CBSE Class 11 Maths Revision Notes Chapter 11

Q.1 Find the centre and radius of the circle x² + y² – 8x + 6y – 2 = 0.

Ans

$\begin{array}{l}\text{The given equation of the circle is}{\mathrm{x}}^2+{\mathrm{y}}^2-8\mathrm{x}+6\mathrm{y}-2=0\\ \text{Comparing it with the general equation of the circle,}\\ {\mathrm{x}}^2+{\mathrm{y}}^2+2\mathrm{gx}+2\mathrm{fy}+\mathrm{c}=0.\text{we get}\mathrm{g}=-4,\mathrm{f}=3\text{and}\mathrm{c}=-2\\ \text{The centre of the circle is given by}(-\mathrm{g},-\mathrm{f})=(4,-3)\text{.}\\ \therefore \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}Radius}=\sqrt{{\mathrm{g}}^2+{\mathrm{f}}^2-\mathrm{c}}=\sqrt{{(-4)}^2+{\left(3\right)}^2+2}\\ =\sqrt{27}=3\sqrt{3}\text{units}\end{array}$

Q.2 Find the equation of the circle whose centre lies at point (2,2) and passes through the centre of the circle x² + y² + 4x + 6y + 2 = 0.

Ans

$\begin{array}{l}\text{The centre of the given circle is given by}(-\mathrm{g},-\mathrm{f})=(-2,-3)\\ \text{Radius of the required circle = distance between points}(2,2)\text{and}(-2,-3)\\ \sqrt{{\left({\mathrm{x}}_1-{\mathrm{x}}_2\right)}^2-{\left({\mathrm{y}}_1-{\mathrm{y}}_2\right)}^2}=\sqrt{{(2-(-2\left)\right)}^2-{(2-(-3\left)\right)}^2}\\ =\sqrt{16+25}=\sqrt{41}\\ \text{The equation of the circle whose centre lies at point}(2,2)\text{and passes through the}(-2,-3)\\ \text{is given by}\\ {\left(\mathrm{x}-2\right)}^2+{\left(\mathrm{y}-2\right)}^2={\left(\sqrt{41}\right)}^2\\ {\mathrm{x}}^2+{\mathrm{y}}^2-4\mathrm{x}-4\mathrm{y}-33=0\end{array}$

Q.3 Find the equation of a circle which is concentric to the given circle x² + y² – 4x – 6y – 3 = 0 and which touches the x axis.

Ans

The centre of the given circle is given by
(–g, –f) = (–1/2 coff. of x, –1/2 coff. of y) = (2,3)
The centre of the given circle is (2,3). The centre of concentric circles are same , therefore the centre of the required circle is (2,3).
Since the circle touches the x-axis therefore the radius of the required circle is 3 units.
The equation of the required circle is
(x – 2)² + (y – 3)² = 3²
⇒ x² + y² – 4x – 6y + 4 = 0

Q.4 Check whether the point (2, 3) lies inside, outside or on the circle x² + y² = 25.

Ans

$\begin{array}{l}\text{The centre of the circle is}(0,0)\text{and radius}=5.\\ \text{Distance between the points}(0,0)\text{and}(2,3)\text{is given by:}\\ \sqrt{{(2-0)}^2+{(3-0)}^2}=\sqrt{4+9}=\sqrt{13}=3.6055\mathrm{approx}.\\ \text{which is clearly inside the circle as the radius of the circle is}5.\end{array}$

Q.5 Find the coordinates of the focus, axis of the parabola, equation of the directrix and the length of the latus rectum for the parabola y² = 16x.

Ans

The given parabola contains y², so the axis of the parabola is the x-axis.
The given parabola is of the form y² = 4ax. Therefore,
a = 4.
Focus is at (4, 0).
Equation of the directrix is : x = – 4.
Length of latus rectum = 4a
= 4(4)
= 16.

Q.6 Find the equation of the parabola with focus (2, 0) and directrix x = –2.

