Question 13 Marks
A man running a racecourse notes that the sum of the distances from the two flag posts from him is always $10 m$ and the distance between the flag posts is $8 m$. Find the equation of the path traced by the man.
AnswerLet A and B be the positions of the two flag posts and P(x, y) be the position of the man.
Accordingly, $PA + PB = 10$
We know that if a point moves in-plane in such a way that the sum of its distance from two fixed points is constant, then the path is an ellipse and this constant value is equal to the length of the major axis of the ellipse.
Therefore, the path described by the man is an ellipse where the length of the major axis is 10m, while points A and B are the foci.
Taking the origin of the coordinate plane as the center of the ellipse, while taking the major axis along the x-axis, 
The equation of the ellipse will be of the form $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$, where a is the semi-major axis.
Accordingly, 2a = 10 $\Rightarrow$ a = 5
Distance between the foci = 2ae = 2c = 8
$\Rightarrow$ c = 4
On using the relation, c = $\sqrt{a^{2}-b^{2}}$, we get,
4 = $\sqrt{25-b^{2}}$
$\Rightarrow 16 = 25 – b^2$
$\Rightarrow b^2 = 25 -1 6 = 9$
$\Rightarrow b = 3$
Put value of a and b in $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$.
$\frac{x^{2}}{5^{2}}+\frac{y^{2}}{3^{2}}=1$ $\Rightarrow$ $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$. View full question & answer→Question 23 Marks
A rod of length $12 m$ moves with its ends always touching the coordinates axes. Determine the equation of the locus of a point P on the rod, which is $3 cm$ from the end in contact with the X-axis.
AnswerLet l be the length of the rod and which at any position meet X-axis at A (a, 0) and also meets the Y-axis at B (0, b), therefore we have
$l^2 = a^2 + b^2$
$\Rightarrow (12)^2 = a^2 + b^2 ...(i)$ [$\because$ l = 12]
Let P be the point on AB which is 3 cm from A and hence 9 cm from B.
This means that the point P divides AB in ratio 3 : 9 i.e., 1 : 3.
If P = (x, y), then by section formula, we have
(x, y) = $\left( \frac { 1 \times 0 + 3 \times a } { 1 + 3 } , \frac { 1 \times b + 3 \times 0 } { 1 + 3 } \right)$
$\Rightarrow$ (x, y) = $\left( \frac { 3 a } { 4 } , \frac { b } { 4 } \right)$
$\Rightarrow$ x = $\frac { 3 a } { 4 }$ , y = $\frac { b } { 4 }$ $\Rightarrow$ a = $\frac { 4 x } { 3 }$ and $b = 4y$
On putting the values of a and b in Equation (i), we get
144 = $\left( \frac { 4 x } { 3 } \right) ^ { 2 } + (4y)^2$
$\Rightarrow$ $\frac { x ^ { 2 } } { 81 } + \frac { y ^ { 2 } } { 9 }$ = 1
which is required equation. View full question & answer→Question 33 Marks
An arc is in the form of a semi-ellipse. It is 8 m wide and 2 m high at the centre. Find the height of the arch at a point 1.5 m from one end.
AnswerHere width of elliptical arch = 8 m.
$\therefore$ AB = 8 m 2a = 8 $\Rightarrow$ a = 4
Height at the centre = 2 m
$\therefore$ OB = 2 $\Rightarrow$ b = 2
The axis is of the ellipse is x-axis.
So the equation of ellipse in standard form is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
$\therefore \frac{{{x^2}}}{{{{(4)}^2}}} + \frac{{{y^2}}}{{{{(2)}^2}}} = 1 \Rightarrow \frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{4} = 1$
Now AP = 1.5 m OP = OA - AP = 4 - 1.5 = 2.5m
Let PQ = h
$\therefore$ Coordinate of Q are (2.5, h)
Since the point Q lies on the ellipse $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{4} = 1$
$\therefore \frac{{{{(2.5)}^2}}}{{16}} + \frac{{{h^2}}}{4} = 1 \Rightarrow \frac{{{h^2}}}{4} ={ 1}-\frac{ 6.25}{{16}}$$ \Rightarrow {h^2} = \frac{{9.75 \times 4}}{{16}} = \frac{{9.75}}{4}$
$\Rightarrow h^2$ = 2.44 $\Rightarrow$ h = $\sqrt{2.44}$ = 1.56 m approx. View full question & answer→Question 43 Marks
The cable of a uniformly loaded suspension bridge hangs in the form of a parabola. The roadway which is horizontal and $100 m$ long is supported by vertical wires attached to the cable, the longest wire being $30 m$ and the shortest being $6 m$. Find the length of a supporting wire attached to the roadway $18 m$ from the middle.
