NCERT Solutions: Class 11 Maths Chapter 10 - Conic Sections

The circle with centre $(h,k)$ and radius $r$ has equation $(x-h)^2+(y-k)^2=r^2$. With centre $(0,2)$ and radius $2$, the equation is $x^2+(y-2)^2=4$.

Using $(x-h)^2+(y-k)^2=r^2$ with centre $(-2,3)$ and $r=4$, we get $(x+2)^2+(y-3)^2=16$.

Here $h=\dfrac12$, $k=\dfrac14$ and $r=\dfrac1{12}$. Hence $(x-h)^2+(y-k)^2=r^2$ gives $\left(x-\dfrac12\right)^2+\left(y-\dfrac14\right)^2=\dfrac{1}{144}$.

With centre $(1,1)$ and radius $\sqrt2$, the equation is $(x-1)^2+(y-1)^2=(\sqrt2)^2=2$.

For centre $(-a,-b)$ and radius $\sqrt{a^2-b^2}$, the equation is $(x+a)^2+(y+b)^2=(\sqrt{a^2-b^2})^2=a^2-b^2$.

Compare with $(x-h)^2+(y-k)^2=r^2$. Here $x+5=x-(-5)$, $y-3=y-3$, and $r^2=36$, so the centre is $(-5,3)$ and the radius is $6$.

Complete squares: $x^2-4x+y^2-8y-45=0$ becomes $(x-2)^2-4+(y-4)^2-16-45=0$. Hence $(x-2)^2+(y-4)^2=65$. The centre is $(2,4)$ and radius is $\sqrt{65}$.

Complete squares: $x^2-8x+y^2+10y-12=0$ becomes $(x-4)^2-16+(y+5)^2-25-12=0$. Hence $(x-4)^2+(y+5)^2=53$. The centre is $(4,-5)$ and radius is $\sqrt{53}$.

Divide by 2: $x^2+y^2-\dfrac{x}{2}=0$. Complete the square in $x$: $\left(x-\dfrac14\right)^2+y^2=\dfrac{1}{16}$. Hence the centre is $\left(\dfrac14,0\right)$ and the radius is $\dfrac14$.

Let the centre be $(h,k)$. Since it lies on $4x+y=16$, $4h+k=16$. Equal distances from $(4,1)$ and $(6,5)$ give $(h-4)^2+(k-1)^2=(h-6)^2+(k-5)^2$, which simplifies to $h+2k=11$. Solving $4h+k=16$ and $h+2k=11$ gives $h=3,k=4$. The radius squared is $(4-3)^2+(1-4)^2=10$. Hence the circle is $(x-3)^2+(y-4)^2=10$.

Let the centre be $(h,k)$. The line condition gives $h-3k-11=0$. Equal distances from $(2,3)$ and $(-1,1)$ give $(h-2)^2+(k-3)^2=(h+1)^2+(k-1)^2$, or $6h+4k=11$. Solving with $h=3k+11$ gives $k=-\dfrac52$ and $h=\dfrac72$. The radius squared is $\left(2-\dfrac72\right)^2+\left(3+\dfrac52\right)^2=\dfrac{65}{2}$. Hence the equation is $\left(x-\dfrac72\right)^2+\left(y+\dfrac52\right)^2=\dfrac{65}{2}$.

Let the centre be $(h,0)$. Since the radius is 5 and the circle passes through $(2,3)$, $(2-h)^2+3^2=25$. Thus $(h-2)^2=16$, so $h=6$ or $h=-2$. The two circles are $(x-6)^2+y^2=25$ and $(x+2)^2+y^2=25$.

A circle through the origin has equation $x^2+y^2+2gx+2fy=0$. If it cuts intercept $a$ on the x-axis, then $(a,0)$ lies on it, so $a^2+2ga=0$ and $2g=-a$. If it cuts intercept $b$ on the y-axis, then $(0,b)$ lies on it, so $b^2+2fb=0$ and $2f=-b$. Therefore the equation is $x^2+y^2-ax-by=0$.

The radius squared is the squared distance from $(2,2)$ to $(4,5)$: $(4-2)^2+(5-2)^2=4+9=13$. Hence the equation is $(x-2)^2+(y-2)^2=13$.

For the point $(-2.5,3.5)$, $x^2+y^2=(-2.5)^2+(3.5)^2=6.25+12.25=18.5$. Since $18.5<25$, the point lies inside the circle.