Ans

Since the focus lies on the x-axis, thus x-axis it self is the axis of the parabola.
Since the directrix is x = –2 and the focus is (2, 0), the parabola is to be of the form
y² = 4ax with a = 2.
Hence the required equation is : y² = 4(2)x = 8x.

Q.7 What is equilateral hyperbola?

Ans

A hyperbola in which a = b is called an equilateral hyperbola.

Q.8 Find the equation of the ellipse, with minor axis is along the x-axis and passing through the points (2, 1) and (1, –3).

Ans

$\begin{array}{l}\text{The standard form of the ellipse is\hspace{0.33em}}{\mathrm{x}}^2/{\mathrm{a}}^2+{\mathrm{y}}^2/{\mathrm{b}}^2=1\\ \text{Since the points}(2,1)\text{and}(1,-3)\text{lie on the ellipse, we have}\\ \frac{4^2}{{\mathrm{a}}^2}+\frac{1^2}{{\mathrm{b}}^2}=1...\left(1\right)\\ \frac{1^2}{{\mathrm{a}}^2}+\frac{3^2}{{\mathrm{b}}^2}=1...\left(2\right)\\ \text{On solving (1) and (2), we get}\\ {\mathrm{a}}^2=\frac{35}{8}\text{and}{\mathrm{b}}^2=\frac{35}{3}\\ \text{Hence, the equation of ellipse is}\\ \frac{8{\mathrm{x}}^2}{35}+\frac{3{\mathrm{y}}^2}{35}=1\end{array}$

Q.9

$\mathbf{\text{Find the equation of the hyperbola with foci}}\mathbf{(}\mathbf{0}\mathbf{,}\mathbf{\pm }\mathbf{4}\mathbf{)}\mathbf{\text{and vertices}}(0,\pm \frac{\sqrt{13}}{2})\mathbf{.}$

Ans

$\begin{array}{l}\text{Since the foci is on}\mathrm{y}\text{-axis, the equation of the hyperbola is of the form,}\\ \frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1.\\ \text{The vertices are}\left(0,\pm \frac{\sqrt{13}}{2}\right)\text{.}\\ \text{So},\mathrm{a}=\frac{\sqrt{13}}{2}\\ \text{Also, since the foci are}(0,\pm 4);\mathrm{c}=4\\ \mathrm{and}{\mathrm{b}}^2={\mathrm{c}}^2-{\mathrm{a}}^2\\ =4^2-{\left(\pm \frac{\sqrt{13}}{2}\right)}^2\\ =\frac{51}{4}\\ \text{Therefore, the equation of the hyperbola is}\\ \frac{{\mathrm{y}}^2}{\frac{13}{4}}-\frac{{\mathrm{x}}^2}{\frac{51}{4}}=1\\ \text{i.e.,}204{\mathrm{y}}^2-52{\mathrm{x}}^2=663.\end{array}$

Q.10 Find the eccentricity of the hyperbola with foci on the x-axis if length of the conjugate axis is 3/4 of the length of its transverse axis.

Ans

The foci of the hyperbola are on the x-axis, so the equation of the hyperbola is :
x²/a² – y²/b² = 1 , where a, b > 0
Transverse axis = 2a and conjugate axis = 2b.
It is given that conjugate axis = (3/4) (length of transverse axis )
2b = (3/4)(2a)
b = (3/4)a

$\mathrm{e}=\frac{\sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}}{\mathrm{a}}=\frac{\sqrt{{\mathrm{a}}^2+\frac{9}{16}{\mathrm{a}}^2}}{\mathrm{a}}=\frac{\frac{5\mathrm{a}}{4}}{\mathrm{a}}=\frac{5}{4}$

Q.11 Define the latus rectum of an ellipse.

Ans

The latus rectum of an ellipse is a line segment perpendicular to the major axis through any of the foci and whose end points lie on the ellipse.

Q.12 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.

