NCERT Solutions · Class 11 Mathematics Conic Sections

The conjugate axis has length \(\displaystyle 2b\).The foci \(\displaystyle (0,\pm 13)\) lie on the \(\displaystyle y\)-axis, so \[\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1,\qquad c^{2}=a^{2}+b^{2}.\]The conjugate axis has length \(\displaystyle 2b\), so \[2b=24\ \Rightarrow\ b=12,\qquad c=13.\]Therefore \[a^{2}=c^{2}-b^{2}=169-144=25.\]\(\displaystyle \dfrac{y^{2}}{25}-\dfrac{x^{2}}{144}=1\)

The latus rectum of a hyperbola has length \(\displaystyle \dfrac{2b^{2}}{a}\).The foci \(\displaystyle (\pm 3\sqrt{5},0)\) lie on the \(\displaystyle x\)-axis, so \[\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,\qquad c^{2}=a^{2}+b^{2}=(3\sqrt5)^{2}=45.\]From the latus rectum, \[\frac{2b^{2}}{a}=8\ \Rightarrow\ b^{2}=4a.\]Substituting in \(\displaystyle a^{2}+b^{2}=45\): \[a^{2}+4a-45=0\ \Rightarrow\ (a+9)(a-5)=0.\]Since \(\displaystyle a>0\), \(\displaystyle a=5\), and then \(\displaystyle b^{2}=4(5)=20\).\(\displaystyle \dfrac{x^{2}}{25}-\dfrac{y^{2}}{20}=1\)

The latus rectum of a hyperbola has length \(\displaystyle \dfrac{2b^{2}}{a}\).The foci \(\displaystyle (\pm 4,0)\) lie on the \(\displaystyle x\)-axis, so \[\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,\qquad c^{2}=a^{2}+b^{2}=16.\]From the latus rectum, \[\frac{2b^{2}}{a}=12\ \Rightarrow\ b^{2}=6a.\]Substituting in \(\displaystyle a^{2}+b^{2}=16\): \[a^{2}+6a-16=0\ \Rightarrow\ (a+8)(a-2)=0.\]Since \(\displaystyle a>0\), \(\displaystyle a=2\), and then \(\displaystyle b^{2}=6(2)=12\).\(\displaystyle \dfrac{x^{2}}{4}-\dfrac{y^{2}}{12}=1\)

Eccentricity gives \(\displaystyle c=ae\).The vertices \(\displaystyle (\pm 7,0)\) lie on the \(\displaystyle x\)-axis, so \[\frac{x^{2}}{a^{2}}-\frac{y^{2}}{b^{2}}=1,\qquad a=7 .\]With \(\displaystyle e=\dfrac{4}{3}\), \[c=ae=7\cdot\frac{4}{3}=\frac{28}{3}.\]Hence \[b^{2}=c^{2}-a^{2}=\frac{784}{9}-49=\frac{784-441}{9}=\frac{343}{9}.\]\(\displaystyle \dfrac{x^{2}}{49}-\dfrac{9y^{2}}{343}=1\)

Impose both conditions on the standard form.The foci \(\displaystyle (0,\pm\sqrt{10})\) lie on the \(\displaystyle y\)-axis, so \[\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1,\qquad a^{2}+b^{2}=c^{2}=10 .\]The curve passes through \(\displaystyle (2,3)\): \[\frac{9}{a^{2}}-\frac{4}{b^{2}}=1 .\]Put \(\displaystyle b^{2}=10-a^{2}\) and write \(\displaystyle t=a^{2}\): \[\frac{9}{t}-\frac{4}{10-t}=1\ \Rightarrow\ 9(10-t)-4t=t(10-t),\] \[90-13t=10t-t^{2}\ \Rightarrow\ t^{2}-23t+90=0\ \Rightarrow\ (t-5)(t-18)=0 .\]So \(\displaystyle a^{2}=5\) or \(\displaystyle a^{2}=18\). The value \(\displaystyle a^{2}=18\) gives \(\displaystyle b^{2}=10-18=-8<0\), which is impossible; hence \(\displaystyle a^{2}=5\) and \(\displaystyle b^{2}=5\).Check: \(\displaystyle \dfrac{9}{5}-\dfrac{4}{5}=1\). ✓\(\displaystyle \dfrac{y^{2}}{5}-\dfrac{x^{2}}{5}=1\)