\displaystyle \textbf{Question 1. }\text{For the ellipse }12x^2+4y^2+24x-16y+25=0
\displaystyle \text{(a) centre is }(-1,2)\qquad \text{(b) lengths of the axes are }\sqrt{3}\text{ and }1
\displaystyle \text{(c) eccentricity}=\sqrt{\frac{2}{3}}\qquad \text{(d) all of these}
\displaystyle \text{Answer:}
\displaystyle 12x^2+4y^2+24x-16y+25=0
\displaystyle 12(x^2+2x)+4(y^2-4y)+25=0
\displaystyle 12(x+1)^2+4(y-2)^2=3
\displaystyle \frac{(x+1)^2}{\frac{1}{4}}+\frac{(y-2)^2}{\frac{3}{4}}=1
\displaystyle \therefore \text{Centre}=(-1,2)
\displaystyle \text{Semi-major axis}=\frac{\sqrt{3}}{2},\quad \text{semi-minor axis}=\frac{1}{2}
\displaystyle \therefore \text{Lengths of axes}=\sqrt{3}\text{ and }1
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}=\sqrt{1-\frac{\frac{1}{4}}{\frac{3}{4}}}
\displaystyle =\sqrt{\frac{2}{3}}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 2. }\text{The equation of the ellipse with focus }(-1,1),\text{ directrix } \\ x-y+3=0\text{ and eccentricity }\frac{1}{2}\text{ is}
\displaystyle \text{(a) }7x^2+2xy+7y^2+10x+10y+7=0
\displaystyle \text{(b) }7x^2+2xy+7y^2+10x-10y+7=0
\displaystyle \text{(c) }7x^2+2xy+7y^2+10x-10y-7=0
\displaystyle \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{For any point }(x,y)\text{ on the ellipse,}
\displaystyle \frac{\text{distance from focus}}{\text{distance from directrix}}=e
\displaystyle \frac{\sqrt{(x+1)^2+(y-1)^2}}{\frac{|x-y+3|}{\sqrt{2}}}=\frac{1}{2}
\displaystyle (x+1)^2+(y-1)^2=\frac{(x-y+3)^2}{8}
\displaystyle 8\{(x+1)^2+(y-1)^2\}=(x-y+3)^2
\displaystyle 8(x^2+y^2+2x-2y+2)=x^2+y^2+9-2xy+6x-6y
\displaystyle 7x^2+2xy+7y^2+10x-10y+7=0
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 3. }\text{The equation of the circle drawn with the two foci of } \\ \frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ as the end-points of a diameter is}
\displaystyle \text{(a) }x^2+y^2=a^2+b^2\qquad \text{(b) }x^2+y^2=a^2
\displaystyle \text{(c) }x^2+y^2=2a^2\qquad \text{(d) }x^2+y^2=a^2-b^2
\displaystyle \text{Answer:}
\displaystyle \text{Foci of the ellipse are }(ae,0)\text{ and }(-ae,0)
\displaystyle \text{Centre of the circle is }(0,0)
\displaystyle \text{Radius}=ae
\displaystyle \therefore \text{Equation of circle is }x^2+y^2=a^2e^2
\displaystyle \text{But }a^2e^2=a^2-b^2
\displaystyle \therefore x^2+y^2=a^2-b^2
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 4. }\text{The eccentricity of the ellipse }\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\text{ if its latus-rectum is equal} \\ \text{ to one half of its minor axis, is}
\displaystyle \text{(a) }\frac{1}{\sqrt{2}}\qquad \text{(b) }\frac{\sqrt{3}}{2}\qquad \text{(c) }\frac{1}{2}\qquad \text{(d) none of these}
\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}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 5. }\text{The eccentricity of the ellipse, if the distance between the foci is equal} \\ \text{to the length of the latus-rectum, is}
\displaystyle \text{(a) }\frac{\sqrt{5}-1}{2}\qquad \text{(b) }\frac{\sqrt{5}+1}{2}\qquad \text{(c) }\frac{\sqrt{5}-1}{4}\qquad \text{(d) none of these}