Ans

$\begin{array}{l}\text{Let the equation of the circle be}{\left(\mathrm{x}-\mathrm{h}\right)}^2+{\left(\mathrm{y}-\mathrm{k}\right)}^2={\mathrm{r}}^2\\ \text{Since the circle passes through}\left(2,-2\right)\text{and}\left(3,4\right),\text{we have}\\ {\left(2-\mathrm{h}\right)}^2+{\left(-2-\mathrm{k}\right)}^2={\mathrm{r}}^2...\left(1\right)\\ \text{and}{\left(3-\mathrm{h}\right)}^2+{\left(4-\mathrm{k}\right)}^2={\mathrm{r}}^2...\left(2\right)\\ \text{From}\left(\text{1}\right)\text{and}\left(\text{2}\right)\text{we get}\\ {\left(2-\mathrm{h}\right)}^2+{\left(-2-\mathrm{k}\right)}^2={\left(3-\mathrm{h}\right)}^2+{\left(4-\mathrm{k}\right)}^2\\ \Rightarrow 2\mathrm{h}+12\mathrm{k}=17...\left(3\right)\\ \text{Also since the centre lies on the line,}\mathrm{x}+\mathrm{y}=2\text{we have}\\ \mathrm{h}+\mathrm{k}=2...\left(4\right)\\ \text{Solving the equations}\left(3\right)\text{and}\left(\text{4}\right)\text{, we get}\\ \mathrm{h}=0.7,\mathrm{k}=1.3\text{and}{\mathrm{r}}^2=12.58\\ \text{Hence, the equation of the required circle is}\\ {\left(\mathrm{x}-0.7\right)}^2+{\left(\mathrm{y}-1.3\right)}^2=12.58\end{array}$

Q.13 Define the eccentricity of an ellipse.

Ans

The eccentricity of an ellipse is the ratio between the distances from the centre of the ellipse to one of the foci and to one of the vertices of the ellipse.

Q.14 Find the equation of the circle passing through the point (2, 4) and has its centre at the point of intersection of lines x – y = 4 and 2x + 3y = –7.

Ans

On solving the equations of the given lines , we get
x = 1 and y = –3.
The centre of the circle is at (1, –3).
The circle passes through (2, 4).

$\begin{array}{l}\text{Thus, radius}=\text{Distance between}\left(1,-3\right)\text{and}\left(2,4\right)\\ =\sqrt{{(2-1)}^2+{(4+3)}^2}=5\sqrt{2}\\ \text{Equation of the required circle:}\\ {\left(\mathrm{x}-1\right)}^2+{\left(\mathrm{y}-\left(-3\right)\right)}^2={\left(5\sqrt{2}\right)}^2\\ {\mathrm{x}}^2+1-2\mathrm{x}+{\mathrm{y}}^2+9+6\mathrm{y}=50\\ {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}+6\mathrm{y}-40=0\end{array}$

Q.15 A beam is supported at its ends by supports which are 24 metres apart. Since the load is concentrated at its centre, there is a deflection of 6 cm at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection 2 cm?

Ans

Let the vertex be at the lowest point and the axis vertical.
Let the coordinate axis be chosen as shown in the figure.
The equation of the parabola is of the form x² = 4ay.
Since it passes through (12, 6/100), we have
144 = 4a(6/100)
⇒ a = 600 m
Let AB be the deflection of the beam which is 2/100 m.
Coordinates of B are (x, 4/100).
Therefore,

$\begin{array}{l}{\mathrm{x}}^2=4\times 600\times \frac{4}{100}=96\\ \Rightarrow \mathrm{x}=\sqrt{96}=4\sqrt{6}\mathrm{m}\end{array}$

Q.16 Find the equation of the hyperbola in the standard form if the distance between the directrices is 4/√3 and passing through the point (2,1).

