NCERT Solutions · Class 11 Mathematics Conic Sections

Standard form of an ellipse. The vertices and foci lie on the \(\displaystyle y\)-axis, so the major axis is along the \(\displaystyle y\)-axis and the equation has the form \[\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,\qquad a>b.\] Here the vertices are \(\displaystyle (0,\pm a)\) and the foci are \(\displaystyle (0,\pm c)\), so \[a=13,\qquad c=5.\] Using \(\displaystyle c^{2}=a^{2}-b^{2}\), \[b^{2}=a^{2}-c^{2}=169-25=144.\]\(\displaystyle \dfrac{x^{2}}{144}+\dfrac{y^{2}}{169}=1\)

Standard form of an ellipse. The vertices \(\displaystyle (\pm 6,0)\) and foci \(\displaystyle (\pm 4,0)\) lie on the \(\displaystyle x\)-axis, so the major axis is the \(\displaystyle x\)-axis and \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a>b.\] Comparing with vertices \(\displaystyle (\pm a,0)\) and foci \(\displaystyle (\pm c,0)\), \[a=6,\qquad c=4.\] Using \(\displaystyle c^{2}=a^{2}-b^{2}\), \[b^{2}=a^{2}-c^{2}=36-16=20.\]\(\displaystyle \dfrac{x^{2}}{36}+\dfrac{y^{2}}{20}=1\)

Standard form of an ellipse. The ends of the major axis are \(\displaystyle (\pm 3,0)\), so the major axis lies along the \(\displaystyle x\)-axis and the equation is \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a>b.\] The ends of the major axis are \(\displaystyle (\pm a,0)\) and the ends of the minor axis are \(\displaystyle (0,\pm b)\), so reading them off directly, \[a=3,\qquad b=2,\] giving \(\displaystyle a^{2}=9\) and \(\displaystyle b^{2}=4\). (No use of \(\displaystyle c\) is needed here — both semi-axes are handed to us.)\(\displaystyle \dfrac{x^{2}}{9}+\dfrac{y^{2}}{4}=1\)

Standard form of an ellipse. The ends of the major axis are \(\displaystyle (0,\pm\sqrt{5})\), so the major axis lies along the \(\displaystyle y\)-axis and the equation is \[\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,\qquad a>b.\] The ends of the major axis are \(\displaystyle (0,\pm a)\) and the ends of the minor axis are \(\displaystyle (\pm b,0)\), so \[a=\sqrt{5},\qquad b=1,\] giving \(\displaystyle a^{2}=5\) and \(\displaystyle b^{2}=1\).\(\displaystyle \dfrac{x^{2}}{1}+\dfrac{y^{2}}{5}=1\)

Standard form of an ellipse. The foci \(\displaystyle (\pm 5,0)\) lie on the \(\displaystyle x\)-axis, so the major axis is the \(\displaystyle x\)-axis and \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a>b.\] The length of the major axis is \(\displaystyle 2a\), so \[2a=26\ \Rightarrow\ a=13,\qquad c=5.\] Using \(\displaystyle c^{2}=a^{2}-b^{2}\), \[b^{2}=a^{2}-c^{2}=169-25=144.\]\(\displaystyle \dfrac{x^{2}}{169}+\dfrac{y^{2}}{144}=1\)

Standard form of an ellipse. The foci \(\displaystyle (0,\pm 6)\) lie on the \(\displaystyle y\)-axis, so the major axis is the \(\displaystyle y\)-axis and \[\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,\qquad a>b.\] The length of the minor axis is \(\displaystyle 2b\), so \[2b=16\ \Rightarrow\ b=8,\qquad c=6.\] Using \(\displaystyle c^{2}=a^{2}-b^{2}\), this time solved for \(\displaystyle a^{2}\), \[a^{2}=b^{2}+c^{2}=64+36=100.\]\(\displaystyle \dfrac{x^{2}}{64}+\dfrac{y^{2}}{100}=1\)

