NCERT Solutions · Class 11 Mathematics Conic Sections

Put the vertex of the axial cross-section at the origin.A parabolic reflector is generated by revolving a parabola about its axis. Take the vertex at the origin and the axis along the \(\displaystyle x\)-axis, so the cross-section is \[y^{2}=4ax .\]The reflector is \(\displaystyle 5\) cm deep, so the rim lies in the plane \(\displaystyle x=5\); its diameter is \(\displaystyle 20\) cm, so at \(\displaystyle x=5\) we have \(\displaystyle y=\pm 10\). Substituting the point \(\displaystyle (5,10)\): \[10^{2}=4a(5)\ \Rightarrow\ 100=20a\ \Rightarrow\ a=5 .\]The focus of \(\displaystyle y^{2}=4ax\) is \(\displaystyle (a,0)\), i.e. \(\displaystyle (5,0)\). Since the rim is also at \(\displaystyle x=5\), the focus happens to lie exactly in the plane of the opening, at its centre.The focus is on the axis, \(\displaystyle 5\) cm from the vertex — the point \(\displaystyle (5,0)\) in this frame.

Take the vertex of the arch at the origin, opening downwards.The axis is vertical and the vertex is the highest point of the arch, so with the vertex at the origin the parabola is \[x^{2}=-4ay,\qquad a>0,\] and points of the arch have \(\displaystyle y\le 0\), with \(\displaystyle |y|\) the depth below the vertex.The arch is \(\displaystyle 10\) m high and \(\displaystyle 5\) m wide at the base, so the base points are \(\displaystyle \left(\pm\tfrac{5}{2},-10\right)\). Substituting \(\displaystyle \left(\tfrac52,-10\right)\): \[\frac{25}{4}=-4a(-10)=40a\ \Rightarrow\ 4a=\frac{25}{40}=\frac{5}{8}.\]So the arch is \(\displaystyle x^{2}=-\dfrac{5}{8}y\).At a point \(\displaystyle 2\) m below the vertex, \(\displaystyle y=-2\): \[x^{2}=-\frac58(-2)=\frac54\ \Rightarrow\ x=\pm\frac{\sqrt5}{2}.\]The width there is \(\displaystyle 2|x|=\sqrt5\).The arch is \(\displaystyle \sqrt{5}\approx 2.24\) m wide at a height \(\displaystyle 2\) m below (i.e. \(\displaystyle 2\) m from) the vertex.

Take the origin at the lowest point of the cable.The cable is a parabola with a vertical axis; its lowest point is at the middle of the span, where the supporting wire is shortest, \(\displaystyle 6\) m. Put the origin at that lowest point of the cable, so \[x^{2}=4ay,\] and the roadway is the horizontal line \(\displaystyle 6\) m below the origin.The roadway is \(\displaystyle 100\) m long, so the cable is fixed at \(\displaystyle x=\pm 50\); there the wire is longest, \(\displaystyle 30\) m, so the cable is \(\displaystyle 30-6=24\) m above the vertex. Substituting \(\displaystyle (50,24)\): \[50^{2}=4a(24)\ \Rightarrow\ 4a=\frac{2500}{24}=\frac{625}{6}.\]So \(\displaystyle x^{2}=\dfrac{625}{6}\,y\).At \(\displaystyle 18\) m from the middle, \(\displaystyle x=18\): \[y=\frac{6x^{2}}{625}=\frac{6(324)}{625}=\frac{1944}{625}=3.1104 .\]The wire runs from the roadway up to the cable, so its length is \(\displaystyle 6+3.1104\).A sketch showing the roadway, the sagging cable and the vertical wires makes the "\(\displaystyle 6\) m plus the rise" step easy to see.The supporting wire is \(\displaystyle 9.11\) m long (exactly \(\displaystyle 9.1104\) m).

Semi-ellipse with its centre at the origin.Put the centre of the semi-ellipse at the origin with the base along the \(\displaystyle x\)-axis: \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad y\ge 0 .\]The arch is \(\displaystyle 8\) m wide, so \(\displaystyle 2a=8\) and \(\displaystyle a=4\); it is \(\displaystyle 2\) m high at the centre, so \(\displaystyle b=2\). Thus \[\frac{x^{2}}{16}+\frac{y^{2}}{4}=1 .\]The ends of the arch are at \(\displaystyle x=\pm 4\), so a point \(\displaystyle 1.5\) m from one end has \(\displaystyle x=4-1.5=2.5\). Then \[\frac{y^{2}}{4}=1-\frac{(2.5)^{2}}{16}=1-\frac{6.25}{16}=\frac{9.75}{16} \ \Rightarrow\ y^{2}=\frac{39}{16}\ \Rightarrow\ y=\frac{\sqrt{39}}{4}.\]The height there is \(\displaystyle \dfrac{\sqrt{39}}{4}\approx 1.56\) m.