AnswerLet AOB be the cable of uniformly loaded suspension bridge. Let AL and BM be the longest wires of length 30 m each. Let OC be the shortest wire of length 6 m and LM be the roadway.
Now AL = BM = 30 m, OC = 6 m and LM = 100 m.
$\therefore$ LC = CM = $\frac{1}{2}$LM= 50 m
Let O be the vertex and axis of the parabola be y-axis. So the equation of parabola in standard form is $x^2 = 4ay$
Coordinates of point B are (50, 24)
Since point B lies on the parabola $x^2 = 4ay$
$\therefore (50)^2 = 4a × 24$ ⇒ a = $\frac{{2500}}{{4 \times 24}} = \frac{{625}}{{24}}$
So equation of parabola is ${x^2} = \frac{{4 \times 625}}{{24}}y \Rightarrow {x^2} = \frac{{625}}{6}y$
Let length of the supporting wire PW at a distance of 18 m be h.
$\therefore$ OR = 18 m and PR = PQ – QP = PQ - OC = h - 6
Coordinates of point P are (18, h - 6)
Since the point P lies on parabola ${x^2} = \frac{{625}}{6}y$
$\therefore (18)^2$ = $\frac{{625}}{6}$ (h - 6) ⇒ 324 $\times$ 6 = 625h - 3750
$\Rightarrow$ 625 h = 1944 + 3750 $\Rightarrow h = \frac{{5694}}{{625}}$ = 9.11 m approx. View full question & answer→Question 53 Marks
An arc is in the form of a parabola with its axis vertical. The arch is 10 m high and 5 m wide at the base. How wide is it 2 m from the vertex of the parabola?
AnswerLet AB be the parabolic arch having O as the vertex and OY as the axis.
The parabola is of the form $x^2 = 4ay$
Now CD = 5 m $\Rightarrow$ OD = 2.5 m
BD = 10 m
$\Rightarrow$ Coordinates of point B are (2.5, 10)
Since the point B lies on the parabola $x^2 = 4ay$
$\therefore {(2.5)^2} = 4a \times 10 \Rightarrow a = \frac{{6.25}}{{40}} = \frac{{625}}{{4000}} = \frac{5}{{32}}$
$\therefore$ Equation of parabola is ${x^2} = 4 \times \frac{5}{{32}}y$
$ \Rightarrow {x^2} = \frac{5}{8}y$
Let PQ = d ⇒ NQ = $\frac{d}{2}$
$\therefore$ Coordinates of Point Q are $\left( {\frac{d}{2},2} \right)$
Since point Q lies on the parabola ${x^2} = \frac{5}{8}y$
$\therefore {\left( {\frac{d}{2}} \right)^2} = \frac{5}{8} \times 2 \Rightarrow \frac{{{d^2}}}{4} = \frac{5}{4} \Rightarrow {d^2} = 5 \Rightarrow d = \sqrt 5$
Thus width of arc $ = \sqrt 5 \;m$ = 2.24m approx. View full question & answer→Question 63 Marks
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$4x^2 + 9y^2 = 36$
AnswerThe equation of given ellipse is $4x^2 + 9y^2 = 36$
i.e. $\frac{{4{x^2}}}{{36}} + \frac{{9{y^2}}}{{36}} = 1 \Rightarrow \frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$
Now 9 > 4 $\Rightarrow$ $a^2 = 9$ and $b^2 = 4$
So the equation of ellipse in standard form is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
$\therefore a^2 = 9 \Rightarrow a = 3 and b^2 = 4 \Rightarrow$ b = 2
We know that $c = \sqrt {{a^2} - {b^2}}$
$c = \sqrt {9 - 4} = \sqrt 5$