Compare $y^2=12x$ with $y^2=4ax$. Then $4a=12$, so $a=3$. Therefore the focus is $(a,0)=(3,0)$, the axis is the x-axis, the directrix is $x=-a=-3$, and the latus rectum length is $4a=12$.

Compare $x^2=6y$ with $x^2=4ay$. Then $4a=6$, so $a=\dfrac32$. Hence the focus is $\left(0,\dfrac32\right)$, the axis is the y-axis, the directrix is $y=-\dfrac32$, and the latus rectum length is $6$.

Write $y^2=-8x$ as $y^2=-4ax$. Then $4a=8$, so $a=2$. The parabola opens to the left, with focus $(-a,0)=(-2,0)$, axis the x-axis, directrix $x=a=2$, and latus rectum length $4a=8$.

Write $x^2=-16y$ as $x^2=-4ay$. Then $4a=16$, so $a=4$. The focus is $(0,-a)=(0,-4)$, the axis is the y-axis, the directrix is $y=a=4$, and the latus rectum length is $16$.

Compare $y^2=10x$ with $y^2=4ax$. Then $a=\dfrac{10}{4}=\dfrac52$. Hence focus $\left(\dfrac52,0\right)$, axis x-axis, directrix $x=-\dfrac52$, and latus rectum length $10$.

Compare $x^2=-9y$ with $x^2=-4ay$. Then $a=\dfrac94$. Thus the focus is $\left(0,-\dfrac94\right)$, the axis is the y-axis, the directrix is $y=\dfrac94$, and the latus rectum length is $9$.

The focus $(6,0)$ and directrix $x=-6$ give vertex at the origin and $a=6$. For a right-opening parabola, $y^2=4ax$, so the equation is $y^2=24x$.

The focus $(0,-3)$ and directrix $y=3$ give vertex at the origin and $a=3$. The parabola opens downward, so its equation is $x^2=-4ay=-12y$.

With vertex at the origin and focus $(3,0)$, the parabola opens to the right with $a=3$. Therefore $y^2=4ax=12x$.

With vertex at the origin and focus $(-2,0)$, the parabola opens to the left with $a=2$. Hence $y^2=-4ax=-8x$.

Since the vertex is at the origin and axis is along the x-axis, take $y^2=4ax$. Passing through $(2,3)$ gives $3^2=4a(2)$, so $9=8a$ and $4a=\dfrac92$. Hence $y^2=\dfrac92x$.

Symmetry with respect to the y-axis and vertex at the origin gives $x^2=4ay$. Since the parabola passes through $(5,2)$, $25=4a(2)$, so $4a=\dfrac{25}{2}$. Hence $x^2=\dfrac{25}{2}y$.

Here $a^2=36$, $b^2=16$, so $a=6$, $b=4$ and $c^2=a^2-b^2=20$. Thus $c=2\sqrt5$, $e=\dfrac ca=\dfrac{\sqrt5}{3}$, and latus rectum length $=\dfrac{2b^2}{a}=\dfrac{32}{6}=\dfrac{16}{3}$.

The larger denominator is under $y^2$, so the major axis is the y-axis. Here $a=5$, $b=2$, and $c^2=25-4=21$. Hence foci are $(0,\pm\sqrt{21})$, vertices $(0,\pm5)$, $e=\dfrac{\sqrt{21}}5$, and latus rectum length $=\dfrac{2b^2}{a}=\dfrac85$.

Here $a=4$, $b=3$ and $c^2=16-9=7$. The major axis is along the x-axis. Thus foci are $(\pm\sqrt7,0)$, vertices $(\pm4,0)$, $e=\dfrac{\sqrt7}{4}$, and latus rectum length $=\dfrac{2\cdot9}{4}=\dfrac92$.

The major axis is along the y-axis. Here $a=10$, $b=5$, and $c^2=100-25=75$, so $c=5\sqrt3$. Therefore $e=\dfrac{5\sqrt3}{10}=\dfrac{\sqrt3}{2}$ and latus rectum length $=\dfrac{2b^2}{a}=5$.

Here $a=7$, $b=6$ and $c^2=49-36=13$. The major axis is horizontal. Thus $e=\dfrac{\sqrt{13}}7$ and latus rectum length $=\dfrac{2\cdot36}{7}=\dfrac{72}{7}$.

