Question 12 Marks
An equilateral triangle is inscribed in the parabola $y^2 = 4ax$ where one vertex is at the vertex of the parabola. Find the length of the side of the triangle.
AnswerThe given equation of parabola is $y^2 = 4ax$ let b be the side of an equilateral $\Delta {\rm O}{\rm A}{\rm B}$ whose one vertex is the vertex of parabola.
Let OC = x
Now AB = b
$\therefore$ AC = BC $ = \frac{1}{2} \times AB = \frac{b}{2}$
Coordinates of point A are $\left( {x,\frac{b}{2}} \right)$
Since point A lies on the parabola $y^2 = 4ax$
$\therefore {\left( {\frac{b}{2}} \right)^2} = 4ax \Rightarrow x = \frac{{{b^2}}}{{4 \times 4a}} \Rightarrow x = \frac{{{b^2}}}{{16a}}$
In right angled $\Delta {\rm O}{\rm A}{\rm C}$
$OA^2 = OC^2 + AC^2$
$\therefore {b^2} = {x^2} + {\left( {\frac{b}{2}} \right)^2} \Rightarrow {b^2} = {\left( {\frac{{{b^2}}}{{16a}}} \right)^2} + \frac{{{b^2}}}{4}$
$\Rightarrow {b^2} = \frac{{{b^4}}}{{256{a^2}}} + \frac{{{b^2}}}{4} \Rightarrow 1 = \frac{{{b^2}}}{{256{a^2}}} + \frac{1}{4}$
$\Rightarrow \frac{{{b^2}}}{{256{a^2}}} = 1 - \frac{1}{4}$$\Rightarrow {b^2} = \frac{3}{4} \times 256{a^2} = {b^2} = 192{a^2}$
$\Rightarrow b = \sqrt {192{a^2}} \Rightarrow b = 8\sqrt 3 a$
Thus the side of equilateral triangle is $8\sqrt 3 a$. View full question & answer→Question 22 Marks
Find the area of the triangle formed by the lines joining the vertex of the parabola $x^2 = 12y$ to the ends of its latus rectum.
AnswerThe given equation of parabola is $x^2 = 12y$ which is of the form $x^2 = 4ay$
4a = 12 $\Rightarrow$ a = 3
Focus of the parabola is (0, 3) $\Rightarrow$ x = $\pm$6
Let AB be the latus rectum if the parabola then y = 3
$\therefore x^2 = 4 \times$ 3 $\times$ 3 = 36
The coordinates of A are (-6, 3) and B are (6, 3)
$\therefore$ Area of $\Delta $AOB$\frac{1}{2}$ [(0 – 0) + (18 + 18) + (0 – 0)]
= $\frac{1}{2}$|36| = 18 sq. units. View full question & answer→Question 32 Marks
If a parabolic reflector is 20 cm in diameter and 5 cm deep, find the focus.
AnswerA parabolic reflector with diameter PR = 20 cm and OQ = 5 cm.
Vertex of the parabola is (0, 0)
Let focus of the parabola be (a, 0).
Now PR = 20 cm $\Rightarrow$ PQ = 10 cm
$\therefore$ Coordinate of point P are (5, 10)
Since the point lies on the parabola $y^2 = 4ax$
$\therefore {(10)^2} = 4a \times 5 \Rightarrow a = \frac{{100}}{{20}} \Rightarrow$ a = 5
Thus required focus of the parabola is (5, 0). View full question & answer→Question 42 Marks
Find the equation of the hyperbola, whose vertices (0, $\pm$3) and foci (0, $\pm$5).
AnswerWe have,
vertices = (0, $\pm$3) = (0, $\pm$a)
$\Rightarrow$ a = 3 and foci = (0, $\pm$c) = (0, $\pm$5)
$\Rightarrow$ c = 5
Also, we know that, $c^2 = a^2 + b^2$
$\Rightarrow 25 = 9 + b^2 [\because$ a = 3]
$\Rightarrow b^2 = 25 - 9 = 16$
Here, the foci and vertices lie on Y-axis,
Therefore equation of hyperbola is of the form
$\frac { y ^ { 2 } } { a ^ { 2 } } - \frac { x ^ { 2 } } { b ^ { 2 } }$ = 1
i.e., $\frac { y ^ { 2 } } { 9 } - \frac { x ^ { 2 } } { 16 }$ = 1 View full question & answer→Question 52 Marks
Find the equation of hyperbola having Vertices (0, $\pm$5), foci (0, $\pm$8)
AnswerThe vertices are (0, $\pm$5) which lie on y-axis.
So the equation of the hyperbola in standard form is $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1$
$\therefore$ the vertices (0, $\pm$a) is (0, $\pm$5)
$\Rightarrow$ a = 5
Foci (0, $\pm$ae) is (0, $\pm$8)
$\Rightarrow$ ae = 8
Now ae = 8
$ \Rightarrow e = \frac{8}{a} \Rightarrow e = \frac{8}{5}$
We know that
$b = a\sqrt {{e^2} - 1} \Rightarrow b = 5\sqrt {\frac{{64}}{{25}} - 1} = 5\frac{{\sqrt {39} }}{5} = \sqrt {39}$
Thus required equation of hyperbola is
$\frac{{{y^2}}}{{{{(5)}^2}}} - \frac{{{x^2}}}{{{{(\sqrt {39} )}^2}}} = 1 \Rightarrow \frac{{{y^2}}}{{25}} - \frac{{{x^2}}}{{39}} = 1$
View full question & answer→Question 62 Marks
Find the equation of hyperbola having Vertices ($\pm$2, 0), foci ($\pm$3, 0)
AnswerThe vertices are ($\pm$2, 0) which lie on x-axis.