$\therefore$ Coordinates of foci are $( \pm c,\;0)$ i.e. $( \pm \sqrt5,\;0)$
Coordinates of vertices are $( \pm a,\;0)$ i.e. $( \pm 3,\;0)$
Length of major axis $ = 2a = 2 \times 3 = 6$
Length of minor axis $2b = 2 \times 2 = 4$
Eccentricity (e) $= \frac{c}{a} = \frac{{\sqrt 5 }}{3}$
Length of latus rectum $= \frac{{2{b^2}}}{a} = \frac{{2 \times 4}}{3} = \frac{8}{3}$
View full question & answer→Question 73 Marks
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$16x^2 + y^2 = 16$
AnswerThe equation of given ellipse is $16x^2 + y^2 = 16$
i.e. $\frac{{16{x^2}}}{{16}} + \frac{{{y^2}}}{{16}} = 1$ $\Rightarrow \frac{{{x^2}}}{1} + \frac{{{y^2}}}{{16}} = 1$
Now 16 > 1 $\Rightarrow$ $a^2 = 16$ and $b^2 = 1$
So the equation of ellipse in standard form is $\frac{{{y^2}}}{{{a^2}}} + \frac{{{x^2}}}{{{b^2}}} = 1$
$a^2 = 16 $\Rightarrow$ a = 4 and b^2 = 1 \Rightarrow b = 1$
We know that $c = \sqrt {{a^2} - {b^2}}$
$\therefore c = \sqrt {16 - 1} = \sqrt {15}$
$\therefore$ Coordinates of foci are $(0,\; \pm c)$ i.e. $(0,\; \pm \sqrt {15} )$
Coordinates of vertices are $(0,\; \pm a)$ i.e. $(0,\; \pm 4)$
Length of major axis = 2 a = $2 \times 4 = 8$
Length of minor axis = 2b = $2 \times 1$ = 2
Eccentricity (e) = $\frac{c}{a} = \frac{{\sqrt {15} }}{4}$
Length of latus rectum $= \frac{{2{b^2}}}{a} = \frac{{2 \times 1}}{4} = \frac{1}{2}$
View full question & answer→Question 83 Marks
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$36x^2 + 4y^2 = 144$
AnswerThe equation of given ellipse is $36x^2 + 4y^2 = 144$
i.e. $\frac{{36{x^2}}}{{144}} + \frac{{4{y^2}}}{{144}} = 1 \Rightarrow \frac{{{x^2}}}{4} + \frac{{{y^2}}}{{36}} = 1$
Now 36 > 4 $\Rightarrow a^2 = 36$ and $b^2 = 4$
So the equation of ellipse in standard form is $\frac{{{y^2}}}{{{a^2}}} + \frac{{{x^2}}}{{{b^2}}} = 1$
$\therefore a^2 = 36 \Rightarrow a = 6$ and $b^2 = 4 \Rightarrow b = 2$
We know that $c = \sqrt {{a^2} - {b^2}}$
$\therefore c = \sqrt {36 - 4} = \sqrt {32} = 4\sqrt 2$
$\therefore$ Coordinates of foci are $(0,\; \pm c)$ i.e. $(0,\; \pm 4\sqrt 2 )$
Coordinates of vertices are $(0,\; \pm a)$ i.e. $(0,\; \pm 6)$
Length of major axis = 2 a = $2 \times 6 = 12$
Length of minor axis $ = 2b = 2 \times 2 = 4$
Eccentricity (e) $ = \frac{c}{a} = \frac{{4\sqrt 2 }}{6} = \frac{{2\sqrt 2 }}{3}$
Length of latus rectum $ = \frac{{a{b^2}}}{a} = \frac{{2 \times 4}}{6} = \frac{4}{3}$
View full question & answer→Question 93 Marks
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$\frac{{{x^2}}}{{100}} + \frac{{{y^2}}}{{400}} = 1$
AnswerThe equation of given ellipse is $\frac{{{x^2}}}{{100}} + \frac{{{y^2}}}{{400}} = 1$
Now 400 > 100 $\Rightarrow a^2 = 400$ and $b^2 = 100$
So the equation of ellipse in standard form is $\frac { y ^ { 2 } } { a ^ { 2 } } + \frac { x ^ { 2 } } { b ^ { 2 } } = 1$