Ans

$\begin{array}{l}\mathrm{Let}\mathrm{the}\mathrm{equation}\mathrm{of}\mathrm{the}\mathrm{hyperbola}\mathrm{be}{\mathrm{x}}^2/{\mathrm{a}}^2-{\mathrm{y}}^2/{\mathrm{b}}^2=1\\ \mathrm{Equations}\mathrm{of}\mathrm{the}\mathrm{directrices}\mathrm{are}\mathrm{x}=\pm \mathrm{a}/\mathrm{e}\\ \text{Distance between the directrix}=\mathrm{a}/\mathrm{e}+\mathrm{a}/\mathrm{e}=2\mathrm{a}/\mathrm{e}\because 2\mathrm{a}/\mathrm{e}=4/\sqrt{3}\\ \therefore \mathrm{a}=2\mathrm{e}/\sqrt{3}\\ \text{Also given that}(2,1)\text{lies on hyperbola.}\\ \text{Therefore}\\ \frac{2^2}{{\mathrm{a}}^2}-\frac{1^2}{{\mathrm{b}}^2}=1\\ \Rightarrow \frac{4}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{b}}^2}=1\\ \Rightarrow \frac{4}{{\mathrm{a}}^2}-\frac{1}{{\mathrm{a}}^2\left({\mathrm{e}}^2-1\right)}=1\\ \Rightarrow 4\left({\mathrm{e}}^2-1\right)-1={\mathrm{a}}^2\left({\mathrm{e}}^2-1\right)\\ \Rightarrow 4\left({\mathrm{e}}^2-1\right)-1={\left(\frac{2}{\sqrt{3}}\mathrm{e}\right)}^2\left({\mathrm{e}}^2-1\right)\\ \Rightarrow 4{\mathrm{e}}^2-5=\frac{4{\mathrm{e}}^2}{3}\left({\mathrm{e}}^2-1\right)\\ \Rightarrow 4{\mathrm{e}}^4-16{\mathrm{e}}^2+15=0\\ {\mathrm{e}}^2=\frac{5}{2},\frac{3}{2}.\\ \text{when}{\mathrm{e}}^2=\frac{5}{2}\\ \mathrm{a}=\sqrt{\frac{5}{2}}\cdot \frac{2}{\sqrt{3}}=\sqrt{\frac{10}{3}}\\ \mathrm{b}=\sqrt{\frac{10}{3}}\cdot \sqrt{\frac{5}{2}-1}=\sqrt{5}\\ \text{The required equation is given by}\\ \frac{{\mathrm{x}}^2}{{\left(\frac{10}{\sqrt{3}}\right)}^2}-\frac{{\mathrm{y}}^2}{\sqrt{5^2}}=1\\ \Rightarrow \frac{3{\mathrm{x}}^2}{10}-\frac{{\mathrm{y}}^2}{5}=1\\ \Rightarrow 3{\mathrm{x}}^2-2{\mathrm{y}}^2=10\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}when}{\mathrm{e}}^2=\frac{3}{2}\\ \mathrm{a}=\sqrt{\frac{3}{2}}\cdot \frac{2}{\sqrt{3}}=\sqrt{2}\\ \mathrm{b}=\sqrt{2}\cdot \sqrt{\frac{3}{2}-1}=1\\ \text{The required equation is given by}\\ \frac{{\mathrm{x}}^2}{{\left(\sqrt{2}\right)}^2}-\frac{{\mathrm{y}}^2}{1^2}=1\\ \Rightarrow \frac{{\mathrm{x}}^2}{2}-\frac{{\mathrm{y}}^2}{1}=1\\ \Rightarrow {\mathrm{x}}^2-2{\mathrm{y}}^2=2\end{array}$

Q.17 Find the lengths of the transverse axis, conjugate axis and coordinates of the foci of the hyperbola x²/9 – y²/25 = 1.

Ans

Here the equation given is x²/9 – y²/25 = 1.
Comparing with the standard equation x²/a² – y²/b² = 1 we get a = 3 and b = 5.
The foci of the hyperbola are on the x-axis.
Transverse axis = 2a = 2(3) = 6 units.
Conjugate axis = 2b = 2(5) = 10 units.
Eccentricity, e = {√(a² + b²)}/a = {√(3² + 5²)}/3
= (√34)/3
Foci : (

$\pm$

ae, 0) = ( √34,0)

Q.18 Find the eccentricity, foci and directrices of the ellipse x²/16 + y²/9 = 1.