\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 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}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 6. }\text{The eccentricity of the ellipse, if the minor axis is equal to the} \\ \text{distance between the foci, is}
\displaystyle \text{(a) }\frac{\sqrt{3}}{2}\qquad \text{(b) }\frac{2}{\sqrt{3}}\qquad \text{(c) }\frac{1}{\sqrt{2}}\qquad \text{(d) }\frac{\sqrt{2}}{3}
\displaystyle \text{Answer:}
\displaystyle \text{Length of minor axis}=2b
\displaystyle \text{Distance between the foci}=2ae
\displaystyle 2b=2ae
\displaystyle b=ae
\displaystyle b^2=a^2e^2
\displaystyle \text{But }b^2=a^2(1-e^2)
\displaystyle a^2(1-e^2)=a^2e^2
\displaystyle 1-e^2=e^2
\displaystyle 2e^2=1
\displaystyle e=\frac{1}{\sqrt{2}}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 7. }\text{The difference between the lengths of the major axis and the} \\ \text{latus-rectum of an ellipse is}
\displaystyle \text{(a) }ae\qquad \text{(b) }2ae\qquad \text{(c) }ae^2\qquad \text{(d) }2ae^2
\displaystyle \text{Answer:}
\displaystyle \text{Length of major axis}=2a
\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a}
\displaystyle \text{Required difference}=2a-\frac{2b^2}{a}
\displaystyle =\frac{2a^2-2b^2}{a}
\displaystyle =\frac{2(a^2-b^2)}{a}
\displaystyle \text{But }a^2-b^2=a^2e^2
\displaystyle \therefore \text{Difference}=\frac{2a^2e^2}{a}=2ae^2
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 8. }\text{The eccentricity of the conic }9x^2+25y^2=225\text{ is}
\displaystyle \text{(a) }\frac{2}{5}\qquad \text{(b) }\frac{4}{5}\qquad \text{(c) }\frac{1}{3}\qquad \text{(d) }\frac{1}{5}
\displaystyle \text{Answer:}
\displaystyle 9x^2+25y^2=225
\displaystyle \frac{x^2}{25}+\frac{y^2}{9}=1
\displaystyle \therefore a^2=25,\quad b^2=9
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}
\displaystyle =\sqrt{1-\frac{9}{25}}
\displaystyle =\sqrt{\frac{16}{25}}
\displaystyle =\frac{4}{5}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 9. }\text{The latus-rectum of the conic }3x^2+4y^2-6x+8y-5=0\text{ is}
\displaystyle \text{(a) }3\qquad \text{(b) }\frac{\sqrt{3}}{2}\qquad \text{(c) }\frac{2}{\sqrt{3}}\qquad \text{(d) none of these}
\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=12
\displaystyle \frac{(x-1)^2}{4}+\frac{(y+1)^2}{3}=1
\displaystyle \therefore a^2=4,\quad b^2=3
\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a}
\displaystyle =\frac{2(3)}{2}=3
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 10. }\text{The equations of the tangents to the ellipse }9x^2+16y^2=144\text{ from} \\ \text{the point}(2,3)\text{ are}
\displaystyle \text{(a) }y=3,\ x=5\qquad \text{(b) }x=2,\ y=3\qquad \\ \text{(c) }x=3,\ y=2\qquad \text{(d) }x+y=5,\ y=3
\displaystyle \text{Answer:}
\displaystyle \text{Given ellipse is }9x^2+16y^2=144
\displaystyle \Rightarrow \frac{x^2}{16}+\frac{y^2}{9}=1
\displaystyle \text{Clearly, }y=3\text{ is a tangent to the ellipse at }(0,3)
\displaystyle \text{Also, line }x+y=5\text{ passes through }(2,3)
\displaystyle \text{Putting }y=5-x\text{ in }9x^2+16y^2=144
\displaystyle 9x^2+16(5-x)^2=144
\displaystyle 25x^2-160x+256=0