$\therefore a^2 = 400 \Rightarrow a = 20$ and $b^2 = 100 \Rightarrow b = 10$
We know that $c = \sqrt {{a^2} - {b^2}}$
$\therefore c = \sqrt {400 - 100} = \sqrt {300} = 10\sqrt 3$
$\therefore$ Coordinates of foci are $(0,\; \pm c)$ i.e. $(0,\; \pm 10\sqrt 3 )$
Coordinates of vertices are $(0,\; \pm a)$ i.e. $(0,\; \pm 20)$
Length of major axis = 2 a $ = 2 \times 20 = 40$
Length of minor axis = 2 b = $2 \times 10$ = 20
Eccentricity (e) $ = \frac{c}{a} = \frac{{10\sqrt 3 }}{{20}} = \frac{{\sqrt 3 }}{2}$
Length of latus rectum $= \frac{{2{b^2}}}{a} = \frac{{2 \times 100}}{{20}} = 10$
View full question & answer→Question 103 Marks
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$\frac{{{x^2}}}{{49}} + \frac{{{y^2}}}{{36}} = 1$
AnswerThe equation of given ellipse is $\frac{{{x^2}}}{{49}} + \frac{{{y^2}}}{{36}} = 1$
Now 49 > 36 $\Rightarrow a^2 = 49$ and $b^2 = 36$
So the equation of ellipse in standard form is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
$\therefore a^2 = 49 \Rightarrow a = 7$ and $b^2 = 36 \Rightarrow b = 6$
We know that $c = \sqrt {{a^2} - {b^2}}$
$\therefore c = \sqrt {49 - 36} = \sqrt {13}$
$\therefore$ Coordinates of foci are $( \pm c,\;0)$ i.e. $( \pm \sqrt {13} ,\;0)$
Coordinates of vertices are $( \pm a,\;0)$ i.e. $( \pm 7,\;0)$
Length of major axis = 2 a = $2 \times 7 = 14$
Length of minor axis = $2b = 2 \times 6 = 12$
Eccentricity (e) $= \frac{c}{a} = \frac{{\sqrt {13} }}{7}$
Length of latus rectum $= \frac{{2{b^2}}}{a} = \frac{{2 \times 36}}{7} = \frac{{72}}{7}$
View full question & answer→Question 113 Marks
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{100}} = 1$
AnswerThe equation of given ellipse is $\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{{100}} = 1$
Now 100 > 25 $\Rightarrow a^2 = 100$ and $b^2 = 25$
So the equation of ellipse in standard form is $\frac{{{y^2}}}{{{a^2}}} + \frac{{{x^2}}}{{{b^2}}} = 1$
$\therefore a^2 = 100 \Rightarrow a = 10$ and $b^2 = 25 \Rightarrow b = 5$
We know that $c = \sqrt {{a^2} - {b^2}}$
$\therefore c = \sqrt {100 - 25} = \sqrt {75} = 5\sqrt 3$
$\therefore$ Coordinates of foci are $(0,\; \pm c)$ i.e. $(0, \pm 5\sqrt 3 )$
Coordinates of vertices are $(0,\; \pm a)$ i.e. $(0,\; \pm 10)$
Length of major axis = 2 a = $2 \times 10$= 20
Length of minor axis = 2 b = $2 \times 5$ = 10
Eccentricity (e) $ = \frac{c}{a} = \frac{{5\sqrt 3 }}{{10}} = \frac{{\sqrt 3 }}{2}$
Length of latus rectum $= \frac{{a{b^2}}}{a} = \frac{{2 \times 25}}{{10}} = 5$
View full question & answer→Question 123 Marks
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$
AnswerThe equation of given ellipse is $\frac{{{x^2}}}{{16}} + \frac{{{y^2}}}{9} = 1$
Now 16 > 9 $\Rightarrow a^2 = 16$ and $b^2 = 9$