So, the equation of hyperbola in standard form is $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$
$\therefore$ the vertices ($\pm$a, 0) is ($\pm$2, 0)
$\Rightarrow$ a = 2
foci ($\pm$ae, 0) is ($\pm$3, 0)
$\Rightarrow$ ae = 3
Now ae = 3
$ \Rightarrow e = \frac{3}{a} \Rightarrow e = \frac{3}{2}$
We know that
$b = a\sqrt {{e^2} - 1} \Rightarrow b = 2\sqrt {\frac{9}{4} - 1} = 2\frac{{\sqrt 5 }}{2} = \sqrt 5$
Thus required equation of hyperbola is
$\frac{{{x^2}}}{{{{(2)}^2}}} - \frac{{{y^2}}}{{{{(\sqrt 5 )}^2}}} = 1 \Rightarrow \frac{{{x^2}}}{4} - \frac{{{y^2}}}{5} = 1$
View full question & answer→Question 72 Marks
Find the coordinates of the foci, and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. $49y^2 - 16x^2 = 784$
AnswerThe given equation of hyperbola is $49y^2 - 16x^2 = 784$
i.e. $\frac{{49{y^2}}}{{784}} - \frac{{16{x^2}}}{{784}} = 1 \Rightarrow \frac{{{y^2}}}{{16}} - \frac{{{x^2}}}{{49}} = 1$ which is of the form $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{a^2}}} = 1$
The foci and vertices of the hyperbola lie on y-axis.
$\therefore a^2 = 16 \Rightarrow a = 4 and b^2 = 49 \Rightarrow b = 7$
Now $c^2 = a^2 + b^2 = 16 + 49 = 65 \Rightarrow$ c = $\sqrt{65}$
$\therefore$ Coordinates of foci are $(0, \pm c)$ i.e. $(0, \pm \sqrt {65} )$
Coordinates of vertices are $(0, \pm a)$ i.e $(0, \pm 4)$
Eccentricity $(e) = \frac{c}{a} = \frac{{\sqrt {65} }}{4}$
Length of latus rectum $ = \frac{{2{b^2}}}{a} = \frac{{2 \times 49}}{4} = \frac{{49}}{2}$
View full question & answer→Question 82 Marks
Find the coordinates of the foci, and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. $5y^2 - 9x^2 = 36$
AnswerThe given equation of hyperbola is $5y^2 - 9x^2 = 36$
i.e. $\frac{{5{y^2}}}{{36}} - \frac{{9{x^2}}}{{36}} = 1 \Rightarrow \frac{{{y^2}}}{{\frac{{36}}{5}}} - \frac{{{x^2}}}{4} = 1$ which is of the form $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1$
The foci and vertices of the hyperbola lie on y-axis.
$\therefore$ ${a^2} = \frac{{36}}{5} \Rightarrow a = \frac{6}{{\sqrt 5 }}$ and $b^2 = 4 \Rightarrow$ b = 2
Now $c^2 = a^2 + b^2 = \frac{{36}}{5} + 4 = \frac{{56}}{5} \Rightarrow c = \sqrt {\frac{{56}}{5}}$
$\therefore$ Coordinates of foci are $(0, \pm c)$ i.e. $\left( {0, \pm \frac{{\sqrt {56} }}{5}} \right)$
Coordinates of vertices are $(0, \pm a)$ i.e. $\left( {0, \pm \frac{6}{{\sqrt 5 }}} \right)$
Eccentricity (e) $= \frac{c}{a} = \frac{{\sqrt {\frac{{56}}{5}} }}{{\frac{6}{{\sqrt 5 }}}} = \frac{{\sqrt {56} }}{6} = \frac{{2\sqrt {14} }}{6} = \frac{{\sqrt {14} }}{3}$
Length of latus rectum $ = \frac{{2{b^2}}}{a} = \frac{{\frac{{2 \times 4}}{6}}}{{\sqrt 5 }} = \frac{{2 \times 4 \times \sqrt 5 }}{6} = \frac{{4\sqrt 5 }}{3}$
View full question & answer→Question 92 Marks
Find the coordinates of the foci, and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. $16x^2 - 9y^2 = 576$
AnswerThe given equation of hyperbola is $16x^2 - 9y^2 = 576$
i.e. $\frac{{16{x^2}}}{{576}} - \frac{{9{y^2}}}{{576}} = 1 \Rightarrow \frac{{{x^2}}}{{36}} - \frac{{{y^2}}}{{64}} = 1$ which is of the form $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$
The foci and vertices of the hyperbola lie on x-axis.
$\therefore a^2 = 36 \Rightarrow a = 6 and b^2 = 64 \Rightarrow b = 8$
Now $c^2 = a^2 + b^2 = 36 + 64 = 100 \Rightarrow$ c = 10
$\therefore$ Coordinates of foci are $( \pm c,0)$ i.e. $( \pm 10,0)$
Coordinates of vertices are $( \pm a,0)$ i.e. $( \pm 6,0)$
Eccentricity (e) $ = \frac{c}{a} = \frac{{10}}{6} = \frac{5}{3}$
Length of latus rectum $= \frac{{2{b^2}}}{a} = \frac{{2 \times 64}}{6} = \frac{{64}}{3}$
View full question & answer→Question 102 Marks
Find the coordinates of the foci, and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas. $9y^2 - 4x^2 = 36$
AnswerThe given equation of hyperbola is $9y^2 - 4x^2 = 36$
i.e. $\frac{{9{y^2}}}{{36}} - \frac{{4{x^2}}}{{36}} = 1 \Rightarrow \frac{{{y^2}}}{4} - \frac{{{x^2}}}{9} = 1$ which is of the form $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1$
The foci and vertices of the hyperbola lie on y-axis.
$\therefore a^2 = 4 \Rightarrow a = 2$ and $(b^2 = 9 \Rightarrow b = 3$
Now $c^2 = a^2 + b^2 = 4 + 9 = 13 \Rightarrow c = \sqrt{13}$
$\therefore$ Coordinates of foci are (0,±c) i.e. (0,±$\sqrt{13}$)
Coordinates of vertices are (0, ± a) i.e. (0, ± 2)
Eccentricity (e) $\frac{c}{a} = \frac{{\sqrt {13} }}{2}$
Length of latus rectum $= \frac{{2{b^2}}}{a} = \frac{{2 \times 9}}{2} = 9$
View full question & answer→Question 112 Marks
Find the coordinates of the foci, and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.
$\frac{{{y^2}}}{9} - \frac{{{x^2}}}{{27}} = 1$
AnswerThe equation of given hyperbola is $\frac{{{y^2}}}{9} - \frac{{{x^2}}}{{27}} = 1$ which is of the form $\frac{{{y^2}}}{a^2} - \frac{{{x^2}}}{{b^2}} = 1$
The foci and vertices of the hyperbola lie on x-axis.