\displaystyle \Rightarrow (5x-16)^2=0
\displaystyle \text{Hence, }x+y=5\text{ is also a tangent.}
\displaystyle \therefore \text{ Required tangents are }x+y=5\text{ and }y=3
\displaystyle \therefore \text{Correct option is (d).}

\displaystyle \textbf{Question 11. }\text{The eccentricity of the ellipse }4x^2+9y^2+8x+36y+4=0\text{ is}
\displaystyle \text{(a) }\frac{5}{6}\qquad \text{(b) }\frac{3}{5}\qquad \text{(c) }\frac{\sqrt{2}}{3}\qquad \text{(d) }\frac{\sqrt{5}}{3}
\displaystyle \text{Answer:}
\displaystyle 4x^2+9y^2+8x+36y+4=0
\displaystyle 4(x^2+2x)+9(y^2+4y)+4=0
\displaystyle 4(x+1)^2+9(y+2)^2=36
\displaystyle \frac{(x+1)^2}{9}+\frac{(y+2)^2}{4}=1
\displaystyle \therefore a^2=9,\quad b^2=4
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}
\displaystyle =\sqrt{1-\frac{4}{9}}
\displaystyle =\frac{\sqrt{5}}{3}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 12. }\text{The eccentricity of the ellipse }4x^2+9y^2=36\text{ is}
\displaystyle \text{(a) }\frac{1}{2\sqrt{3}}\qquad \text{(b) }\frac{1}{\sqrt{3}}\qquad \text{(c) }\frac{\sqrt{5}}{3}\qquad \text{(d) }\frac{\sqrt{5}}{6}
\displaystyle \text{Answer:}
\displaystyle 4x^2+9y^2=36
\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1
\displaystyle \therefore a^2=9,\quad b^2=4
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}
\displaystyle =\sqrt{1-\frac{4}{9}}
\displaystyle =\frac{\sqrt{5}}{3}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 13. }\text{The eccentricity of the ellipse }5x^2+9y^2=1\text{ is}
\displaystyle \text{(a) }\frac{2}{3}\qquad \text{(b) }\frac{3}{4}\qquad \text{(c) }\frac{4}{5}\qquad \text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle 5x^2+9y^2=1
\displaystyle \frac{x^2}{\frac{1}{5}}+\frac{y^2}{\frac{1}{9}}=1
\displaystyle \therefore a^2=\frac{1}{5},\quad b^2=\frac{1}{9}
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}
\displaystyle =\sqrt{1-\frac{\frac{1}{9}}{\frac{1}{5}}}
\displaystyle =\sqrt{1-\frac{5}{9}}=\frac{2}{3}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 14. }\text{For the ellipse }x^2+4y^2=9
\displaystyle \text{(a) the eccentricity is }\frac{1}{2}\qquad \text{(b) the latus-rectum is }\frac{3}{2}
\displaystyle \text{(c) a focus is }(3\sqrt{3},0)\qquad \text{(d) a directrix is }x=-2\sqrt{3}
\displaystyle \text{Answer:}
\displaystyle x^2+4y^2=9
\displaystyle \frac{x^2}{9}+\frac{y^2}{\frac{9}{4}}=1
\displaystyle \therefore a=3,\quad b=\frac{3}{2}
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}
\displaystyle =\sqrt{1-\frac{\frac{9}{4}}{9}}=\sqrt{\frac{3}{4}}=\frac{\sqrt{3}}{2}
\displaystyle \text{Length of latus-rectum}=\frac{2b^2}{a}=\frac{2\cdot \frac{9}{4}}{3}=\frac{3}{2}
\displaystyle \text{Focus}=(\pm ae,0)=\left(\pm\frac{3\sqrt{3}}{2},0\right)
\displaystyle \text{Directrices are }x=\pm\frac{a}{e}=\pm 2\sqrt{3}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 15. }\text{If the latus-rectum of an ellipse is one half of its minor axis, then} \\ \text{its eccentricity is}
\displaystyle \text{(a) }\frac{1}{2}\qquad \text{(b) }\frac{1}{\sqrt{2}}\qquad \text{(c) }\frac{\sqrt{3}}{2}\qquad \text{(d) }\frac{\sqrt{3}}{4}