On the equation of ellipse in standard form is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
$a^2 = 16 \Rightarrow a = 4$ and $b^2 = 9 \Rightarrow b = 3$
We know that $c = \sqrt {{a^2} - {b^2}}$
$\therefore c = \sqrt {16 - 9} = \sqrt 7$
$\therefore$ Coordinates of foci are $( \pm c,\;0)$ i.e. $( \pm \sqrt 7 ,\;0)$
Coordinates of vertices are $( \pm a,\;0)$ i.e. $( \pm 4,\;0)$
Length of major axis = 2 a $= 2 \times 4 = 8$
Length of minor axis = 2b $ = 2 \times 3$= 6
Eccentricity (e) $ = \frac{c}{a} = \frac{{\sqrt 7 }}{4}$
Length of latus rectum $ = \frac{{a{b^2}}}{a} = \frac{{2 \times 9}}{4} = \frac{9}{2}$
View full question & answer→Question 133 Marks
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{25}} = 1$
AnswerThe equation of given ellipse is $\frac{{{x^2}}}{4} + \frac{{{y^2}}}{{25}} = 1$
Now 25 > 4 $\Rightarrow a^2 = 25$ and $b^2 = 4$
So the equation of ellipse in standard form is $\frac{{{y^2}}}{{{a^2}}} + \frac{{{x^2}}}{{{b^2}}} = 1$
$\therefore a^2 = 25 \Rightarrow a = 5$ and $b^2 = 4 \Rightarrow b = 2$
We know that $c = \sqrt {{a^2} - {b^2}}$
$\therefore c = \sqrt {25 - 4} = \sqrt {21}$
$\Rightarrow$ Coordinates of foci are $(0,\; \pm c)$ i.e. $(0 \pm \sqrt {21} )$
Coordinates of vertices are $(0,\; \pm a)$ i.e. $(0,\; \pm 5)$
Length of major axis $= 2a = 2 \times 5 = 10$
Length of minor axis = 2b $= 2 \times 2 = 4$
Eccentricity (e) $\frac{c}{a} = \frac{{\sqrt {21} }}{5}$
Length of latus rectum $= \frac{{a{b^2}}}{a} = \frac{{2 \times 4}}{5} = \frac{8}{5}$
View full question & answer→Question 143 Marks
Find the coordinates of the foci, the vertices, the length of major axis, the minor axis, the eccentricity and the length of the latus rectum of the ellipse.
$\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{16}} = 1$
AnswerThe equation of given ellipse is $\frac{{{x^2}}}{{36}} + \frac{{{y^2}}}{{16}} = 1$
Now 36 > 16 $\Rightarrow {a^2} = 36$ and $b^2 = 16$
So the equation of ellipse in standard form is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
$a^2 = 36 \Rightarrow a = 6$ and $b^2 = 16 \Rightarrow b = 4$
We know that $c = \sqrt {{a^2} - {b^2}}$
$c = \sqrt {36 - 16} = \sqrt {20} = 2\sqrt 5$
$\therefore$ Coordinates of foci are $( \pm c,\;0)$ i.e. $( \pm 2\sqrt 5 ,\;0)$
Coordinates of vertices are $( \pm a,\;0)$ i.e. $( \pm 6,\;0)$
Length of major axis = 2a $= 2 \times 6 = 12$
Length of minor axis = 2b = $2 \times 4$ = 8
Eccentricity (e) $= \frac{c}{a} = \frac{{2\sqrt 5 }}{6} = \frac{{\sqrt 5 }}{3}$
Length of latus rectum $ = \frac{{2{b^2}}}{a} = \frac{{2 \times 16}}{6} = \frac{{16}}{3}$
View full question & answer→Question 153 Marks
Find the equation of the circle with radius 5 whose centre lies on x-axis and passes through the point (2, 3).