Ans

The given equation is x²/16 + y²/9 = 1.
Comparing with standard equation x²/a² + y²/b² = 1, we get
a = 4 and b = 3.

$\begin{array}{l}\text{So,}\mathrm{e}=\frac{\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}}{\mathrm{a}}=\frac{\sqrt{16-9}}{4}=\frac{\sqrt{7}}{4}\\ \text{Coordinates of the foci are}(\pm \mathrm{ae},0)=(\pm \sqrt{7},0)\\ \text{Directrices are given by}\\ \mathrm{X}=\pm \frac{\mathrm{a}}{\mathrm{e}}\\ \text{So,\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{x}=\pm \frac{4}{\frac{\sqrt{7}}{4}}\\ \Rightarrow \sqrt{7}\mathrm{x}=\pm 16\end{array}$

Q.19 For the parabola x² = – 4ay, state whether the following given informations are correct; if not, correct them:
(i) Vertex: (0, 1)
(ii) Focus : (0,a)
(iii) Directrix : y – a = 0
(iv) latus rectum : 4a

Ans

(i) Incorrect. Vertex: (0, 0)
(ii) Incorrect. Focus: (0, – a)
(iii) Correct.
(iv) Correct.

Q.20 Find the equation of the ellipse satisfying the following conditions:
vertices at (

$\mathbf{\pm }$

4, 0) and foci (

$\mathbf{\pm }$

3, 0).

Ans

The foci are at (

$\pm$

3, 0) These are on the x axis.
Let the equation of the ellipse be x²/a² + y²/b² = 1, where b = a√(1 – e²)
The vertices are (4,0) and (–4,0), a = 4
Also , foci are (3,0) and (–3,0) , ae = 3 or 4e = 3 or e = 3/4.
b = a √(1 – e²) = 4 √{1 – (3/4)²} = √7
The required equation is

$\begin{array}{l}\frac{{\mathrm{x}}^2}{4^2}+\frac{{\mathrm{y}}^2}{{\left(\sqrt{7}\right)}^2}=1\\ \frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{7}=1\end{array}$

Q.21 Check whether the following are correct or incorrect:

(i) Standard equation for ellipse is x²/a² + y²/b² = 1
(ii) Foci : ( 0,

$\pm$

ae) for an ellipse having foci on y-axis
(iii) Directrices : y =

$\pm$

a/e for an ellipse having foci on x-axis

Ans

(i) It is correct. x²/a² + y²/b² = 1 is the standard equation for an ellipse
(ii) It is correct.
(iii) It is incorrect. : y =

$\pm$

a/e is the equation of the directrices for an ellipse having foci on y-axis.

Q.22 Write the equation of the circle given that the (3,2) and (–1,6) are the end points of diameter.

Ans

Let A (3,2) and B(-1,6) are the end points of the given diameter.
Let P(x,y) be a general point on a circle.
∴ PA is perpendicular to PB.
Slope of PA Slope of PB = –1.
This is required equation of the circle.

$\begin{array}{l}\frac{\mathrm{y}-2}{\mathrm{x}-3}\times \frac{\mathrm{y}-6}{\mathrm{x}+1}=-1\\ \Rightarrow {\mathrm{y}}^2-2\mathrm{y}-6\mathrm{y}+12=-\left({\mathrm{x}}^2-3\mathrm{x}+\mathrm{x}-3\right)\\ \Rightarrow {\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-8\mathrm{y}+9=0\end{array}$

Q.23 Find the length of latus rectum of the parabola 5y² = 16x.

Ans

5y² = 16x ⇒ y² = (16/5 ) x
Comparing with standard equation y² = 4ax we get
4a = 16/5.
Therefore, the length of latus rectum = 16/5

Q.24 Find the radius of the circle x² + y² – 4x + 2y + 1 = 0.