\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}
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 16. }\text{An ellipse has its centre at }(1,-1)\text{ and semi-major axis}=8 \\ \text{ and it passes through the point }(1,3).\text{ The equation }   \text{of the ellipse is}
\displaystyle \text{(a) }\frac{(x+1)^2}{64}+\frac{(y+1)^2}{16}=1\qquad \text{(b) }\frac{(x-1)^2}{64}+\frac{(y+1)^2}{16}=1
\displaystyle \text{(c) }\frac{(x-1)^2}{16}+\frac{(y+1)^2}{64}=1\qquad \text{(d) }\frac{(x+1)^2}{64}+\frac{(y-1)^2}{16}=1
\displaystyle \text{Answer:}
\displaystyle \text{Centre}=(1,-1)
\displaystyle \therefore \text{Equation is of the form }\frac{(x-1)^2}{a^2}+\frac{(y+1)^2}{b^2}=1
\displaystyle \text{Semi-major axis}=8
\displaystyle \therefore a^2=64
\displaystyle \text{Since }(1,3)\text{ lies on the ellipse,}
\displaystyle \frac{(1-1)^2}{64}+\frac{(3+1)^2}{b^2}=1
\displaystyle \frac{16}{b^2}=1
\displaystyle b^2=16
\displaystyle \therefore \text{Equation is }\frac{(x-1)^2}{64}+\frac{(y+1)^2}{16}=1
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 17. }\text{The sum of the focal distances of any point on the ellipse } \\ 9x^2+16y^2=144\text{ is}
\displaystyle \text{(a) }32\qquad \text{(b) }18\qquad \text{(c) }16\qquad \text{(d) }8
\displaystyle \text{Answer:}
\displaystyle 9x^2+16y^2=144
\displaystyle \frac{x^2}{16}+\frac{y^2}{9}=1
\displaystyle \therefore a^2=16,\quad a=4
\displaystyle \text{Sum of focal distances of any point on an ellipse}=2a
\displaystyle =2(4)=8
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 18. }\text{If }(2,4)\text{ and }(10,10)\text{ are the ends of a latus-rectum} \\ \text{of an ellipse with eccentricity }\frac{1}{2},\text{ then the length of } \text{semi-major axis is}
\displaystyle \text{(a) }\frac{20}{3}\qquad \text{(b) }\frac{15}{3}\qquad \text{(c) }\frac{40}{3}\qquad \text{(d) none of these}
\displaystyle \text{Answer:}
\displaystyle \text{Length of latus-rectum}=\sqrt{(10-2)^2+(10-4)^2}
\displaystyle =\sqrt{8^2+6^2}=\sqrt{100}=10
\displaystyle \text{For an ellipse, length of latus-rectum}=\frac{2b^2}{a}
\displaystyle \text{Also }b^2=a^2(1-e^2)
\displaystyle e=\frac{1}{2}
\displaystyle \therefore b^2=a^2\left(1-\frac{1}{4}\right)=\frac{3a^2}{4}
\displaystyle \therefore \frac{2b^2}{a}=\frac{2\cdot \frac{3a^2}{4}}{a}=\frac{3a}{2}
\displaystyle \frac{3a}{2}=10
\displaystyle a=\frac{20}{3}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 19. }\text{The equation }\frac{x^2}{2-\lambda}+\frac{y^2}{\lambda-5}+1=0\text{ represents an ellipse, if}
\displaystyle \text{(a) }\lambda<5\qquad \text{(b) }\lambda<2\qquad \text{(c) }2<\lambda<5\qquad \text{(d) }\lambda<2\text{ or }\lambda>5
\displaystyle \text{Answer:}
\displaystyle \frac{x^2}{2-\lambda}+\frac{y^2}{\lambda-5}+1=0
\displaystyle \frac{x^2}{\lambda-2}+\frac{y^2}{5-\lambda}=1
\displaystyle \text{For an ellipse, both denominators must be positive}
\displaystyle \lambda-2>0,\quad 5-\lambda>0
\displaystyle \lambda>2,\quad \lambda<5
\displaystyle \therefore 2<\lambda<5
\displaystyle \therefore \text{Correct option is (c)}
\\

\displaystyle \textbf{Question 20. }\text{The eccentricity of the ellipse } \\ 9x^2+25y^2-18x-100y-116=0,\text{ is}