Standard form of an ellipse. The foci \(\displaystyle (\pm 3,0)\) lie on the \(\displaystyle x\)-axis, so the major axis is the \(\displaystyle x\)-axis and \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a>b.\] We are given \(\displaystyle a=4\) and \(\displaystyle c=3\). Using \(\displaystyle c^{2}=a^{2}-b^{2}\), \[b^{2}=a^{2}-c^{2}=16-9=7.\]\(\displaystyle \dfrac{x^{2}}{16}+\dfrac{y^{2}}{7}=1\)

Standard form of an ellipse. The centre is the origin and the foci are on the \(\displaystyle x\)-axis, so the major axis is the \(\displaystyle x\)-axis and \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a>b.\] Given \(\displaystyle b=3\) and \(\displaystyle c=4\), use \(\displaystyle c^{2}=a^{2}-b^{2}\) in the form \[a^{2}=b^{2}+c^{2}=9+16=25.\] So \(\displaystyle a=5>b=3\), consistent with the major axis being the \(\displaystyle x\)-axis.\(\displaystyle \dfrac{x^{2}}{25}+\dfrac{y^{2}}{9}=1\)

Substituting the given points. The centre is \(\displaystyle (0,0)\) and the major axis is the \(\displaystyle y\)-axis, so the equation has the form \[\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1,\qquad a>b.\] Both points must satisfy it. Substituting \(\displaystyle (3,2)\) and \(\displaystyle (1,6)\), \[\frac{9}{b^{2}}+\frac{4}{a^{2}}=1,\qquad \frac{1}{b^{2}}+\frac{36}{a^{2}}=1.\] Put \(\displaystyle u=\dfrac{1}{b^{2}}\) and \(\displaystyle v=\dfrac{1}{a^{2}}\) to make these linear: \[9u+4v=1,\qquad u+36v=1.\] From the second, \(\displaystyle u=1-36v\); substituting into the first, \[9(1-36v)+4v=1\ \Rightarrow\ 9-320v=1\ \Rightarrow\ v=\frac{1}{40}.\] Then \(\displaystyle u=1-\dfrac{36}{40}=\dfrac{1}{10}\). Hence \[b^{2}=10,\qquad a^{2}=40,\] and indeed \(\displaystyle a^{2}>b^{2}\), so the major axis is along the \(\displaystyle y\)-axis as required.Check: \(\displaystyle \dfrac{9}{10}+\dfrac{4}{40}=1\) and \(\displaystyle \dfrac{1}{10}+\dfrac{36}{40}=1\). ✓\(\displaystyle \dfrac{x^{2}}{10}+\dfrac{y^{2}}{40}=1\)

Substituting the given points. The major axis is the \(\displaystyle x\)-axis (and the centre is the origin), so \[\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1,\qquad a>b.\] Substituting the points \(\displaystyle (4,3)\) and \(\displaystyle (6,2)\), \[\frac{16}{a^{2}}+\frac{9}{b^{2}}=1,\qquad \frac{36}{a^{2}}+\frac{4}{b^{2}}=1.\] Put \(\displaystyle u=\dfrac{1}{a^{2}}\) and \(\displaystyle v=\dfrac{1}{b^{2}}\): \[16u+9v=1,\qquad 36u+4v=1.\] Multiply the first by \(\displaystyle 4\) and the second by \(\displaystyle 9\): \[64u+36v=4,\qquad 324u+36v=9.\] Subtracting, \(\displaystyle 260u=5\), so \(\displaystyle u=\dfrac{1}{52}\). Then \(\displaystyle 9v=1-\dfrac{16}{52}=\dfrac{9}{13}\), so \(\displaystyle v=\dfrac{1}{13}\). Hence \[a^{2}=52,\qquad b^{2}=13,\] and \(\displaystyle a^{2}>b^{2}\), consistent with the major axis being the \(\displaystyle x\)-axis.Check: \(\displaystyle \dfrac{16}{52}+\dfrac{9}{13}=\dfrac{4}{13}+\dfrac{9}{13}=1\) and \(\displaystyle \dfrac{36}{52}+\dfrac{4}{13}=\dfrac{9}{13}+\dfrac{4}{13}=1\). ✓\(\displaystyle \dfrac{x^{2}}{52}+\dfrac{y^{2}}{13}=1\)