Write the ends of the rod as intercepts, then eliminate them.Let the rod be \(\displaystyle AB\) with \(\displaystyle A=(\alpha,0)\) on the \(\displaystyle x\)-axis and \(\displaystyle B=(0,\beta)\) on the \(\displaystyle y\)-axis. Since the rod has length \(\displaystyle 12\), \[\alpha^{2}+\beta^{2}=144 .\tag{1}\]\(\displaystyle P\) is \(\displaystyle 3\) cm from the end touching the \(\displaystyle x\)-axis, so \(\displaystyle AP=3\) and \(\displaystyle PB=12-3=9\); that is, \(\displaystyle P\) divides \(\displaystyle AB\) in the ratio \(\displaystyle 1:3\) measured from \(\displaystyle A\). By the section formula, \[P=\left(\frac{1\cdot 0+3\cdot\alpha}{1+3},\ \frac{1\cdot\beta+3\cdot 0}{1+3}\right)=\left(\frac{3\alpha}{4},\ \frac{\beta}{4}\right).\]So if \(\displaystyle P=(x,y)\), then \[x=\frac{3\alpha}{4}\ \Rightarrow\ \alpha=\frac{4x}{3},\qquad y=\frac{\beta}{4}\ \Rightarrow\ \beta=4y .\]Substituting in ($\displaystyle 1$): \[\frac{16x^{2}}{9}+16y^{2}=144 .\]Dividing by \(\displaystyle 144\):A small diagram of the rod sliding with its ends on the axes, with \(\displaystyle P\) marked \(\displaystyle 3\) cm from the \(\displaystyle x\)-axis end, would help here.\(\displaystyle \dfrac{x^{2}}{81}+\dfrac{y^{2}}{9}=1\) — the locus is an ellipse.

Find the ends of the latus rectum, then use \(\displaystyle \tfrac12\times\text{base}\times\text{height}\).For \(\displaystyle x^{2}=4ay\) the vertex is \(\displaystyle (0,0)\), the focus is \(\displaystyle (0,a)\) and the latus rectum is the chord through the focus perpendicular to the axis.Here \(\displaystyle x^{2}=12y\), so \(\displaystyle 4a=12\) and \(\displaystyle a=3\); the focus is \(\displaystyle (0,3)\).Putting \(\displaystyle y=3\) in \(\displaystyle x^{2}=12y\) gives \(\displaystyle x^{2}=36\), so \(\displaystyle x=\pm 6\). The ends of the latus rectum are \[L(-6,3)\quad\text{and}\quad L'(6,3).\]The triangle \(\displaystyle OLL'\) has base \(\displaystyle LL'=12\) (along the line \(\displaystyle y=3\)) and height equal to the distance from \(\displaystyle O(0,0)\) to that line, which is \(\displaystyle 3\). Hence \[\text{Area}=\frac12\times 12\times 3=18 .\]Area \(\displaystyle =18\) square units.

The defining property of an ellipse.The sum of the distances of the man from the two flag posts is constant, so by definition his path is an ellipse whose foci are the two flag posts.Take the mid-point of the posts as the origin and the line joining them as the \(\displaystyle x\)-axis: \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad \text{foci }(\pm c,0),\ b^{2}=a^{2}-c^{2}.\]The constant sum is \(\displaystyle 2a\) and the distance between the foci is \(\displaystyle 2c\), so \[2a=10\ \Rightarrow\ a=5,\qquad 2c=8\ \Rightarrow\ c=4 .\]Hence \[b^{2}=a^{2}-c^{2}=25-16=9 .\]The path traced is the ellipse \(\displaystyle \dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}=1\).

Use the symmetry of the parabola about its axis.Let the triangle be \(\displaystyle OPQ\) with \(\displaystyle O=(0,0)\), the vertex of \(\displaystyle y^{2}=4ax\). Since \(\displaystyle OP=OQ\) and the parabola is symmetric about the \(\displaystyle x\)-axis, \(\displaystyle P\) and \(\displaystyle Q\) are mirror images in the axis: \[P=(x_{1},y_{1}),\qquad Q=(x_{1},-y_{1}),\qquad y_{1}^{2}=4ax_{1},\ y_{1}>0 .\]Then \(\displaystyle PQ=2y_{1}\) and \(\displaystyle OP=\sqrt{x_{1}^{2}+y_{1}^{2}}\). For an equilateral triangle \(\displaystyle OP=PQ\): \[x_{1}^{2}+y_{1}^{2}=4y_{1}^{2}\ \Rightarrow\ x_{1}^{2}=3y_{1}^{2}\ \Rightarrow\ x_{1}=\sqrt3\,y_{1}\quad (x_{1}>0).\]Substituting into \(\displaystyle y_{1}^{2}=4ax_{1}\): \[y_{1}^{2}=4a\sqrt3\,y_{1}\ \Rightarrow\ y_{1}=4\sqrt3\,a,\qquad x_{1}=\sqrt3\,y_{1}=12a .\]So the side is \(\displaystyle PQ=2y_{1}=8\sqrt3\,a\); as a check, \(\displaystyle OP=\sqrt{(12a)^{2}+(4\sqrt3 a)^{2}}=\sqrt{192a^{2}}=8\sqrt3\,a\). ✓The side of the triangle is \(\displaystyle 8\sqrt{3}\,a\).