$a^2 = 9 \Rightarrow a = 3$ and $b^2 = 27 \Rightarrow b = 3\sqrt 3$
Now $c^2 = a^2 + b^2 = 9 + 27 = 36 \Rightarrow c = 6$
$\therefore$ Coordinates of foci are (0, ± c) i.e. (0, ± 6)
Coordinates of vertices are (0, ± a) i.e. (0, ± 3)
Eccentricity (e) = $= \frac{c}{a} = \frac{6}{3} = 2$
Length of latus rectum $= \frac{{a{b^2}}}{a} = \frac{{2 \times 27}}{3} = 18$
View full question & answer→Question 122 Marks
Find the equation of hyperbola which has Foci $(0, \pm \sqrt {10} )$ and passing through $(2, 3)$
AnswerHere foci $(0, \pm \sqrt {10} )$ which lie on y-axis
So the equation of hyperbola in standard form is $\frac{{{y^2}}}{{{a^2}}} - \frac{{{x^2}}}{{{b^2}}} = 1$
$\therefore$ foci $(0, \pm a)$ is $(0, \pm \sqrt {10} ) \Rightarrow a = \sqrt {10}$
We know that $c^2 = a^2 + b^2$
$\therefore {(\sqrt {10} )^2} = {a^2} + {b^2} \Rightarrow b^2 = 10 - a^2$
Since the hyperbola passes through (2, 3)
$\therefore \frac{9}{{{a^2}}} - \frac{4}{{{b^2}}} = 1 \Rightarrow \frac{9}{{{a^2}}} - \frac{4}{{10 - {a^2}}} = 1$
$\Rightarrow \frac{{9(10 - {a^2}) - 4{a^2}}}{{{a^2}(10 - {a^2})}} =a^2{(10-a^2)}\Rightarrow a^4 - 23a^2 + 90 = 0$
$\Rightarrow a^4 - 18a^2 - 5a^{2$+$}90 = 0 \Rightarrow (a^2 - 18)(a^2 - 5) = 0\Rightarrow a^2=5 ,18$
When $a^2 = 18$ then $b^2 = 10 - 18 = -8$ (which is not possible)
When $a^2 = 5$ then $b^2 = 10 - 5 = 5$
Thus required equation of hyperbola is
$\frac{{{y^2}}}{5} - \frac{{{x^2}}}{5} = 1$
View full question & answer→Question 132 Marks
Find the equation of hyperbola which has Vertices $( \pm 7,0),e = \frac{4}{3}$
AnswerHere vertices are (± 7, 0) which lie on x-axis.
So the equation of hyperbola in standard form is $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$
$\therefore$ Vertices (± a, 0) is (± 7, 0) ⇒ a = 7
Now $e = \frac{4}{3} \Rightarrow \frac{c}{a} = \frac{4}{3} \Rightarrow \frac{c}{7} = \frac{4}{3} \Rightarrow c = \frac{{28}}{3}$
We know that $c^2 = a^2 + b^2$
$\therefore {\left( {\frac{{28}}{3}} \right)^2} = {(7)^2} + {b^2} \Rightarrow {b^2} = \frac{{784}}{9} - 49 = \frac{{343}}{9}$
Thus required equation of hyperbola is
$\frac{{{x^2}}}{{{{(7)}^2}}} - \frac{{{y^2}}}{{\frac{{343}}{9}}} = 1 \Rightarrow \frac{{{x^2}}}{{49}} - \frac{{{9y^2}}}{{343}} = 1$
View full question & answer→Question 142 Marks
Find the equation of hyperbola having Foci ($\pm$4, 0) and the latus rectum is of length $12$.
AnswerHere foci are $( \pm 4,0)$ which lie on $x$-axis.
So the equation of hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
$\therefore \text { foci }( \pm \mathrm{c}, 0) \text { is } \pm 4,0)$
$\Rightarrow \mathrm{c}=4$
Length of latus rectum $\frac{2 b^2}{a}=12 \Rightarrow b^2=6 a$
We know that $\mathrm{c}^2=\mathrm{a}^2+\mathrm{b}^2$
$\therefore(4)^2=a^2+6 a$
$\Rightarrow a^2+6 a-16=0$
$\Rightarrow(a+8)(a-2)=0$
$\Rightarrow a=2(\because a=-8 \text { is not possible })$
$\Rightarrow a^2=4$
Also $b^2=6 \times 2=12$
Thus required equation of hyperbola is
$\frac{x^2}{4}-\frac{y^2}{12}=1$
View full question & answer→Question 152 Marks
Find the equation of hyperbola having Foci ($\pm$3$\sqrt{5}$, 0), the latus rectum is of length $8$.
AnswerHere foci are $( \pm 3 \sqrt{5}, 0)$ which lie on $x$-axis.
So the equation of hyperbola in standard form is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$ $\therefore$ foci $( \pm c, 0)$ is $( \pm 3 \sqrt{5}, 0)$
$\Rightarrow c=3 \sqrt{5}$
Length of latus rectum $\frac{2 b^2}{a}=8 \Rightarrow b^2=4 a$
We know that $\mathrm{c}^2=\mathrm{a}^2+\mathrm{b}^2$
$\therefore(3 \sqrt{5})^2=a^2+4 a$
$\Rightarrow a^2+4 a-45=0$
$\Rightarrow(a+9)(a-5)=0$
$\Rightarrow a=5(\because a=-9 \text { is not possible })$
Also $a=5$
$\Rightarrow b^2=4 \times 5=20$
Thus required equation of hyperbola is
$\frac{x^2}{25}-\frac{y^2}{20}=1$
View full question & answer→Question 162 Marks
Find the equation of hyperbola having Foci (0, $\pm$13) and the conjugate axis is of length 24.
AnswerHere foci are $(0, \pm 13)$ which lie on $y$-axis.
So the equation of hyperbola in standard form is $\frac{y^2}{a^2}-\frac{x^2}{b^2}=1$
$\therefore(13)^2=a^2+(12)^2$
$\Rightarrow a^2=169-144=25$
Thus required equation of hyperbola is
$\frac{y^2}{25}-\frac{x^2}{(12)^2}=1 \Rightarrow \frac{y^2}{25}-\frac{x^2}{144}=1$
View full question & answer→Question 172 Marks
Find the equation of hyperbola, when foci are at ($\pm$5, 0) and transverse axis is of length 8.
