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

$\displaystyle \textbf{Question 1. }\text{If the lengths of semi-major and semi-minor axes of an ellipse are }2\text{ and }\sqrt{3}\text{ and their corresponding}$
$\displaystyle \text{equations are }y-5=0\text{ and }x+3=0,\text{ then write the equation of the ellipse.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Semi-major axis}=2,\quad \text{semi-minor axis}=\sqrt{3}$
$\displaystyle \text{Major axis is }y-5=0\text{ and minor axis is }x+3=0$
$\displaystyle \therefore \text{Centre}=(-3,5)$
$\displaystyle \therefore \text{Equation of ellipse is }$
$\displaystyle \frac{(x+3)^2}{4}+\frac{(y-5)^2}{3}=1$
$\displaystyle 3x^2+4y^2+18x-40y+115=0$
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$\displaystyle \textbf{Question 2. }\text{Write the eccentricity of the ellipse }9x^2+5y^2-18x-2y-16=0.$
$\displaystyle \text{Answer:}$
$\displaystyle 9x^2+5y^2-18x-2y-16=0$
$\displaystyle 9(x^2-2x)+5\left(y^2-\frac{2}{5}y\right)=16$
$\displaystyle 9(x-1)^2+5\left(y-\frac{1}{5}\right)^2=16+9+\frac{1}{5}$
$\displaystyle 9(x-1)^2+5\left(y-\frac{1}{5}\right)^2=\frac{126}{5}$
$\displaystyle \frac{(x-1)^2}{\frac{14}{5}}+\frac{\left(y-\frac{1}{5}\right)^2}{\frac{126}{25}}=1$
$\displaystyle \therefore a^2=\frac{126}{25},\quad b^2=\frac{14}{5}$
$\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1-\frac{\frac{14}{5}}{\frac{126}{25}}}$
$\displaystyle =\sqrt{1-\frac{5}{9}}=\frac{2}{3}$
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$\displaystyle \textbf{Question 3. }\text{Write the centre and eccentricity of the ellipse }3x^2+4y^2-6x+8y-5=0.$
$\displaystyle \text{Answer:}$
$\displaystyle 3x^2+4y^2-6x+8y-5=0$
$\displaystyle 3(x^2-2x)+4(y^2+2y)=5$
$\displaystyle 3(x-1)^2+4(y+1)^2=5+3+4$
$\displaystyle 3(x-1)^2+4(y+1)^2=12$
$\displaystyle \frac{(x-1)^2}{4}+\frac{(y+1)^2}{3}=1$
$\displaystyle \therefore \text{Centre}=(1,-1)$
$\displaystyle a^2=4,\quad b^2=3$
$\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{3}{4}}=\frac{1}{2}$
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$\displaystyle \textbf{Question 4. }PSQ\text{ is a focal chord of the ellipse }4x^2+9y^2=36\text{ such that }SP=4.\text{ If }S'\text{ is the another focus, write the value}$
$\displaystyle \text{of }S'Q.$
$\displaystyle \text{Answer:}$
$\displaystyle 4x^2+9y^2=36$
$\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1$
$\displaystyle \therefore a=3,\quad b=2$
$\displaystyle \text{For any point on an ellipse, sum of distances from the two foci}=2a$
$\displaystyle \therefore SP+S'P=2a=6$
$\displaystyle SP=4$
$\displaystyle \therefore S'P=2$
$\displaystyle \text{Since }PSQ\text{ is a focal chord, }P,S,Q\text{ are collinear}$
$\displaystyle \text{For the two endpoints of a focal chord, distances from the other focus are interchanged}$
$\displaystyle \therefore S'Q=SP=4$
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$\displaystyle \textbf{Question 5. }\text{Write the eccentricity of an ellipse whose latus-rectum is one half of the minor axis.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a}$
$\displaystyle \text{Length of minor axis}=2b$
$\displaystyle \frac{2b^2}{a}=\frac{1}{2}(2b)$
$\displaystyle \frac{2b^2}{a}=b$
$\displaystyle 2b=a$
$\displaystyle \frac{b}{a}=\frac{1}{2}$
$\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1-\frac{1}{4}}=\frac{\sqrt{3}}{2}$
