Examples (Revised) - Chapter 10 Conic Sections class 11 ncert solutions Maths - SaraNextGen [2024-2025]

Updated On May 15, 2024

By SaraNextGen

Chapter 10: Conic Sections Class 11 NCERT Solutions - Maths

Example 1

Find an equation of the circle with centre at $(0,0)$ and radius $r$.
Solution

Here $h=k=0$. Therefore, the equation of the circle is $x^2+y^2=r^2$.
Example 2

Find the equation of the circle with centre $(-3,2)$ and radius 4 .
Solution

Here $h=-3, k=2$ and $r=4$. Therefore, the equation of the required circle is
$
(x+3)^2+(y-2)^2=16
$

Example 3

Find the centre and the radius of the circle $x^2+y^2+8 x+10 y-8=0$
Solution

The given equation is
$
\left(x^2+8 x\right)+\left(y^2+10 y\right)=8
$

Now, completing the squares within the parenthesis, we get
$
\left(x^2+8 x+16\right)+\left(y^2+10 y+25\right)=8+16+25
$
i.e.
$
(x+4)^2+(y+5)^2=49
$
i.e.
$
\{x-(-4)\}^2+\{y-(-5)\}^2=7^2
$

Therefore, the given circle has centre at $(-4,-5)$ and radius 7 .

Example 4

Find the equation of the circle which passes through the points $(2,-2)$, and $(3,4)$ and whose centre lies on the line $x+y=2$.
Solution

Let the equation of the circle be $(x-h)^2+(y-k)^2=r^2$.
Since the circle passes through $(2,-2)$ and $(3,4)$, we have
$
(2-h)^2+(-2-k)^2=r^2
$
and $(3-h)^2+(4-k)^2=r^2$
Also since the centre lies on the line $x+y=2$, we have
$
h+k=2
$

Solving the equations (1), (2) and (3), we get
$
h=0.7, k=1.3 \text { and } r^2=12.58
$

Hence, the equation of the required circle is
$
(x-0.7)^2+(y-1.3)^2=12.58 .
$

Example 5

Find the coordinates of the focus, axis, the equation of the directrix and latus rectum of the parabola $y^2=8 x$.
Solution

The given equation involves $y^2$, so the axis of symmetry is along the $x$-axis.
The coefficient of $x$ is positive so the parabola opens to the right. Comparing with the given equation $y^2=4 a x$, we find that $a=2$.

Thus, the focus of the parabola is $(2,0)$ and the equation of the directrix of the parabola is $x=-2$ (Fig 10.19).
Length of the latus rectum is $4 a=4 \times 2=8$.

Example 6

Find the equation of the parabola with focus $(2,0)$ and directrix $x=-2$.
Solution

Since the focus $(2,0)$ lies on the $x$-axis, the $x$-axis itself is the axis of the parabola. Hence the equation of the parabola is of the form either $y^2=4 a x$ or $y^2=-4 a x$. Since the directrix is $x=-2$ and the focus is $(2,0)$, the parabola is to be of the form $y^2=4 a x$ with $a=2$. Hence the required equation is $y^2=4(2) x=8 x$

Example 7

Find the equation of the parabola with vertex at $(0,0)$ and focus at $(0,2)$.
Solution

Since the vertex is at $(0,0)$ and the focus is at $(0,2)$ which lies on $y$-axis, the $y$-axis is the axis of the parabola. Therefore, equation of the parabola is of the form $x^2=4 a y$. thus, we have
$
x^2=4(2) y \text {, i.e., } x^2=8 y \text {. }
$

Example 8

Find the equation of the parabola which is symmetric about the $y$-axis, and passes through the point $(2,-3)$.
Solution

Since the parabola is symmetric about $y$-axis and has its vertex at the origin, the equation is of the form $x^2=4 a y$ or $x^2=-4 a y$, where the sign depends on whether the parabola opens upwards or downwards. But the parabola passes through $(2,-3)$ which lies in the fourth quadrant, it must open downwards. Thus the equation is of the form $x^2=-4 a y$.
Since the parabola passes through $(2,-3)$, we have
$
2^2=-4 a(-3) \text {, i.e., } a=\frac{1}{3}
$

Therefore, the equation of the parabola is
$
x^2=-4\left(\frac{1}{3}\right) y \text {, i.e., } 3 x^2=-4 y \text {. }
$

Example 9

Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the latus rectum of the ellipse
$
\frac{x^2}{25}+\frac{y^2}{9}=1
$

Solution

Since denominator of $\frac{x^2}{25}$ is larger than the denominator of $\frac{y^2}{9}$, the major

axis is along the $x$-axis. Comparing the given equation with $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, we get
$
\begin{aligned}
a & =5 \text { and } b=3 . \text { Also } \\
c & =\sqrt{a^2-b^2}=\sqrt{25-9}=4
\end{aligned}
$

Therefore, the coordinates of the foci are $(-4,0)$ and $(4,0)$, vertices are $(-5,0)$ and $(5,0)$. Length of the major axis is 10 units length of the minor axis $2 b$ is 6 units and the eccentricity is $\frac{4}{5}$ and latus rectum is $\frac{2 b^2}{a}=\frac{18}{5}$.
Example 10

Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse $9 x^2+4 y^2=36$.
Solution

The given equation of the ellipse can be written in standard form as
$
\frac{x^2}{4}+\frac{y^2}{9}=1
$

Since the denominator of $\frac{y^2}{9}$ is larger than the denominator of $\frac{x^2}{4}$, the major axis is along the $y$-axis. Comparing the given equation with the standard equation
$
\frac{x^2}{b^2}+\frac{y^2}{a^2}=1 \text {, we have } b=2 \text { and } a=3 .
$

Also
$
c=\sqrt{a^2-b^2}=\sqrt{9-4}=\sqrt{5}
$
and
$
e=\frac{c}{a}=\frac{\sqrt{5}}{3}
$

Hence the foci are $(0, \sqrt{5})$ and $(0,-\sqrt{5})$, vertices are $(0,3)$ and $(0,-3)$, length of the major axis is 6 units, the length of the minor axis is 4 units and the eccentricity of the ellipse is $\frac{\sqrt{5}}{3}$.
Example 11

Find the equation of the ellipse whose vertices are $( \pm 13,0)$ and foci are $( \pm 5,0)$.
Solution Since the vertices are on $x$-axis, the equation will be of the form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a$ is the semi-major axis.