The major axis is along the y-axis. Here $a=20$, $b=10$, and $c^2=400-100=300$, so $c=10\sqrt3$. Hence $e=\dfrac{\sqrt3}{2}$ and latus rectum length $=\dfrac{2\cdot100}{20}=10$.

Divide by 144 to get $\dfrac{x^2}{4}+\dfrac{y^2}{36}=1$. Thus $a=6$, $b=2$, $c^2=36-4=32$, so $c=4\sqrt2$. The major axis is vertical, $e=\dfrac{4\sqrt2}{6}=\dfrac{2\sqrt2}{3}$, and latus rectum length $=\dfrac{2\cdot4}{6}=\dfrac43$.

Divide by 16 to get $x^2+\dfrac{y^2}{16}=1$. Thus $a=4$, $b=1$, $c^2=16-1=15$. The major axis is vertical, and latus rectum length $=\dfrac{2b^2}{a}=\dfrac12$.

Divide by 36 to get $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$. Hence $a=3$, $b=2$, $c^2=9-4=5$, $e=\dfrac{\sqrt5}{3}$, and latus rectum length $=\dfrac{2\cdot4}{3}=\dfrac83$.

The major axis is along the x-axis with $a=5$ and $c=4$. Hence $b^2=a^2-c^2=25-16=9$. The ellipse is $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$.

The major axis is along the y-axis with $a=13$ and $c=5$. Then $b^2=169-25=144$. Therefore the equation is $\dfrac{x^2}{144}+\dfrac{y^2}{169}=1$.

Here $a=6$ and $c=4$, with major axis along the x-axis. Thus $b^2=a^2-c^2=36-16=20$. The equation is $\dfrac{x^2}{36}+\dfrac{y^2}{20}=1$.

The semi-major axis is $a=3$ along the x-axis, and the semi-minor axis is $b=2$ along the y-axis. Hence the equation is $\dfrac{x^2}{9}+\dfrac{y^2}{4}=1$.

The major axis is along the y-axis with $a=\sqrt5$, so $a^2=5$. The minor axis has $b=1$. Therefore the equation is $\dfrac{x^2}{1}+\dfrac{y^2}{5}=1$, i.e. $x^2+\dfrac{y^2}{5}=1$.

The major axis is along the x-axis and $2a=26$, so $a=13$. The foci give $c=5$. Thus $b^2=a^2-c^2=169-25=144$. Hence $\dfrac{x^2}{169}+\dfrac{y^2}{144}=1$.

The foci are on the y-axis, so the major axis is vertical. The minor axis length is 16, so $b=8$. Also $c=6$, hence $a^2=b^2+c^2=64+36=100$. The equation is $\dfrac{x^2}{64}+\dfrac{y^2}{100}=1$.

The foci lie on the x-axis, so $c=3$ and $a=4$. Then $b^2=a^2-c^2=16-9=7$. Hence the equation is $\dfrac{x^2}{16}+\dfrac{y^2}{7}=1$.

Since the foci are on the x-axis, the major axis is horizontal. Given $b=3$ and $c=4$, $a^2=b^2+c^2=9+16=25$. Therefore the ellipse is $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$.

With major axis on the y-axis, write the ellipse as $\dfrac{x^2}{b^2}+\dfrac{y^2}{a^2}=1$. Let $A=\dfrac1{b^2}$ and $B=\dfrac1{a^2}$. Substituting $(3,2)$ and $(1,6)$ gives $9A+4B=1$ and $A+36B=1$. Solving, $A=\dfrac1{10}$ and $B=\dfrac1{40}$. Hence $\dfrac{x^2}{10}+\dfrac{y^2}{40}=1$.

With major axis on the x-axis, write $\dfrac{x^2}{a^2}+\dfrac{y^2}{b^2}=1$. Let $A=\dfrac1{a^2}$ and $B=\dfrac1{b^2}$. Substituting $(4,3)$ and $(6,2)$ gives $16A+9B=1$ and $36A+4B=1$. Solving gives $A=\dfrac1{52}$ and $B=\dfrac1{13}$. Therefore $\dfrac{x^2}{52}+\dfrac{y^2}{13}=1$.