Ans

The given circle is x² + y² – 4x + 2y + 1 = 0.
Comparing with x² + y² + 2gx + 2fy + C = 0 we get
g = –2, f = 1 and C = 1
Radius = √(g² + f ² – C) = √(4 + 1 – 1) = 2 units.

Q.25 Show that the line x + y = 5 touches the circle x² y² – 2x – 4y + 3 = 0. Also, find the point of contact.

Ans

$\begin{array}{l}\text{The given equation of line is}\mathrm{x}+\mathrm{y}=5\text{and that of the circle is}{\mathrm{x}}^2+{\mathrm{y}}^2-2\mathrm{x}-4\mathrm{y}+3=0.\\ \text{Solving\hspace{0.33em}them,\hspace{0.33em}we\hspace{0.33em}get}\\ {\mathrm{x}}^2+{\left(5-\mathrm{x}\right)}^2-2\mathrm{x}-4\left(5-\mathrm{x}\right)+3=0\\ \Rightarrow 2{\mathrm{x}}^2-8\mathrm{x}+8=0\\ \Rightarrow {\mathrm{x}}^2-4\mathrm{x}+4=0\\ \Rightarrow \mathrm{x}=2,2\\ \therefore \mathrm{y}=5-2=3\\ \text{Thus, the given line and the tangent intersect at coincident points}(2,3)\text{and}(2,3)\text{.}\\ \text{Therefore, the line touches the circle and the point of contact is}(2,3)\text{.}\end{array}$

Q.26 Find the eccentricity of the elllipse 7x² + 16y² = 112.

Ans

$\begin{array}{l}7{\mathrm{x}}^2+16{\mathrm{y}}^2=112\\ \text{or\hspace{0.33em}}\frac{{\mathrm{x}}^2}{16}+\frac{{\mathrm{y}}^2}{7}=1\\ \text{Comparing with standard equation of ellipse}\\ \frac{{\mathrm{x}}^2}{{\mathrm{a}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{b}}^2}=1\text{we get}{\mathrm{a}}^2=16\text{and}{\mathrm{b}}^2=7\\ \text{The ecentricity of ellipse is given by}\\ {\mathrm{e}}^2=\frac{{\mathrm{a}}^2-{\mathrm{b}}^2}{{\mathrm{a}}^2}=\frac{9}{16}\\ \therefore \mathrm{e}=\frac{3}{4}\end{array}$

Q.27 Find the equation of a circle with radius 7 cm and whose centre lies on the point (3, 5).

Ans

$\begin{array}{l}\text{Equation of circle\hspace{0.33em}is}\\ {\left(\mathrm{x}-\mathrm{h}\right)}^2+{\left(\mathrm{y}-\mathrm{k}\right)}^2={\mathrm{r}}^2\\ \text{Where}\mathrm{h}=3,\mathrm{k}=5\text{and}\mathrm{r}=7\\ {\left(\mathrm{x}-3\right)}^2+{\left(\mathrm{y}-5\right)}^2=7^2\\ {\mathrm{x}}^2-6\mathrm{x}+9+{\mathrm{y}}^2-10\mathrm{y}+25=49\\ {\mathrm{x}}^2-6\mathrm{x}+{\mathrm{y}}^2-10\mathrm{y}=49-34\\ {\mathrm{x}}^2-6\mathrm{x}+{\mathrm{y}}^2-10\mathrm{y}=15.\end{array}$

Q.28 Where does the point (3, 5) lie in the circle x² + y² = 36.

Ans

$\begin{array}{l}\text{We\hspace{0.33em}have,}\\ {\mathrm{x}}^2+{\mathrm{y}}^2=36\\ {\mathrm{x}}^2+{\mathrm{y}}^2=6^2...\left(1\right)\\ \text{Putting}\mathrm{x}=3\text{and}\mathrm{y}=5\text{in LHS}\\ {\mathrm{x}}^2+{\mathrm{y}}^2={\left(3\right)}^2+{\left(5\right)}^2\\ =9+25\\ =34<36\\ \text{So, the point lies inside the circle.}\end{array}$

Q.29 Find the coordinates of focus and latus rectum of the parabola y² = 12x.