AnswerHere, foci are at ($\pm$5, 0)
$\therefore$ ($\pm$c, 0) = ($\pm$5,0)
$\Rightarrow$ c = 5
And length of transverse
axis = 2a = 8 $\Rightarrow$ a = 4
Also, we know that, $c^2 = a^2 + b^2$
$\Rightarrow 25 = 16 + b^2 [\because$ a = 4, c = 5]
$\Rightarrow b^2 = 9$
Since, the foci lie on X-axis. Therefore, the equation of hyperbola is of the form
$\frac { x ^ { 2 } } { a ^ { 2 } } - \frac { y ^ { 2 } } { b ^ { 2 } }$ = 1
On putting the values of $a^2$ and $b^2$, we get
$\frac { x ^ { 2 } } { 16 } - \frac { y ^ { 2 } } { 9 }$ = 1
which is the required equation of hyperbola. View full question & answer→Question 182 Marks
Find the coordinates of the foci, and the vertices, the eccentricity and the length of the latus rectum of the hyperbolas.
$\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$
AnswerThe equation of given hyperbola is $\frac{{{x^2}}}{{16}} - \frac{{{y^2}}}{9} = 1$ which is of the form $\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1$
The foci and vertices of the hyperbola lie on x-axis.
$\therefore a^2 = 16 \Rightarrow$ a = 4 and $b^2 = 9 \Rightarrow$ b = 3
Now $c^2 = a^2 + b^2 = 16 + 9 = 25 \Rightarrow c = 5$
$\therefore$ Coordinates of foci are (± c, 0) i.e. (± 5, 0)
Coordinates of vertices are (± a, 0) i.e. (± 4, 0)
Eccentricity $(e) = \frac{c}{a} = \frac{5}{4}$
Length of latus rectum $ = \frac{{2{b^2}}}{a} = \frac{{2 \times 9}}{4} = \frac{9}{2}$
View full question & answer→Question 192 Marks
Find the equation of ellipse having Major axis on the x-axis and passes through the points (4, 3) and (6, 2)
AnswerSince the major axis is along x-axis.
So the equation of ellipse in standard form is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
Since the ellipse passes through point (4, 3)
$\therefore \frac{{16}}{{{a^2}}} + \frac{9}{{{b^2}}} = 1$. . . (i)
Also the ellipse passes through point (6, 2)
$\therefore \frac{{36}}{{{a^2}}} + \frac{4}{{{b^2}}} = 1$....(ii)
Solving (i) and (ii), we have
$a^2 = 52$ and $b^2 = 13$
Thus equation of required ellipse is
$\frac{{{x^2}}}{{52}} + \frac{{{y^2}}}{{13}} = 1$
View full question & answer→Question 202 Marks
Find the equation of ellipse having Centre at (0, 0) major axis on the y-axis and passes through the points (3, 2) and (1, 6) .
AnswerSince the major axis is along y-axis
So the equation of ellipse in standard form is $\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1$
Since the ellipse passes through the point (3, 2)
$\therefore \frac{9}{{{b^2}}} + \frac{4}{{{a^2}}} = 1$
Also the ellipse passes through point (1, 6)
$\therefore \frac{1}{{{b^2}}} + \frac{{36}}{{{a^2}}} = 1$
Solving (i) and (ii), we have
$a^2 = 40$ and $b^2 = 10$
Thus equation of required ellipse is
$\frac{{{x^2}}}{{10}} + \frac{{{y^2}}}{{40}} = 1$
View full question & answer→Question 212 Marks
Find the equation of ellipse having b = 3, c= 4, centre at origin, foci on the x-axis.
AnswerThe foci lie on x-axis
So the equation of ellipse in standard form is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
We know that $c^2=a^2-b^2$
$\therefore(4)^2=a^2-(3)^2 \Rightarrow a^2=16+9=25$
Thus equation of required ellipse is
$\frac{{{x^2}}}{{25}} + \frac{{{y^2}}}{9} = 1$
View full question & answer→Question 222 Marks
Find the equation of ellipse having Foci ($\pm$3, 0), a = 4.
AnswerThe foci $( \pm 3,0)$ lie on $x$-axis
So the equation of ellipse in standard form is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Now foci $( \pm c, 0)$ is $( \pm 3,0)$
$\Rightarrow c=3$
We know that $\mathrm{c}^2=\mathrm{a}^2-\mathrm{b}^2$
$\therefore(3)^2=(4)^2-b^2$
$\Rightarrow b^2=16-9=7$
Thus equation of required ellipse is
$\frac{x^2}{16}+\frac{y^2}{7}=1$
View full question & answer→Question 232 Marks
Find the equation of ellipse having Length of minor axis 16, foci (0,$\pm$6)
AnswerThe foci (0,$\pm$6) lie on y-axis.
So the equation of ellipse in standard form is $\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1]$
Now length of minor axis 2b = 16 $\Rightarrow$ b = 8
foci (0, ±c) is (0,± 6) ⇒ c = 6
We know that ${c^2} = {a^2} - {b^2}$
$\therefore (6)^2 = a^2 - (8)^2 \Rightarrow a^2 = 36 + 64 = 100$
Thus equation of required ellipse is
$\frac{{{x^2}}}{{64}} + \frac{{{y^2}}}{{100}} = 1$
View full question & answer→Question 242 Marks
Find the equation of ellipse having Length of major axis 26, foci ($\pm$5, 0)
AnswerThe foci $( \pm 5,0)$ lie on $x$-axis
So the equation of ellipse in standard form is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Now length of major axis $2 a=26$
$\Rightarrow \mathrm{a}=13$
$\text { foci }( \pm \mathrm{c}, 0) \text { is }( \pm 5,0)$
$\Rightarrow \mathrm{c}=5$
We know that $\mathrm{c}^2=\mathrm{a}^2-\mathrm{b}^2$
$\therefore(5)^2=(13)^2-b^2$
$\Rightarrow b^2=169-25=144$
Thus equation of required ellipse is
$\frac{x^2}{169}+\frac{y^2}{144}=1$
View full question & answer→Question 252 Marks
Find the equation of ellipse having Ends of major axis $(0, \pm \sqrt 5 )$ ,ends of minor axis $( \pm 1,0)$
AnswerEnds of major axis $(0, \pm \sqrt 5 )$ lies on y-axis
So the equation of ellipse in standard form is $\frac{{{x^2}}}{{{b^2}}} + \frac{{{y^2}}}{{{a^2}}} = 1$
Now ends of major axis (0, ± a) is (0, ± $\sqrt 5$) ⇒ a = $\sqrt 5$
Ends of minor axis (± b, 0) is (±1, 0) ⇒ b = 1
Thus equation of required ellipse is
$\frac{{{x^2}}}{1} + \frac{{{y^2}}}{5} = 1$
View full question & answer→Question 262 Marks
Find the equation of ellipse having Ends of major axis $( \pm 3,0)$, ends of minor axis $(0, \pm 2)$
AnswerEnds of major axis $( \pm 3,0)$ lie on x-axis.