Here $a^2=16$, $b^2=9$, so $a=4$, $b=3$ and $c^2=a^2+b^2=25$. Thus $c=5$, $e=\dfrac ca=\dfrac54$, and latus rectum length $=\dfrac{2b^2}{a}=\dfrac{18}{4}=\dfrac92$.

This is a vertical hyperbola with $a^2=9$ and $b^2=27$. Hence $a=3$ and $c^2=a^2+b^2=36$, so $c=6$. Therefore $e=\dfrac ca=2$ and latus rectum length $=\dfrac{2b^2}{a}=18$.

Divide by 36 to get $\dfrac{y^2}{4}-\dfrac{x^2}{9}=1$. Thus $a=2$, $b=3$, and $c^2=4+9=13$. Hence foci are $(0,\pm\sqrt{13})$, vertices $(0,\pm2)$, $e=\dfrac{\sqrt{13}}2$, and latus rectum length $=\dfrac{2\cdot9}{2}=9$.

Divide by 576 to get $\dfrac{x^2}{36}-\dfrac{y^2}{64}=1$. Hence $a=6$, $b=8$, $c^2=36+64=100$, so $c=10$. Thus $e=\dfrac{10}{6}=\dfrac53$ and latus rectum length $=\dfrac{2\cdot64}{6}=\dfrac{64}{3}$.

Write the equation as $\dfrac{y^2}{36/5}-\dfrac{x^2}{4}=1$. Thus $a^2=\dfrac{36}{5}$ and $b^2=4$. Then $c^2=a^2+b^2=\dfrac{36}{5}+4=\dfrac{56}{5}$. Hence $e=\dfrac ca=\sqrt{\dfrac{56/5}{36/5}}=\dfrac{\sqrt{14}}3$, and latus rectum length $=\dfrac{2b^2}{a}=\dfrac{8}{6/\sqrt5}=\dfrac{4\sqrt5}{3}$.

Divide by 784 to get $\dfrac{y^2}{16}-\dfrac{x^2}{49}=1$. Hence $a=4$, $b=7$, $c^2=16+49=65$. Thus the foci are $(0,\pm\sqrt{65})$, vertices $(0,\pm4)$, $e=\dfrac{\sqrt{65}}4$, and latus rectum length $=\dfrac{2\cdot49}{4}=\dfrac{49}{2}$.

The transverse axis is along the x-axis with $a=2$ and $c=3$. Since $c^2=a^2+b^2$, $b^2=9-4=5$. Therefore the equation is $\dfrac{x^2}{4}-\dfrac{y^2}{5}=1$.

The transverse axis is along the y-axis with $a=5$ and $c=8$. Then $b^2=c^2-a^2=64-25=39$. Hence the equation is $\dfrac{y^2}{25}-\dfrac{x^2}{39}=1$.

Here $a=3$ and $c=5$, with transverse axis along the y-axis. Thus $b^2=25-9=16$, giving $\dfrac{y^2}{9}-\dfrac{x^2}{16}=1$.

The foci are on the x-axis, so $c=5$. The transverse axis length is $2a=8$, hence $a=4$. Then $b^2=c^2-a^2=25-16=9$. Therefore the equation is $\dfrac{x^2}{16}-\dfrac{y^2}{9}=1$.

The foci are on the y-axis, so $c=13$. The conjugate axis length is $2b=24$, hence $b=12$. Since $c^2=a^2+b^2$, $a^2=169-144=25$. The equation is $\dfrac{y^2}{25}-\dfrac{x^2}{144}=1$.

The foci are on the x-axis, so $c=3\sqrt5$ and $c^2=45$. Let the hyperbola be $\dfrac{x^2}{a^2}-\dfrac{y^2}{b^2}=1$. The latus rectum length is $\dfrac{2b^2}{a}=8$, so $b^2=4a$. Also $c^2=a^2+b^2$, so $45=a^2+4a$. This gives $a=5$ and $b^2=20$. Hence $\dfrac{x^2}{25}-\dfrac{y^2}{20}=1$.

Here $c=4$, so $c^2=16$. The latus rectum length $\dfrac{2b^2}{a}=12$ gives $b^2=6a$. Since $c^2=a^2+b^2$, $16=a^2+6a$, so $a=2$ and $b^2=12$. Thus $\dfrac{x^2}{4}-\dfrac{y^2}{12}=1$.

