a\bigl(t+\dfrac{2}{t}\bigr)^2,-2a\bigl(t+\dfrac{2}{t}\bigr)\bigr)\).
Normal: \(y+tx=2at+at^3\). Intersect with \(y^2=4ax\) again. Parametric: let second point have parameter \(m\). Then the normal at \(t\) is also the chord joining \(t\) and \(m\), so comparing with the chord formula (or solving) yields \(m=-t-\dfrac{2}{t}\).
Hence \(Q=\bigl(a m^2,2am\bigr)=\bigl(a\bigl(t+\dfrac{2}{t}\bigr)^2,-2a\bigl(t+\dfrac{2}{t}\bigr)\bigr)\).
Find the locus of the midpoint of a chord of the parabola \(y^2=4ax\) which subtends a right angle at the vertex.
Let ends have parameters \(t_1,t_2\). Midpoint \(\bigl(a\dfrac{t_1^2+t_2^2}{2},a(t_1+t_2)\bigr)\). Chord subtends right angle at vertex: directions from origin to the points are perpendicular, so \(at_1^2\cdot at_2^2+2at_1\cdot2at_2=0\Rightarrow t_1t_2(t_1t_2+4)=0\Rightarrow t_1t_2=-4\).
Also \(t_1^2+t_2^2=(t_1+t_2)^2-2t_1t_2\). Let \(h=a(t_1^2+t_2^2)/2\), \(k=a(t_1+t_2)\). Then \(t_1+t_2=k/a\), \(t_1t_2=-4\), so \(2h/a=(k/a)^2+8\Rightarrow 2ah=k^2+8a^2\Rightarrow\) locus \(y^2=2ax-8a^2\) (or \(y^2=2a(x-4a)\)).
The ellipse \(\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1\) has eccentricity \(e\). Show that the equation of the tangent at the point \((a\cos\theta,b\sin\theta)\) is \(\dfrac{x\cos\theta}{a}+\dfrac{y\sin\theta}{b}=1\). Find the locus of the point of intersection of perpendicular tangents.
Differentiate implicitly: \(\dfrac{2x}{a^2}+\dfrac{2y}{b^2}y'=0\Rightarrow y'=-\dfrac{b^2x}{a^2y}\). At the point, gradient \(=-\dfrac{b}{a}\cot\theta\). Equation of tangent follows by point-gradient form and simplification using the ellipse equation.
Perpendicular tangents: the director circle. For an ellipse the locus is \(x^2+y^2=a^2+b^2\).
A variable tangent to the parabola \(y^2=4ax\) meets the coordinate axes at \(A\) and \(B\). Find the locus of the midpoint of \(AB\).
Tangent \(ty=x+at^2\). Meets \(x\)-axis (\(y=0\)): \(x=-at^2\), so \(A(-at^2,0)\). Meets \(y\)-axis (\(x=0\)): \(y=at\), so \(B(0,at)\).
Midpoint \(\bigl(-\dfrac{1}{2}at^2,\dfrac{1}{2}at\bigr)\). Let \(h=-\dfrac{1}{2}at^2\), \(k=\dfrac{1}{2}at\Rightarrow t=\dfrac{2k}{a}\). Then \(h=-\dfrac{1}{2}a\cdot\dfrac{4k^2}{a^2}=-\dfrac{2k^2}{a}\). Locus: \(x=-\dfrac{2y^2}{a}\), or \(y^2=-\dfrac{1}{2}ax\).