So the equation of ellipse in standard form is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
Now ends of major axis (± a, 0) is (± 3, 0) ⇒ a = 3
Ends of minor axis $(0, \pm b)$ is $(0, \pm 2) \Rightarrow$ b = 2
Thus equation of required ellipse is
$\frac{{{x^2}}}{9} + \frac{{{y^2}}}{4} = 1$
View full question & answer→Question 272 Marks
Find the equation of ellipse which has Vertices ($\pm$6, 0), foci ($\pm$4, 0)
AnswerThe foci $( \pm 4,0)$ lie on $x$-axis.
So the equation of ellipse in standard form is $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Now vertices $( \pm \mathrm{a}, 0)$ is $( \pm 6,0)$
$\Rightarrow \mathrm{a}=6$
$\text { foci }( \pm \mathrm{c}, 0) \text { is }( \pm 4,0)$
$\Rightarrow \mathrm{c}=4$
We know that $c^2=a^2-b^2$
$\therefore(4)^2=(6)^2-b^2$
$\Rightarrow b^2=36-16=20$
Thus equation of required ellipse is
$\frac{x^2}{36}+\frac{y^2}{20}=1$
View full question & answer→Question 282 Marks
Find the equation of the ellipse which has Vertices (0, $\pm$13), foci(0, $\pm$5)
AnswerThe foci $(0, \pm 5)$ lie on $y$-axis
So the equation of ellipse in standard form is $\frac{y^2}{a^2}+\frac{x^2}{b^2}=1$
Now vertices $(0, \pm a)$ is $(0, \pm 13)$
$\Rightarrow a=13$
Foci $(0, \pm \mathrm{c})$ is $(0, \pm 5)$
$\Rightarrow c=5$
We know that $c^2=a^2-b^2$
$\therefore(5)^2=(13)^2-b^2$
$\Rightarrow b^2=169-25=144$
Thus equation of required ellipse is
$\frac{x^2}{144}+\frac{y^2}{169}=1$
View full question & answer→Question 292 Marks
Find the equation of ellipse which has vertices ($\pm$ 5, 0), foci ($\pm$4, 0)
AnswerVertices $( \pm 5,0)$ and foci $( \pm 4,0)$
Here, the vertices are on the x-axis.a
Therefore, the equation of the ellipse will be of form $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, where $a$ is the semi-major axis.
Accordingly, $\mathrm{a}=5$ and $\mathrm{c}=\mathrm{ae}=4$.
It is known that $a^2=b^2+c^2$.
$\therefore 5^2=b^2+4^2$
$\Rightarrow 25=b^2+16$
$\Rightarrow b^2=25-16$
$\Rightarrow b=\sqrt{9}=3$
Thus, the equation of the ellipse is $\frac{x^2}{5^2}+\frac{y^2}{3^2}=1$ or $\frac{x^2}{25}+\frac{y^2}{9}=1$.
View full question & answer→Question 302 Marks
Find the equation of the parabola that satisfies the given conditions: Vertex (0, 0) Focus (3, 0)
AnswerThe vertex of the parabola is at (0, 0) and focus is at (3, 0),
$\Rightarrow$ y = 0 $\Rightarrow$ The axis of parabola is along x-axis
So the parabola is of the form $y^2 = 4ax$.
The required equation of parabola is
$y^2=4 \times 3 x \Rightarrow y^2=12 x$
View full question & answer→Question 312 Marks
Find the equation of the parabola that satisfies the given conditions: Focus (0, - 3) directrix y = 3
AnswerSince the focus $(0,-3)$ lies on the $y$-axis, therefore $y$-axis is the axis of parabola. Also the directrix is $y=3$ i.e. $y=a$ and focus $(0,-3)$ i.e. $(0,-a)$. So the parabola is of the form $x^2=-4 a y$.
The required equation of parabola is
$x^2=-4 \times 3 y \Rightarrow x^2=-12 x$
View full question & answer→Question 322 Marks
Find the equation of the parabola that satisfies the given conditions: Focus (6, 0) directrix x = -6
AnswerThe required equation of parabola is
$y^2=4 \times 6 x \Rightarrow y^2=24 x$
View full question & answer→Question 332 Marks
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum: $x^2 = - 9y$
AnswerThe given equation of parabola is $x^2 = -9y$ which is of the form $x^2 = -4ay$
$\therefore$ 4a = 9 $\Rightarrow a = \frac{9}{4}$
$\therefore$ Coordinates of focus are $\left( {0,\;\frac{{ - 9}}{4}} \right)$
Axis of parabola is x = 0
Equation of the directrix is $y = \frac{9}{4} \Rightarrow$ 4y - 9 = 0
Length of latus rectum $= 4 \times \frac{9}{4} = 9$
View full question & answer→Question 342 Marks
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum: $y^2 = 10x$
AnswerThe given equation of parabola is $y^2 = 10x$ which is of the form $y^2 = 4ax$
$\therefore$ 4a = 10 $\Rightarrow a = \frac{{10}}{4} \Rightarrow a = \frac{5}{2}$
$\therefore$ Coordinates of focus are $\left( {\frac{5}{2},\;0} \right)$
Axis of parabola is y = 0
Equation of the directrix is $x = \frac{{ - 5}}{2} \Rightarrow$ 2 x + 5 = 0
Length of latus rectum $ = \frac{{4 \times 5}}{2} = 10$
View full question & answer→Question 352 Marks
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum: $x^2 = -16y$
AnswerThe given equation of parabola is $x^2 = 16y$ which is of the form $x^2 = -4ay$
$\therefore$ 4a = 16 $\Rightarrow$ a = 4
$\therefore$ Coordinates of focus are (0, -4)
Axis of parabola is x = 0
Equation of the directrix is y = 4 $\Rightarrow$y - 4 = 0
Length of latus rectum = 4 $\times$ 4 = 16
View full question & answer→Question 362 Marks
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum: $y^2 = -8x$
AnswerThe given equation of parabola is $y^2 = -8x$ which is of the form $y^2 = -4 ax$
$\therefore$ 4a = 8 $\Rightarrow$ a = 2
$\therefore$ Coordinates of focus are (-2, 0)
Axis of parabola is y = 0
Equation of the directrix is x = 2 $\Rightarrow$ x - 2= 0
Length of latus rectum = 4 $\times$ 2 = 8
View full question & answer→Question 372 Marks
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum: $x^2 = 6y$
AnswerThe given equation of parabola is $x^2 = 6y$ which is of the form $x^2 = 4ay$
$\therefore$ 4a = 6 $\Rightarrow a = \frac{6}{4} \Rightarrow a = \frac{3}{2}$
$\therefore$ Coordinates of focus are $\left( {0,\;\frac{3}{2}} \right)$
Axis of parabola is x = 0
Equation of the directrix is $y = \frac{{ - 3}}{2} \Rightarrow$ 2y + 3 =0
Length of latus rectum $= \frac{{4 \times 3}}{2}$ = 6
View full question & answer→Question 382 Marks
Find the equation of the parabola that satisfies the given conditions: Vertex (0, 0) passing through (5, 2) and symmetric with respect to y-axis.