The vertices give $a=7$. Since $e=\dfrac ca=\dfrac43$, $c=\dfrac{28}{3}$. Hence $b^2=c^2-a^2=\dfrac{784}{9}-49=\dfrac{343}{9}$. The equation is $\dfrac{x^2}{49}-\dfrac{y^2}{343/9}=1$, i.e. $\dfrac{x^2}{49}-\dfrac{9y^2}{343}=1$.

The foci are on the y-axis, so use $\dfrac{y^2}{a^2}-\dfrac{x^2}{b^2}=1$ with $c^2=10$ and $b^2=c^2-a^2=10-a^2$. Since $(2,3)$ lies on it, $\dfrac{9}{a^2}-\dfrac{4}{10-a^2}=1$. Let $A=a^2$. Then $\dfrac9A-\dfrac4{10-A}=1$, which gives $A^2-23A+90=0$. Since $A<c^2=10$, $A=5$. Thus $a^2=5$ and $b^2=5$, so $\dfrac{y^2}{5}-\dfrac{x^2}{5}=1$.

Take the vertex as the origin and the axis along the positive x-axis. The parabola is $y^2=4ax$. The reflector has depth $5$ cm and diameter $20$ cm, so the point $(5,10)$ lies on it. Thus $10^2=4a(5)$, giving $a=5$. Hence the focus is $5$ cm from the vertex.

Place the vertex at the origin and let the parabola open downward: $x^2=-4ay$. At the base, $y=-10$ and $x=\pm\dfrac52$, so $\left(\dfrac52\right)^2=40a$, giving $a=\dfrac5{32}$. At 2 m from the vertex, $y=-2$, so $x^2=8a=\dfrac54$. The width is $2|x|=2\cdot\dfrac{\sqrt5}{2}=\sqrt5$ m.

Take the origin at the lowest point of the cable above the middle of the roadway. The cable has equation $y=ax^2$. At the ends, $x=\pm50$ and the cable is $30-6=24$ m above the lowest point, so $24=a(50)^2$, giving $a=\dfrac6{625}$. At $x=18$, $y=\dfrac6{625}\cdot18^2=\dfrac{1944}{625}$. Adding the shortest wire length, the wire is $6+\dfrac{1944}{625}=\dfrac{5694}{625}$ m.

Place the centre at the origin, with the arch as the upper half of $\dfrac{x^2}{16}+\dfrac{y^2}{4}=1$. A point 1.5 m from one end is at horizontal coordinate $x=4-1.5=2.5$ in magnitude. Then $y=2\sqrt{1-\dfrac{(2.5)^2}{16}}=2\sqrt{\dfrac{39}{64}}=\dfrac{\sqrt{39}}4$ m.

Let the end on the x-axis be $A$ and the end on the y-axis be $B$. Given $AB=12$ and $AP=3$, so $PB=9$. If the rod makes an angle $\theta$ with the x-axis, then the coordinates of $P$ satisfy $x=9\cos\theta$ and $y=3\sin\theta$. Therefore $\dfrac{x^2}{81}+\dfrac{y^2}{9}=\cos^2\theta+\sin^2\theta=1$.

For $x^2=12y$, $4a=12$, so $a=3$. The latus rectum endpoints are $(\pm2a,a)=(\pm6,3)$, and the vertex is $(0,0)$. The base between the endpoints has length $12$, and its perpendicular distance from the vertex is $3$. Hence the area is $\dfrac12\cdot12\cdot3=18$ square units.

The path is an ellipse with the flag posts as foci. The constant sum of distances is $2a=10$, so $a=5$. The distance between foci is $2c=8$, so $c=4$. Hence $b^2=a^2-c^2=25-16=9$. Taking the foci on the x-axis, the equation is $\dfrac{x^2}{25}+\dfrac{y^2}{9}=1$.

Let the vertex of the parabola be $O(0,0)$ and the other two vertices be symmetric points $(x,y)$ and $(x,-y)$ on $y^2=4ax$. The base length is $2y$, and the side from $O$ to $(x,y)$ has length $\sqrt{x^2+y^2}$. For an equilateral triangle, $\sqrt{x^2+y^2}=2y$, so $x^2=3y^2$. Since $y^2=4ax$, we get $x^2=12ax$, hence $x=12a$. Then $y^2=48a^2$, so $y=4\sqrt3a$. The side length is $2y=8\sqrt3a$.