Ans

$\begin{array}{l}\text{Parabola}{\mathrm{y}}^2=12\mathrm{x}\text{has axis of symmetry along}\mathrm{x}\text{– axis,}\\ \text{So, comparing it with}{\mathrm{y}}^2=4\mathrm{ax}\text{, we get}\\ \mathrm{a}=3\\ \text{Thus coordinates of focus is}(3,0)\text{\hspace{0.33em}and latus rectum}=4\mathrm{a}\\ =4\left(3\right)\\ =12.\end{array}$

Q.30 Find the eccentricity and the latus rectum of the ellipse x²/49 + y²/25 = 1.

Ans

$\begin{array}{l}\text{The equation of ellipse is}\\ \frac{{\mathrm{x}}^2}{49}+\frac{{\mathrm{y}}^2}{25}=1\\ \frac{{\mathrm{x}}^2}{7^2}+\frac{{\mathrm{y}}^2}{5^2}=1\\ \text{Here}\mathrm{a}=7\text{and}\mathrm{b}=5\\ \text{Since,}\\ {\mathrm{c}}^2={\mathrm{a}}^2-{\mathrm{b}}^2\\ =7^2-5^2\\ =49-25\\ =24\Rightarrow \mathrm{c}=\pm 2\sqrt{6}\\ \text{Thus, coordinates of foci are}(2\sqrt{6},0)\text{and}(-2\sqrt{6},0)\text{.}\\ \text{Latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2\left(5^2\right)}{7}\\ =\frac{50}{7}\end{array}$

Q.31 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola 5y² – 9x² = 36.

Ans

$\begin{array}{l}\text{The given equation is}5{\mathrm{y}}^2-9{\mathrm{x}}^2=36\\ \Rightarrow \frac{{\mathrm{y}}^2}{\left(\frac{36}{5}\right)}-\frac{{\mathrm{x}}^2}{4}=1\\ \Rightarrow \frac{{\mathrm{y}}^2}{{\left(\frac{6}{\sqrt{5}}\right)}^2}-\frac{{\mathrm{x}}^2}{2^2}=1\text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}…}\left(1\right)\\ \text{On comparing equation}\left(\text{1}\right)\text{with}\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1,\text{we have}\\ \mathrm{a}=\frac{6}{\sqrt{5}}\mathrm{b}=2\\ \text{We know that}{\mathrm{a}}^2+{\mathrm{b}}^2={\mathrm{c}}^2\\ \therefore {\mathrm{c}}^2=\frac{36}{5}+4=\frac{56}{5}\\ \Rightarrow \mathrm{c}=\frac{2\sqrt{14}}{\sqrt{5}}\\ \text{Therefore, the coordinates of the foci are}\left(0,\pm \frac{2\sqrt{14}}{\sqrt{5}}\right)\\ \text{The coordinates of the vertices are}\left(0,\pm \frac{6}{\sqrt{5}}\right)\\ \text{Eccentricity,}\mathrm{e}=\frac{\mathrm{c}}{\mathrm{a}}=\frac{\left(\frac{2\sqrt{14}}{\sqrt{5}}\right)}{\left(\frac{6}{\sqrt{5}}\right)}=\frac{\sqrt{14}}{3}.\end{array}$

Q.32 Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse x²/9 + y²/25 = 1.

Ans

$\begin{array}{l}\text{The given equation is}\\ \frac{{\mathrm{x}}^2}{3^2}+\frac{{\mathrm{y}}^2}{5^2}=1\\ \text{Here, denominator of}{\mathrm{y}}^2>\text{denominator of}{\mathrm{x}}^2\text{\hspace{0.33em}so ellipse is along}\mathrm{Y}\text{– axis.}\\ \text{Comparing with}\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1,\text{we\hspace{0.33em}get}\\ \mathrm{a}=5\text{and}\mathrm{b}=3\\ \mathrm{c}=\sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}=\sqrt{25-9}=4\\ \text{Therefore, the coordinates of foci are}(0,4)\text{and}(0,-4)\\ \text{The coordinates of the vertices are}(0,5)\text{and}(0,-5)\\ \text{The length of major axis}=2\mathrm{a}=2\left(5\right)=10\\ \text{The length of minor axis}=2\mathrm{b}=2\left(3\right)=6\\ \text{Eccentricity}=\frac{\mathrm{c}}{\mathrm{a}}=\frac{4}{5}\\ \text{Length of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}=\frac{2{\left(3\right)}^2}{5}=\frac{18}{5}\end{array}$