AnswerSince the centre of the circle lies on x-axis, the coordinates of centres is (h, 0)
Now the circle passes through the point (2, 3)
$\therefore$ Radius of circle $ = \sqrt {{{(h - 2)}^2} + {{(0 - 3)}^2}} = \sqrt {{h^2} + 4 - 4h + 9} $ $= \sqrt {{h^2} - 4h + 13}$
But radius of circle = 5
$\therefore \;\sqrt {{h^2} - 4h + 13} = 5$ $\Rightarrow h^2 - 4h + 13 = 25 \Rightarrow h^2 - 4h - 12 = 0$
$\Rightarrow$ (h - 6)(h + 2) = 0 $\Rightarrow$ h = 6 or h = -2
When h = 6
Equation of required circle is
$(x - 6)^2 + (y - 0)^2 = (5)^2 \Rightarrow x^2 + 36 - 12x + y^2 = 25$
$\Rightarrow x^2 + y^2 -12x + 11 = 0$
When h = -2
Equation of required circle is
$(x + 2)^2 + (y - 0)^2 = (5)^2 \Rightarrow x^2 + 4 + 4x + y^2 = 25$
$\Rightarrow x^2 + y^2 + 4x - 21 = 0$
View full question & answer→Question 163 Marks
Find the equation of the circle passing through the points $(2, 3)$ and $(-1, 1)$ and whose centre is on the line $x - 3y - 11 = 0.$
AnswerThe equation of the circle is
$(x-h)^2+(y-k)^2=r^2 \ldots(1)$ Since the circle passes through $(-1,1)$,
$\therefore(-1-h)^2+(1-k)^2=r^2$
$\Rightarrow 1+h^2+2 h+1+k^2-2 k=r^2$
$\Rightarrow h^2+k^2+2 h-2 k+2=r^2 \ldots \ldots(2)$
Since the circle passes through point $(2,3)$
$\therefore(2-h)^2+(3-k)^2=r^2 \Rightarrow 4+h^2-4 h+9+k^2-6 k=r^2$
$\Rightarrow h^2+k^2-4 h-6 k+13=r^2 \ldots \ldots(3)$
From (2) and (3), we have
$h^2+k^2-4 h-6 k+13=h^2+k^2+2 h-2 k+2$
$\Rightarrow-6 h-4 k=-11 \Rightarrow 6 h+4 k=11 \ldots .(4)$
Since the centre $(h, k)$ of the circle lies on the line $(x-3 y-11=0)$
$\therefore \mathrm{h}-3 \mathrm{k}-11=0 \Rightarrow \mathrm{~h}-3 \mathrm{k}=11 \ldots \text { (5) }$
Solving (4) and (5), we have,
$h=\frac{7}{2} \text { and } k=\frac{-5}{2}$
Putting these values of $h$ and $k$ in (3), we have
$\left(\frac{7}{2}\right)^2+\left(\frac{-5}{2}\right)^2-\frac{4 \times 7}{2}-6 \times \frac{-5}{2}+13=r^2$
$\Rightarrow \frac{49}{4}+\frac{25}{4}-14+15+13=r^2 \Rightarrow r^2=\frac{65}{2}$
Thus equation of required circle is
$\left(x-\frac{7}{2}\right)^2+\left(y+\frac{5}{2}\right)^2=\frac{65}{2} \Rightarrow x^2+\frac{49}{4}-7 x+y^2+\frac{25}{4}+5 y=\frac{65}{2}$
$\Rightarrow 4 x^2+49-28 x+4 y^2+25+20 y=130$
$\Rightarrow 4 x^2+4 y^2-28 x+20 y-56=0$
$\Rightarrow 4\left(x^2+y^2-7 x+5 y-14\right)=0$
$\Rightarrow x^2+y^2-7 x+5 y-14=0$
View full question & answer→Question 173 Marks
Find the equation of the circle passing through the points (4, 1) and (6, 5) and whose centre is on the line 4x + y = 16.
