Class 8-12 Mathematics Practice Questions, Solutions | ICSE ISC CBSE Math Mathematics Made Easy for School Students

$\displaystyle \textbf{Question 1. }\text{Write the eccentricity of the hyperbola }9x^2-16y^2=144.$
$\displaystyle \text{Answer:}$
$\displaystyle 9x^2-16y^2=144$
$\displaystyle \Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1$
$\displaystyle a^2=16,\quad b^2=9$
$\displaystyle e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1+\frac{9}{16}}$
$\displaystyle =\sqrt{\frac{25}{16}}=\frac{5}{4}$
$\\$

$\displaystyle \textbf{Question 2. }\text{Write the eccentricity of the hyperbola whose latus-rectum is half} \\ \text{of its transverse axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For the hyperbola }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,$
$\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a}$
$\displaystyle \text{Length of transverse axis}=2a$
$\displaystyle \text{Given that latus-rectum is half of transverse axis}$
$\displaystyle \therefore \frac{2b^2}{a}=\frac{1}{2}(2a)=a$
$\displaystyle \Rightarrow 2b^2=a^2$
$\displaystyle \Rightarrow \frac{b^2}{a^2}=\frac{1}{2}$
$\displaystyle \text{Now, eccentricity of hyperbola is }e=\sqrt{1+\frac{b^2}{a^2}}$
$\displaystyle \therefore e=\sqrt{1+\frac{1}{2}}=\sqrt{\frac{3}{2}}=\frac{\sqrt{6}}{2}$
$\displaystyle \therefore \text{Required eccentricity }=\frac{\sqrt6}{2}$

$\displaystyle \textbf{Question 3. }\text{Write the coordinates of the foci of the hyperbola } \\ 9x^2-16y^2=144.$
$\displaystyle \text{Answer:}$
$\displaystyle 9x^2-16y^2=144$
$\displaystyle \Rightarrow \frac{x^2}{16}-\frac{y^2}{9}=1$
$\displaystyle a^2=16,\quad b^2=9$
$\displaystyle c^2=a^2+b^2=16+9=25$
$\displaystyle c=5$
$\displaystyle \therefore \text{Foci}=(\pm5,0)$
$\\$

$\displaystyle \textbf{Question 4. }\text{Write the equation of the hyperbola of eccentricity }\sqrt2,\text{ if it} \\ \text{is known that the distance between its foci is }16.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Distance between foci}=2ae=16$
$\displaystyle ae=8$
$\displaystyle e=\sqrt2$
$\displaystyle a=\frac{8}{\sqrt2}=4\sqrt2$
$\displaystyle a^2=32$
$\displaystyle e^2=1+\frac{b^2}{a^2}$
$\displaystyle 2=1+\frac{b^2}{32}$
$\displaystyle b^2=32$
$\displaystyle \therefore \text{Equation of hyperbola is }\frac{x^2}{32}-\frac{y^2}{32}=1$
$\displaystyle \Rightarrow x^2-y^2=32$
$\\$

$\displaystyle \textbf{Question 5. }\text{If the foci of the ellipse }\frac{x^2}{16}+\frac{y^2}{b^2}=1\text{ and the hyperbola }$
$\displaystyle \frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25} \text{coincide, write the value of }b^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \frac{x^2}{144}-\frac{y^2}{81}=\frac{1}{25}$
$\displaystyle \Rightarrow \frac{x^2}{\frac{144}{25}}-\frac{y^2}{\frac{81}{25}}=1$
$\displaystyle a^2=\frac{144}{25},\quad b^2=\frac{81}{25}$
$\displaystyle c^2=a^2+b^2=\frac{144}{25}+\frac{81}{25}=9$
$\displaystyle \therefore \text{Foci of hyperbola are }(\pm3,0)$
$\displaystyle \text{For ellipse }\frac{x^2}{16}+\frac{y^2}{b^2}=1,\text{ foci are }(\pm3,0)$
$\displaystyle \therefore c^2=9$
$\displaystyle c^2=16-b^2$
$\displaystyle 9=16-b^2$
$\displaystyle b^2=7$
$\\$