AnswerThe vertex of the parabola is at (0, 0) and it is symmetrical about the y-axis.
$\therefore$ axis of parabola is Y-axis
So the parabola is of the form $x^2 = 4ay$
Since the parabola passes through point (5, 2)
$\therefore {(5)^2} = 4a \times 2 \Rightarrow 25 = 8a \Rightarrow a = \frac{{25}}{8}$
The required equation of parabola is
${x^2} = \frac{{4 \times 25}}{8}y \Rightarrow {x^2} = \frac{{25}}{2}y \Rightarrow 2{x^2} = 25y$
View full question & answer→Question 392 Marks
Find the equation of the parabola that satisfies the given conditions: Vertex (0, 0) passing through (2, 3) and axis is along x-axis.
AnswerThe vertex of the parabola is at (0, 0) and the axis is along x-axis.
So the parabola is of the form $y^2 = 4ax$
Since the parabola passes through point (2, 3)
$\therefore (3)^2 = 4a \times 2 \Rightarrow$ 9 = 8 a $\Rightarrow a = \frac{9}{8}$
The required equation of parabola is
${y^2} = \frac{{4 \times 9}}{8}x \Rightarrow {y^2} = \frac{9}{2}x \Rightarrow 2{y^2} = 9x$
View full question & answer→Question 402 Marks
Find the equation of the parabola that satisfies the given conditions: Vertex (0, 0) Focus (-2, 0)
AnswerThe vertex of the parabola is at $(0,0)$ and focus is at $(-2,0)^{\prime}$
$\Rightarrow \mathrm{y}=0 \Rightarrow$ The axis of parabola is along x -axis
So the parabola is of the form $y^2=4 a x$.
The required equation of parabola is
$y^2=4 x-2 x \Rightarrow y^2=-8 x$.
View full question & answer→Question 412 Marks
Find the coordinates of the focus, axis of the parabola, the equation of the directrix and the length of the latus rectum: $y^2 = 12x$
AnswerThe given equation of parabola is $y^2 = 12x$ which is of the form $y^2 = 4ax$.
$\therefore$ 4a = 12 $\Rightarrow$ a = 3
$\therefore$ Coordinates of focus are (3, 0)
Axis of parabola is y = 0
Equation of the directrix is x = -3 $\Rightarrow$ x + 3 = 0
Length of latus rectum = 4$\times$3 = 12
View full question & answer→Question 422 Marks
Find the centre and radius of the circle $2x^2 + 2y^2- x = 0$.
AnswerThe given equation of circle is
$2x^2 + 2y^2 - x = 0 \Rightarrow x^2 + y^2 - \frac { x } { 2 }$ = 0
$\Rightarrow$ $\left( x ^ { 2 } - \frac { x } { 2 } \right)$ $+ y^2 = 0$
On adding $\frac { 1 } { 16 }$ to make perfect squares, we get
$\left( x ^ { 2 } - \frac { x } { 2 } + \frac { 1 } { 16 } \right) + y^2 = \frac { 1 } { 16 }$
$\Rightarrow$ $\left( x - \frac { 1 } { 4 } \right) ^ { 2 } + (y - 0)^2 = \left( \frac { 1 } { 4 } \right) ^ { 2 }$
On comparing with $(x - h)^2 + (y - k)^2 = r^2$, we get
h = $\frac { 1 } { 4 }$, k = 0 and r = $\frac { 1 } { 4 }$
$\therefore$ Centre = (h, k) = $\left( \frac { 1 } { 4 } , 0 \right)$
and Radius = $\frac { 1 } { 4 }$
View full question & answer→Question 432 Marks
Find the centre and radius of the circle. $x^2+y^2-8 x-10 y-12=0$
AnswerThe given equation of circle is
$x^2+y^2-8 x-10 y-12=0$
$\therefore\left(x^2-8 x\right)+\left(y^2+10 y\right)=12$
Completing the square
$\Rightarrow\left[x^2-8 x+(4)^2\right]+\left[y^2+10 y+(5)^2\right]$
$=12+(4)^2+(5)^2$
$\Rightarrow(x-4)^2+(y+5)^2=12+16+25$
$\Rightarrow(x-4)^2+(y+5)^2=53$
$\Rightarrow(x-4)^2+(y+5)^2=(\sqrt{53})^2$
Comparing it with $(x-h)^2+(y-k)^2=r^2$, we have
$h=4, k=-5 \text { and } r=\sqrt{53}$
Thus coordinates of the centre is $(4,-5)$ and radius is $\sqrt{53}$.
View full question & answer→Question 442 Marks
Find the centre and radius of the circle. $x^2+y^2-4 x-8 y-45=0$
AnswerThe given equation of circle is
$x^2+y^2-4 x-8 y-45=0$
$\therefore\left(x^2-4 x\right)+\left(y^2-8 y\right)=45$
$\Rightarrow\left[x^2-4 x+(2)^2\right]+\left[y^2-8 y+(4)^2\right]$
$=45+(2)^2+(4)^2$
$\Rightarrow(x-2)^2+(y-4)^2=45+4+16$
$\Rightarrow(x-2)^2+(y-4)^2=65$
$\Rightarrow(x-2)^2+(y-4)^2=(\sqrt{65})^2$
Comparing it with $(x-h)^2+(y-k)^2=r^2$, we have
$h=2, k=4 \text { and } r=\sqrt{65}$
Thus coordinates of the centre is $(2,4)$ and radius is $\sqrt{65}$.