AnswerThe equation of the circle is
$(x-h)^2+(y-k)^2=r^2 \ldots\text { (i) }$
Since the circle passes through point $(4,1)$
$\therefore(4-h)^2+(1-k)^2=r^2 \Rightarrow 16+h^2-8 h+1+k^2-2 k=r^2$
$\Rightarrow h^2+k^2-8 h-2 k+17=r^2 \ldots . . \text { (ii) }$
Also the circle passes through point $(6,5)$
$\therefore(6-h)^2+(5-k)^2=r^2 \Rightarrow 36+h^2-12 h+25+k^2-10 k=r^2$
$\Rightarrow h^2+k^2-12 h-10 k+61=r^2 \ldots . . \text { (iii) }$
From (ii) and (iii), we have
$h^2+k^2-8 h-2 k+17=h^2+k^2-12 h-10 k+61$
$\Rightarrow 4 h+8 k=44 \Rightarrow h+2 k=11 \ldots . . \text { (iv) }$
Since the centre ( $h, k$ ) of the circle lies on the line $4 x+y=16$
$\therefore 4 h+k=16 \ldots(v)$
Solving (iv) and (v), we have
$\mathrm{h}=3 \text { and } \mathrm{k}=4$
Putting value of $h$ and $k$ in (ii), we have
$(3)^2+(4)^2-8 \times 3-2 \times 4+17=r^2$
$r^2=10$
Thus equation of required circle is
$(x-3)^2+(y-4)^2=10 \Rightarrow x^2+9-6 x+y^2+16-8 y=10$
$\Rightarrow x^2+y^2-6 x-8 y+15=0$
View full question & answer→Question 183 Marks
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 $\frac{x^{2}}{25}+\frac{y^{2}}{9}=1$
AnswerSince the 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
a = 5 and b = 3. Also $c=\sqrt{a^{2}-b^{2}}=\sqrt{25-9}=4$
Therefore, the coordinates of the foci are (– 4,0) and (4,0), vertices are (– 5, 0) and (5, 0).
Length of the major axis 2a is 10 units length of the minor axis 2b is 6 units
Eccentricity is $\frac{4}{5}$
Length of latus rectum is $\frac{2 b^{2}}{a}=\frac{18}{5}$.
View full question & answer→Question 193 Marks
A beam is supported at its ends by supports which are $12$ metres apart. Since the load is concentrated at its centre, there is a deflection of $3$ cm at the centre and the deflected beam is in the shape of a parabola. How far from the centre is the deflection $1$ cm?
Answer
The equation of the parabola takes the form $x^2 = 4ay$.
Since it passes through
$\left(6, \frac{3}{100}\right)$ we have $(6)^{2}=4 a\left(\frac{3}{100}\right)$ ,
i.e., $a=\frac{36 \times 100}{12}=300 \mathrm{m}$ $\Rightarrow a = 300m$
Let AB be the deflection of the beam which is $\frac{1}{100} \mathrm{m}$ Coordinates of B are $\left(x, \frac{2}{100}\right)$
Therefore $x^{2}=4 \times 300 \times \frac{2}{100}=24$
i.e. $x=\sqrt{24}=2 \sqrt{6}$ m View full question & answer→Question 203 Marks
The focus of a parabolic mirror as shown in is at a distance of $5$ cm from its vertex. If the mirror is $45$ cm deep, find the distance $AB$
AnswerSince the distance from the focus to the vertex is 5 cm . We have, $a=5$. If the origin is taken at the vertex and the axis of the mirror lies along the positive $x$-axis, the equation of the parabolic section is $y^2=4(5) x=20 x==>$ required eqution of parabola $y^2=20 x$
Note that $x=45$. Thus
$y^2=900$
Therefore $y= \pm 30$
Hence $A B=2 y=2 \times 30=60 \mathrm{~cm}$
View full question & answer→Question 213 Marks
Find the equation of the hyperbola where foci are (0, $\pm$12) and the length of the latus rectum is $36$.
AnswerWe have given foci are $(0, \pm 12)$
it follows that $\mathrm{c}=12$
Length of the latus rectum $=\frac{2 b^2}{a}=36$ or $b^2=18 a$
Therefore $c^2=a^2+b^2$; gives
$144=a^2+18 a$
i.e., $a^2+18 a-144=0$,
So $\mathrm{a}=-24,6$.