\displaystyle \text{(a) }\frac{25}{16}\qquad \text{(b) }\frac{4}{5}\qquad \text{(c) }\frac{16}{25}\qquad \text{(d) }\frac{5}{4}
\displaystyle \text{Answer:}
\displaystyle 9x^2+25y^2-18x-100y-116=0
\displaystyle 9(x^2-2x)+25(y^2-4y)=116
\displaystyle 9(x-1)^2+25(y-2)^2=116+9+100
\displaystyle 9(x-1)^2+25(y-2)^2=225
\displaystyle \frac{(x-1)^2}{25}+\frac{(y-2)^2}{9}=1
\displaystyle \therefore a^2=25,\quad b^2=9
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}
\displaystyle =\sqrt{1-\frac{9}{25}}
\displaystyle =\frac{4}{5}
\displaystyle \therefore \text{Correct option is (b)}
\\

\displaystyle \textbf{Question 21. }\text{If the major axis of an ellipse is three times the minor axis, then} \\ \text{its eccentricity is equal to}
\displaystyle \text{(a) }\frac{1}{3}\qquad \text{(b) }\frac{1}{\sqrt{3}}\qquad \text{(c) }\frac{1}{\sqrt{2}}\qquad \text{(d) }\frac{2\sqrt{2}}{3}
\displaystyle \text{Answer:}
\displaystyle \text{Major axis}=3\times \text{minor axis}
\displaystyle 2a=3(2b)
\displaystyle a=3b
\displaystyle \frac{b}{a}=\frac{1}{3}
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}
\displaystyle =\sqrt{1-\frac{1}{9}}
\displaystyle =\sqrt{\frac{8}{9}}
\displaystyle =\frac{2\sqrt{2}}{3}
\displaystyle \therefore \text{Correct option is (d)}
\\

\displaystyle \textbf{Question 22. }\text{The eccentricity of the ellipse }25x^2+16y^2=400\text{ is}
\displaystyle \text{(a) }\frac{3}{5}\qquad \text{(b) }\frac{1}{3}\qquad \text{(c) }\frac{2}{5}\qquad \text{(d) }\frac{1}{5}
\displaystyle \text{Answer:}
\displaystyle 25x^2+16y^2=400
\displaystyle \frac{x^2}{16}+\frac{y^2}{25}=1
\displaystyle \therefore a^2=25,\quad b^2=16
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}
\displaystyle =\sqrt{1-\frac{16}{25}}
\displaystyle =\sqrt{\frac{9}{25}}
\displaystyle =\frac{3}{5}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 23. }\text{The eccentricity of the ellipse }5x^2+9y^2=1\text{ is}
\displaystyle \text{(a) }\frac{2}{3}\qquad \text{(b) }\frac{3}{4}\qquad \text{(c) }\frac{4}{5}\qquad \text{(d) }\frac{1}{2}
\displaystyle \text{Answer:}
\displaystyle 5x^2+9y^2=1
\displaystyle \frac{x^2}{\frac{1}{5}}+\frac{y^2}{\frac{1}{9}}=1
\displaystyle \therefore a^2=\frac{1}{5},\quad b^2=\frac{1}{9}
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}
\displaystyle =\sqrt{1-\frac{\frac{1}{9}}{\frac{1}{5}}}
\displaystyle =\sqrt{1-\frac{5}{9}}
\displaystyle =\frac{2}{3}
\displaystyle \therefore \text{Correct option is (a)}
\\

\displaystyle \textbf{Question 24. }\text{The eccentricity of the ellipse }4x^2+9y^2=36\text{ is}
\displaystyle \text{(a) }\frac{1}{2\sqrt{3}}\qquad \text{(b) }\frac{1}{\sqrt{3}}\qquad \text{(c) }\frac{\sqrt{5}}{3}\qquad \text{(d) }\frac{\sqrt{5}}{6}
\displaystyle \text{Answer:}
\displaystyle 4x^2+9y^2=36
\displaystyle \frac{x^2}{9}+\frac{y^2}{4}=1
\displaystyle \therefore a^2=9,\quad b^2=4
\displaystyle e=\sqrt{1-\frac{b^2}{a^2}}
\displaystyle =\sqrt{1-\frac{4}{9}}
\displaystyle =\frac{\sqrt{5}}{3}
\displaystyle \therefore \text{Correct option is (c)}
\\

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