View full question & answer→Question 452 Marks
Find the centre and radius of the circle. $(x+5)^2+(y-3)^2=36$
AnswerThe given equation of circle is
$(x+5)^2+(y-3)^2=36 \Rightarrow(x+5)^2+(y-3)^2=(16)^2$
Comparing it with $(x-h)^2+(y-k)^2=r^2$ we have
$\mathrm{h}=-5, \mathrm{k}=3 \text { and } \mathrm{r}=6$
Thus the coordinates of the centre is $(-5,3)$ and radius is 6 .
View full question & answer→Question 462 Marks
Find the equation of the circle with centre $(-a, -b)$ and radius $\sqrt {{a^2} - {b^2}}$
AnswerHere $h =- a , k =- b$ and $r =\sqrt{a^2-b^2}$
The equation of circle is
$(x-h)^2+(y-k)^2=r^2$
$\therefore(x+a)^2+(y+b)^2=\left(\sqrt{a^2-b^2}\right)^2$
$\Rightarrow x^2+a^2+2 a x+y^2+b^2+2 b y=a^2-b^2$
$\Rightarrow x^2+y^2+2 a x+2 b y+2 b^2=0$
Which is required equation of circle.
View full question & answer→Question 472 Marks
Find the equation of the circle with centre (1, 1) and radius $\sqrt2$
AnswerHere $h=1, k=1$ and $r=\sqrt{2}$
The equation of circle is
$(x-h)^2+(y-k)^2=r^2$
$\therefore(x-1)^2+(y-1)^2=(\sqrt{2})^2$
$\Rightarrow x^2+1-2 x+y^2+1-2 y=2$
$\Rightarrow x^2+y^2-2 x-2 y=0$
Which is required equation of circle.
View full question & answer→Question 482 Marks
Find the equation of the circle with centre $\left( \frac { 1 } { 2 } , \frac { 1 } { 4 } \right)$ and radius $\frac { 1 } { 12 }$.
AnswerGiven centre is $\left( \frac { 1 } { 2 } , \frac { 1 } { 4 } \right)$
$\therefore$ h = $\frac { 1 } { 2 }$, k = $\frac { 1 } { 4 }$ and radius, r = $\frac { 1 } { 12 }$
On putting these values in equation of circle
$(x-h)^2+(y-k)^2=r^2 \text {, we get }$
$\left(x-\frac{1}{2}\right)^2+\left(y-\frac{1}{4}\right)^2=\left(\frac{1}{12}\right)^2$
$\Rightarrow x^2+\frac{1}{4}-x+y^2+\frac{1}{16}-\frac{y}{2}=\frac{1}{144}$
$\Rightarrow x^2+y^2-x-\frac{y}{2}+\frac{1}{4}+\frac{1}{16}-\frac{1}{144}=0$
$\Rightarrow x^2+y^2-x-\frac{y}{2}+\frac{11}{36}=0$
$\Rightarrow 36 x^2+36 y^2-36 x-18 y+11=0$
which is the required equation of circle. View full question & answer→Question 492 Marks
Find the equation of the circle with centre (-2, 3) and radius 4
AnswerHere $h=-2, k=3$ and $r=4$
The equation of circle is
$(x-h)^2+(y-k)^2=r^2$
$\therefore(x+2)^2+(y-3)^2=(4)^2$
$\Rightarrow x^2+4+4 x+y^2+9-6 y=16$
$\Rightarrow x^2+y^2+4 x-6 y-3=0$
Which is required equation of circle.
View full question & answer→Question 502 Marks
Does the point (-2.5, 3.5) lie inside, outside or on the circle $x^2 + y^2 = 25$?
AnswerThe equation of given circle is
$x^2 + y^2 = 25$
$\Rightarrow (x - 0)^2 + (y - 0)^2 = (5)^2$
Comparing it with $(x - h)^2 + (y - k)^2 = r^2$, we have
h = 0 , k = 0 and r = 5
Now distance of the point (-2.5, 3.5) from the centre (0, 0)
$= \sqrt {{{(0 + 2.5)}^2} + {{(0 - 3.5)}^2}} = \sqrt {6.25 + 12.25} = \sqrt {18.5}$ = 4.3 < 5
Thus the point (-2.5, 3.5)lies inside the circle.
View full question & answer→Question 512 Marks
Find the equation of a circle with centre (2, 2) and passes through the point (4, 5).
AnswerThe equation of circle is
$(x - h)^2 + (y - k)^2 = r^2 . . . (i)$
Since the circle passes through point (4, 5) and co-ordinates of centre are (2, 2).
$\therefore$ radius of circle $= \sqrt {{{(4 - 2)}^2} + {{(5 - 2)}^2}} = \sqrt {4 + 9} = \sqrt {13}$
Now the equation of required circle is
$(x - 2)^2 + (y - 2)^2= (\sqrt{13})^2 \Rightarrow x^2 + 4 - 4x + y^2 + 4 - 4y = 13$
$\Rightarrow x^2 + y^2 - 4x - 4y - 5 = 0$
View full question & answer→Question 522 Marks
Find the equation of the circle passing through (0, 0) and making intercepts a and b on the coordinate axes.
AnswerThe circle makes intercepts a with x -axis and b with y -axis.
$\therefore \mathrm{OA}=\mathrm{a} \text { and } \mathrm{OB}=\mathrm{b}$
So the co-ordinates of $A$ are $(a, 0)$ and $B$ are $(0, b)$
Now the circle passes through three points $O(0,0), A(a, 0)$ and $B(0, b)$
Putting the co-ordinates of three points in the equation of circle.
$\mathrm{x}^2+\mathrm{y}^2+2 \mathrm{gx}+2 \mathrm{fy}+\mathrm{c}=0 \ldots$(i)
$\mathrm{c}=0$
$\mathrm{a}^2+2 \mathrm{ga}=0 \Rightarrow \mathrm{a}(\mathrm{a}+2 \mathrm{~g})=0 \Rightarrow g=\frac{-1}{2} a$
$\mathrm{~b}^2+2 \mathrm{fb}=0 \Rightarrow \mathrm{~b}(\mathrm{~b}+2 \mathrm{f})=0 \Rightarrow f=\frac{-1}{2} b$
Putting these values of $\mathrm{g}, \mathrm{f}$ and c in (i) we have
$x^2+y^2+2 \times \frac{-1}{2} a x+2 \times \frac{-1}{2} b y+0=0$
$\Rightarrow x^2+y^2-a x-b y=0$
which is required equation of circle.