Since $a$ cannot be negative, we take $a=6$ and so $b^2=108$
No, the equation of the hyperbola is $\frac{y^{2}}{36}-\frac{x^{2}}{108}=1, \text { i.e., } 3 y^{2}-x^{2}=108$
Hence, the equation of the hyperbola = $ 3 y^{2}-x^{2}=108$
View full question & answer→Question 223 Marks
Find the coordinates of the foci and the vertices, the eccentricity, the length of the latus rectum of the hyperbola: $y^2 – 16x^2 = 16$
AnswerDividing 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 $c=\sqrt{a^{2}+b^{2}}=\sqrt{16+1}=\sqrt{17}$ ==> c = ae = $\sqrt 17$
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 length of latus rectum $=\frac{2 b^{2}}{a}=\frac{1}{2}$
View full question & answer→Question 233 Marks
Find the coordinates of the foci and the vertices, the eccentricity, the length of the latus rectum of the hyperbolas: $\frac{x^{2}}{9}-\frac{y^{2}}{16}=1$
AnswerComparing 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 = $\sqrt{a^{2}+b^{2}}=\sqrt{9+16}=5$ ==> c = ae = 5
Therefore, the coordinates of the foci are ($\pm$ 5, 0) and that of vertices are (± 3, 0). Also,
The eccentricity $e=\frac{c}{a}=\frac{5}{3}$. The length oflatus rectum $=\frac{2 b^{2}}{a}=\frac{32}{3}$
View full question & answer→Question 243 Marks
Find the equation of the ellipse, with major axis along the x-axis and passing through the points (4, 3) and (– 1,4).
AnswerThe 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
$\frac{16}{a^{2}}+\frac{9}{b^{2}}=1$ ... equation(1)
and $\frac{1}{a^{2}}+\frac{16}{b^{2}}=1$ ….equation(2)
Solving equations (1) and (2), we find that $a^{2}=\frac{247}{7}$ and $b^{2}=\frac{247}{15}$.
Hence the required equation is $\frac{x^{2}}{\left(\frac{247}{7}\right)}+\frac{y^{2}}{(\frac{247}{15})}=1$, i.e., $7x^2 + 15y^2 = 247$.
View full question & answer→Question 253 Marks
Find the equation of the ellipse, whose length of the major axis is $20$ and foci are (0, $\pm$ 5).
AnswerSince the foci are on the y-axis, the major axis is along the y-axis. So, the equation of the ellipse is of the form $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$
Given, length of the major axis is $20$ and foci are $(0, ± 5). ==> 2a = 20$ and $c = ae = 5$
a = semi-major axis = $\frac{20}{2}$= 10
and the relation $c^2 = a^2 – b^2$ gives
$5^2 = 10^2 – b^2 i.e., b^2 = 75$
Therefore, the equation of the ellipse is $\frac{x^{2}}{75}+\frac{y^{2}}{100}=1$
View full question & answer→Question 263 Marks
Find the equation of the ellipse whose vertices are ($\pm$ 13, 0) and foci are ($\pm$ 5, 0).
AnswerSince 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 vertices are ($\pm$ 13, 0) and foci are ($\pm$ 5, 0) ==>> a = 13, c = ae = $\pm$5
Therefore, from the relation $c^2 = a^2 – b^2$ , we get
$25 = 169 – b^2$ , i.e., $b = 12$
Hence the equation of the ellipse is $\frac{x^{2}}{169}+\frac{y^{2}}{144}=1$
View full question & answer→Question 273 Marks
Find the coordinates of the foci, the vertices, the lengths of major and minor axes and the eccentricity of the ellipse $9x^2 + 4y^2 = 36.$
AnswerThe 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 $\frac{x^{2}}{b^{2}}+\frac{y^{2}}{a^{2}}=1$, we have b = 2 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,$(0, \sqrt{5})$ & $(0,-\sqrt{5})$ vertices are (0,3) & (0, –3),length of the major axis = 2a = 6 units
the length of the minor axis = 2b = 4 units and
the eccentricity of the ellipse = $\frac{\sqrt{5}}{3}$.
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