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$\displaystyle \textbf{Question 6. }\text{If the distance between the foci of an ellipse is equal to the length of the latus-rectum, write the eccentricity of}$
$\displaystyle \text{the ellipse.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Distance between the foci}=2ae$
$\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a}$
$\displaystyle 2ae=\frac{2b^2}{a}$
$\displaystyle ae=\frac{b^2}{a}$
$\displaystyle a^2e=b^2$
$\displaystyle \text{But }b^2=a^2(1-e^2)$
$\displaystyle a^2e=a^2(1-e^2)$
$\displaystyle e=1-e^2$
$\displaystyle e^2+e-1=0$
$\displaystyle e=\frac{-1+\sqrt{5}}{2}$
$\displaystyle \therefore e=\frac{\sqrt{5}-1}{2}$
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$\displaystyle \textbf{Question 7. }\text{If }S\text{ and }S'\text{ are two foci of the ellipse }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ and }B\text{ is an end of the minor axis such that }\triangle BSS'$
$\displaystyle \text{is equilateral, then write the eccentricity of the ellipse.}$
$\displaystyle \text{Answer:}$
$\displaystyle S=(ae,0),\quad S'=(-ae,0),\quad B=(0,b)$
$\displaystyle SS'=2ae$
$\displaystyle BS=\sqrt{a^2e^2+b^2}$
$\displaystyle \text{But }b^2=a^2(1-e^2)$
$\displaystyle \therefore BS=\sqrt{a^2e^2+a^2(1-e^2)}=a$
$\displaystyle \triangle BSS'\text{ is equilateral}$
$\displaystyle \therefore SS'=BS$
$\displaystyle 2ae=a$
$\displaystyle e=\frac{1}{2}$
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$\displaystyle \textbf{Question 8. }\text{If the minor axis of an ellipse subtends an equilateral triangle with vertex at one end of major axis, then write}$
$\displaystyle \text{the eccentricity of the ellipse.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the ellipse be }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\displaystyle \text{Ends of minor axis are }(0,b)\text{ and }(0,-b)$
$\displaystyle \text{One end of major axis is }(a,0)$
$\displaystyle \text{Since the triangle is equilateral,}$
$\displaystyle \text{minor axis}=2b=\sqrt{a^2+b^2}$
$\displaystyle 4b^2=a^2+b^2$
$\displaystyle a^2=3b^2$
$\displaystyle \frac{b^2}{a^2}=\frac{1}{3}$
$\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}$
$\displaystyle =\sqrt{1-\frac{1}{3}}$
$\displaystyle =\sqrt{\frac{2}{3}}$
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$\displaystyle \textbf{Question 9. }\text{If a latus-rectum of an ellipse subtends a right angle at the centre of the ellipse, then write the eccentricity of}$
$\displaystyle \text{the ellipse.}$
$\displaystyle \text{Answer:}$
$\displaystyle \text{Let the ellipse be }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$\displaystyle \text{One latus-rectum has endpoints }\left(ae,\frac{b^2}{a}\right)\text{ and }\left(ae,-\frac{b^2}{a}\right)$
$\displaystyle \text{It subtends a right angle at the centre }(0,0)$
$\displaystyle \therefore \text{Product of slopes of lines from centre to endpoints}=-1$
$\displaystyle \frac{\frac{b^2}{a}}{ae}\times \frac{-\frac{b^2}{a}}{ae}=-1$
$\displaystyle -\frac{b^4}{a^4e^2}=-1$
$\displaystyle b^4=a^4e^2$
$\displaystyle \left(\frac{b^2}{a^2}\right)^2=e^2$
$\displaystyle (1-e^2)^2=e^2$
$\displaystyle 1-e^2=e$
$\displaystyle e^2+e-1=0$
$\displaystyle e=\frac{\sqrt{5}-1}{2}$
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