Given that $a=13, c= \pm 5$.
Therefore, from the relation $c^2=a^2-b^2$, we get
$
25=169-b^2 \text {, i.e., } b=12
$

Hence the equation of the ellipse is $\frac{x^2}{169}+\frac{y^2}{144}=1$.
Example 12

Find the equation of the ellipse, whose length of the major axis is 20 and foci are $(0, \pm 5)$.

Solution

Since the foci are on $y$-axis, the major axis is along the $y$-axis. So, equation of the ellipse is of the form $\frac{x^2}{b^2}+\frac{y^2}{a^2}=1$.

Given that
$
a=\text { semi-major axis }=\frac{20}{2}=10
$
and the relation
$
\begin{aligned}
& c^2=a^2-b^2 \text { gives } \\
& 5^2=10^2-b^2 \text { i.e., } b^2=75
\end{aligned}
$

Therefore, the equation of the ellipse is
$
\frac{x^2}{75}+\frac{y^2}{100}=1
$

Example 13

Find the equation of the ellipse, with major axis along the $x$-axis and passing through the points $(4,3)$ and $(-1,4)$.

Solution

The standard form of the ellipse is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$. Since the points $(4,3)$ and $(-1,4)$ lie on the ellipse, we have
$
\frac{16}{a^2}+\frac{9}{b^2}=1
$

and
$
\frac{1}{a^2}+\frac{16}{b^2}=1
$

Solving equations (1) and (2), we find that $a^2=\frac{247}{7}$ and $b^2=\frac{247}{15}$. Hence the required equation is

$\frac{x^2}{\left(\frac{247}{7}\right)}+\frac{y^2}{\frac{247}{15}}=1 \text {, i.e., } 7 x^2+15 y^2=247 \text {. }$

Example 14

Find the coordinates of the foci and the vertices, the eccentricity,the length of the latus rectum of the hyperbolas:
(i) $\frac{x^2}{9}-\frac{y^2}{16}=1$,
(ii) $y^2-16 x^2=16$

Solution

(i) Comparing the equation $\frac{x^2}{9}-\frac{y^2}{16}=1$ with the standard equation
$
\frac{x^2}{a^2}-\frac{y^2}{b^2}=1
$

Here, $a=3, b=4$ and $c=\sqrt{a^2+b^2}=\sqrt{9+16}=5$
Therefore, the coordinates of the foci are $( \pm 5,0)$ and that of vertices are $( \pm 3,0)$.Also, The eccentricity $e=\frac{c}{a}=\frac{5}{3}$. The latus rectum $=\frac{2 b^2}{a}=\frac{32}{3}$
(ii) Dividing the equation by 16 on both sides, we have $\frac{y^2}{16}-\frac{x^2}{1}=1$

Comparing the equation with the standard equation $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$, we find that $a=4, b=1$ and $c=\sqrt{a^2+b^2}=\sqrt{16+1}=\sqrt{17}$.

Therefore, the coordinates of the foci are $(0, \pm \sqrt{17})$ and that of the vertices are $(0, \pm 4)$. Also,
The eccentricity $e=\frac{c}{a}=\frac{\sqrt{17}}{4}$. The latus rectum $=\frac{2 b^2}{a}=\frac{1}{2}$.
Example 15

Find the equation of the hyperbola with foci $(0, \pm 3)$ and vertices $\left(0, \pm \frac{\sqrt{11}}{2}\right)$.

Solution

Since the foci is on y-axis, the equation of the hyperbola is of the form $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
Since vertices are $\left(0, \pm \frac{\sqrt{11}}{2}\right), a=\frac{\sqrt{11}}{2}$
Also, since foci are $(0, \pm 3) ; c=3$ and $b^2=c^2-a^2=\frac{25}{4}$.
Therefore, the equation of the hyperbola is
$
\frac{y^2}{\left(\frac{11}{4}\right)}-\frac{x^2}{\left(\frac{25}{4}\right)}=1 \text {, i.e., } 100 y^2-44 x^2=275 \text {. }
$

Example 16

Find the equation of the hyperbola where foci are $(0, \pm 12)$ and the length of the latus rectum is 36 .

Solution

Since foci are $(0, \pm 12)$, it follows that $c=12$.
Length of the latus rectum $=\frac{2 b^2}{a}=36$ or $b^2=18 a$

Therefore
$
\begin{aligned}
& c^2=a^2+b^2 ; \text { gives } \\
& 144=a^2+18 a \\
& a^2+18 a-144=0, \\
& a=-24,6 .
\end{aligned}
$

So
Since $a$ cannot be negative, we take $a=6$ and so $b^2=108$.
Therefore, the equation of the required hyperbola is $\frac{y^2}{36}-\frac{x^2}{108}=1$, i.e., $3 y^2-x^2=108$