View full question & answer→Question 532 Marks
Find the equation of the circle with centre (0, 2) and radius 2
AnswerHere $h=0, k=2$ and $r=2$
The equation of circle is
$(x-h)^2+(y-k)^2=r^2$
$\therefore(x-0)^2+(y-2)^2=(2)^2$
$\Rightarrow x^2+y^2+4-4 y=4$
$x^2+y^2-4 y=0$
Which is required equation of circle.
View full question & answer→Question 542 Marks
Find the equation of the parabola which is symmetric about the y-axis, and passes through the point (2, -3).
AnswerSince the parabola is symmetric about the y-axis and has its vertex at the origin, the equation is of the form $x^2 = 4ay$ or $x^2 = -4ay$,
But the parabola passes through (2,–3) which lies in the fourth quadrant, it must open downwards.
Thus the equation is of the form $x^2 = -4ay$
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, $ i.e., $3 x^2 = -4y$
View full question & answer→Question 552 Marks
Find the equation of the parabola with vertex at (0, 0) and focus at (0, 2).
AnswerGiven, the vertex is $(0,0)$ and focus is at $(0,2)$ which lies on $Y$-axis.
The $Y$ axis is the axis of parabola.
Therefore, equation of parabola is of the form
$x^2=4 a y$
$x^2=4(2) y \text { i.e., } x^2=8 y$
View full question & answer→Question 562 Marks
Find the equation of the parabola with focus (2, 0) and directrix x = – 2.
AnswerGiven that the directrix is $x=-2$ and the focus is $(2,0)$,
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 $y^2=4(2) x=8 x$
View full question & answer→Question 572 Marks
Find the coordinates of focus, axis, the equation of directrix and latus rectum of parabola $\mathrm{y}^2=8 \mathrm{x}$.
AnswerWe have, equation of parabola is $y^2=8 x$.
The given equation involves $y^2$, so the axis of symmetry is along X -axis. The coefficient of x is positive, so the parabola opens to right.
On comparing with the given equation $y^2=4 a x$, we get,
$a=2$
Thus, focus $=(2,0)$
Equation of directrix, $x=-2$
Length of latus rectum is $4 a=4 \times 2=8$.
View full question & answer→Question 582 Marks
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.
AnswerLet the equation of circle with centre $(h, k)$ and radius $r$ be $(x-h)^2+(y-k)^2=r^2 \ldots(i)$
Since, circle passes through the points $(2,-2)$ and $(3,4)$, so the points $(2,-2)$ and $(3,4)$ will lie on Eq. (i).
$\therefore(2-h)^2+(-2-k)^2=r^2 \ldots \text { (iii) }$
$\text { and }(3-h)^2+(4-k)^2=r^2 \ldots \text { (iii) }$
Now, from Eqs. (ii) and (iii), we get
$(2-h)^2+(-2-k)^2=(3-h)^2+(4-k)^2$
$\Rightarrow 4+h^2-4 h+4+k^2+4 k=9+h^2-6 h+16+k^2-8 k$
$\Rightarrow 2 h+12 k=17 \ldots \text { (iv) }$
Also, given that centre ( $h, k$ ) lies on $x+y=2$. So, it will satisfy it.
$\therefore h+k=2 \ldots(v)$
On solving Eqs. (iv) and (v), we get
$\mathrm{h}=0.7, \mathrm{k}=1.3$
Now, $r^2=(2-0.7)^2+(-2-1.3)^2=1.69+10.89=12.58$
On putting $h=0.7, k=13$ and $r^2=12.58$ in Eq. (i), we get
$(x-0.7)^2+(y-1.3)^2=12.58$
which is the required equation of circle.
View full question & answer→Question 592 Marks
Find the centre and the radius of the circle: $x^2+y^2+8 x+10 y-8=0$
AnswerThe 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. $\{\mathrm{x}-(-4)\}^2+\{\mathrm{y}-(-5)\}^2=49$ [comparing with $\{\mathrm{x}-(\mathrm{h})\}^2+\{\mathrm{y}-(\mathrm{k})\}^2=\mathrm{r}^2$, centre $(-\mathrm{h},-\mathrm{k})$ and r radius]
Therefore, the given circle has centre at $(-4,-5)$ and radius 7 .
View full question & answer→Question 602 Marks
A rod AB of length 15 cm rests in between two coordinate axes in such a way that the end point A lies on x-axis and end point B lies on y-axis. A point P(x, y) is taken on the rod in such a way that AP = 6 cm. Show that the locus of P is an ellipse.
AnswerLet AB be the rod making an angle θ with OX as shown in figure and P (x, y) the point on it such that AP = 6 cm
Since AB = 15 cm, we have
PB = 9 cm.
From P draw PR and PQ perpendiculars on x-axis and y-axis, respectively.
From $\Delta \mathrm{PBQ}, \cos \theta=\frac{x}{9}$
From $\Delta \mathrm{PRA}, \sin \theta=\frac{y}{6}$
Since $\cos ^2 \theta+\sin ^2 \theta=1$
$\left(\frac{x}{9}\right)^{2}+\left(\frac{y}{6}\right)^{2}=1$
or $\frac{x^{2}}{81}+\frac{y^{2}}{36}=1$
Therefore, locus of P is an ellipse. View full question & answer→Question 612 Marks
Find the equation of the hyperbola with foci (0, $\pm$ 3) and vertices $\left(0, \pm \frac{\sqrt{11}}{2}\right)$
AnswerSince the foci is on the y-axis, the equation of the hyperbola is of the form $\frac{y^{2}}{a^{2}}-\frac{x^{2}}{b^{2}}=1$given, foci (0, $\pm$ 3) and vertices $\left(0, \pm \frac{\sqrt{11}}{2}\right)$
Since vertices are $\left(0, \pm \frac{\sqrt{11}}{2}\right), \quad a=\frac{\sqrt{11}}{2}$
Also, since foci are (0, ± 3); c= ae = 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, i.e., $100 y^2 – 44 x^2 = 275$
View full question & answer→