$\displaystyle \textbf{Question 6. }\text{Write the length of the latus-rectum of the hyperbola } \\ 16x^2-9y^2=144.$
$\displaystyle \text{Answer:}$
$\displaystyle 16x^2-9y^2=144$
$\displaystyle \Rightarrow \frac{x^2}{9}-\frac{y^2}{16}=1$
$\displaystyle a^2=9,\quad b^2=16$
$\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a}$
$\displaystyle =\frac{2(16)}{3}$
$\displaystyle =\frac{32}{3}$
$\\$

$\displaystyle \textbf{Question 7. }\text{If the latus-rectum through one focus of a hyperbola subtends a right}$
$\displaystyle \text{angle at the farther vertex, then write the eccentricity of the hyperbola.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the hyperbola be }\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\displaystyle \text{Focus }S=(ae,0)\text{ and farther vertex }A'=(-a,0)$
$\displaystyle \text{Ends of latus-rectum through }S\text{ are }\left(ae,\pm\frac{b^2}{a}\right)$
$\displaystyle \text{Since these ends subtend a right angle at }A',$
$\displaystyle \left(ae+a,\frac{b^2}{a}\right)\cdot\left(ae+a,-\frac{b^2}{a}\right)=0$
$\displaystyle a^2(e+1)^2-\frac{b^4}{a^2}=0$
$\displaystyle a(e+1)=\frac{b^2}{a}$
$\displaystyle b^2=a^2(e+1)$
$\displaystyle \text{But }b^2=a^2(e^2-1)$
$\displaystyle a^2(e^2-1)=a^2(e+1)$
$\displaystyle e^2-1=e+1$
$\displaystyle e^2-e-2=0$
$\displaystyle (e-2)(e+1)=0$
$\displaystyle \therefore e=2$
$\\$

$\displaystyle \textbf{Question 8. }\text{Write the distance between the directrices of the hyperbola } \\ x=8\sec\theta,\ y=8\tan\theta.$
$\displaystyle \text{Answer:}$
$\displaystyle x=8\sec\theta,\quad y=8\tan\theta$
$\displaystyle \frac{x^2}{64}-\frac{y^2}{64}=\sec^2\theta-\tan^2\theta=1$
$\displaystyle \therefore a^2=64,\quad b^2=64$
$\displaystyle e=\sqrt{1+\frac{b^2}{a^2}}=\sqrt2$
$\displaystyle \text{Directrices are }x=\pm\frac{a}{e}$
$\displaystyle \therefore \text{Distance between directrices}=\frac{2a}{e}$
$\displaystyle =\frac{16}{\sqrt2}=8\sqrt2$
$\\$

$\displaystyle \textbf{Question 9. }\text{Write the equation of the hyperbola whose vertices are }(\pm3,0) \\ \text{ and foci are }(\pm5,0).$
$\displaystyle \text{Answer:}$
$\displaystyle a=3,\quad c=5$
$\displaystyle c^2=a^2+b^2$
$\displaystyle 25=9+b^2$
$\displaystyle b^2=16$
$\displaystyle \therefore \text{Equation of hyperbola is }\frac{x^2}{9}-\frac{y^2}{16}=1$
$\\$

$\displaystyle \textbf{Question 10. }\text{If }e_1\text{ and }e_2\text{ are respectively the eccentricities of the ellipse }$
$\displaystyle \frac{x^2}{18}+\frac{y^2}{4}=1 \text{ and the hyperbola }\frac{x^2}{9}-\frac{y^2}{4}=1,\text{ then write the value of }2e_1^2+e_2^2.$
$\displaystyle \text{Answer:}$
$\displaystyle \text{For ellipse, }a^2=18,\quad b^2=4$
$\displaystyle e_1^2=1-\frac{b^2}{a^2}=1-\frac{4}{18}=\frac{7}{9}$
$\displaystyle \text{For hyperbola, }a^2=9,\quad b^2=4$
$\displaystyle e_2^2=1+\frac{b^2}{a^2}=1+\frac{4}{9}=\frac{13}{9}$
$\displaystyle \therefore 2e_1^2+e_2^2=2\left(\frac{7}{9}\right)+\frac{13}{9}$
$\displaystyle =\frac{27}{9}=3$
$\\$