Q.33 Find the coordinates of the foci, the vertices, the length of the major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse 49x² + 25y² = 1225.

Ans

$\begin{array}{l}\text{We have}\\ 49{\mathrm{x}}^2+25{\mathrm{y}}^2=1225\\ \Rightarrow \frac{49{\mathrm{x}}^2}{1225}+\frac{25{\mathrm{y}}^2}{1225}=1\\ \Rightarrow \frac{{\mathrm{x}}^2}{25}+\frac{{\mathrm{y}}^2}{49}=1\\ \Rightarrow \frac{{\mathrm{x}}^2}{5^2}+\frac{{\mathrm{y}}^2}{7^2}=1\\ \text{Comparing with}\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}+\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}=1,\text{we\hspace{0.33em}get}\\ \mathrm{b}=5\text{and}\mathrm{a}=7\\ \therefore \mathrm{c}=\pm \sqrt{{\mathrm{a}}^2-{\mathrm{b}}^2}=\pm \sqrt{49-25}\\ =\pm \sqrt{24}=\pm 2\sqrt{6}\\ \text{Therefore,}\\ \text{The coordinates of the foci are}(0,\pm 2\sqrt{6})\\ \text{The coordinates of the foci are}(0,\pm 7)\\ \text{Length of major axis}=2\mathrm{a}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=2(7)=14}\\ \text{Length of minor axis}=2\mathrm{b}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}=2(5)=10}\\ \text{Eccentricity,\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}}\mathrm{e}=\frac{\mathrm{c}}{\mathrm{a}}\\ =\frac{2\sqrt{6}}{7}\\ \text{Length of latus rectu}\mathrm{m}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(5\right)}^2}{7}\\ =\frac{50}{7}\end{array}$

Q.34 Find the coordinates of the foci and the vertices, the eccentricity and the length of the latus rectum of the hyperbola Y²/144 – x²/225 = 1.

Ans

$\begin{array}{l}\text{We\hspace{0.33em}have,}\\ \frac{{\mathrm{y}}^2}{144}-\frac{{\mathrm{x}}^2}{225}=1\\ \Rightarrow \frac{{\mathrm{y}}^2}{{12}^2}-\frac{{\mathrm{x}}^2}{{15}^2}=1\\ \text{Comparing it with}\frac{{\mathrm{y}}^2}{{\mathrm{a}}^2}-\frac{{\mathrm{x}}^2}{{\mathrm{b}}^2}=1,\text{we\hspace{0.33em}get}\\ \mathrm{a}=12\text{and}\mathrm{b}=15\\ \mathrm{c}=\pm \sqrt{{\mathrm{a}}^2+{\mathrm{b}}^2}\\ =\pm \sqrt{144+225}\\ =\pm \sqrt{369}=\pm 3\sqrt{41}\\ \text{Therefore,}\\ \text{\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}\hspace{0.33em}coordinates of foci}=(0,\pm \mathrm{c})=(0,\pm 3\sqrt{41})\\ \text{coordinates of vertices}=(0,\mathrm{a})=(0,\pm 12)\\ \text{Eccentricity,}\mathrm{e}=\frac{\mathrm{c}}{\mathrm{a}}=\frac{3\sqrt{41}}{12}=\frac{\sqrt{41}}{4}\\ \text{Length of latus rectum}=\frac{2{\mathrm{b}}^2}{\mathrm{a}}\\ =\frac{2{\left(15\right)}^2}{4}\\ =\frac{450}{4}\end{array}$