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 $(\pm4, 0)$ 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 $\pm 4, 0)$
$\Rightarrow c = 4$
Length of latus rectum $\frac{{2{b^2}}}{a} = 12 \Rightarrow b^2 = 6a$
We know that $c^2 = a^2 + b^2$
$\therefore (4)^2 = a^2 + 6a$
$\Rightarrow a^2 + 6a – 16 = 0$
$\Rightarrow (a + 8) (a – 2) = 0$
$\Rightarrow a = 2 (\because a = -8$ 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 $\frac{{2{b^2}}}{a} = 8 \Rightarrow b^2 = 4a$
We know that $c^2 = a^2 + b^2$
$\therefore (3\sqrt{5})^2 = a^2 + 4a$
$\Rightarrow a^2 + 4a – 45 = 0$
$\Rightarrow (a + 9) (a – 5) = 0$
$\Rightarrow a = 5 (\because a = -9$ 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 $(\pm5, 0)$ and transverse axis is of length $8.$
AnswerHere, foci are at $(\pm5, 0)$
$\therefore (\pm c, 0) = (\pm5,0)$
$\Rightarrow c = 5$
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 $(0,± 6) \Rightarrow c = 6$
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 $2a = 26$
$\Rightarrow a = 13$
foci $(\pm c, 0)$ is $(\pm 5, 0)$
$\Rightarrow c = 5$
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
$\frac{{{x^2}}}{{169}} + \frac{{{y^2}}}{{144}} = 1$
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 a, 0)$ is $(\pm 6, 0)$
$\Rightarrow a = 6$
foci $(± c, 0)$ is $(\pm 4, 0)$
$\Rightarrow c = 4$
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 $(0, \pm c)$ 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, $a = 5$ and $c = ae = 4.$
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
$y^2 = 4 \times 3x \Rightarrow y^2 = 12x$
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 $x ^2 = - 4ay.$
The required equation of parabola is
$x^2 = -4 \times 3y \Rightarrow x^2 = - 12x$
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 6x $
$\Rightarrow y^2 = 24x$
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 $(-2, 0)'$
$\Rightarrow y = 0 \Rightarrow$ The axis of parabola is along x-axis
So the parabola is of the form $y^2= 4ax.$
The required equation of parabola is
$y^2 = 4x - 2x \Rightarrow y^2 = -8x.$
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. $x ^2 + y ^2 - 8x - 10y - 12 = 0$
AnswerThe given equation of circle is
$x ^2 +y ^2 - 8x - 10y - 12 = 0$
$\therefore (x ^2 - 8x) + (y ^2 + 10y) = 12$
Completing the square
$\Rightarrow [x ^2 - 8x + (4) ^2] + [y ^2 + 10y + (5) ^2]$
$= 12 + (4) ^2 + (5) ^2$
$\Rightarrow (x - 4) ^2 + (y + 5) ^2 = 12 + 16 + 25$
$\Rightarrow (x - 4) ^2 + (y + 5) ^2 = 53$
$\Rightarrow (x - 4) ^2 + (y + 5) ^2 = (\sqrt{53})^2$
Comparing it with $(x - h) ^2 + (y - k) ^2 = r ^2,$ we have
$h = 4, k = -5$ 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. $x^2 + y^2 - 4x - 8y - 45 = 0$
AnswerThe given equation of circle is
$x^2 + y^2 - 4x - 8y - 45 = 0$
$\therefore (x^2 - 4x) + (y^2 - 8y) = 45$
$\Rightarrow [x^2 - 4x + (2)^2] + [y^2 - 8y + (4)^2]$
$= 45 + (2)^2 + (4)^2$
$\Rightarrow (x - 2)^2 + (y - 4)^2 = 45 + 4 + 16$
$\Rightarrow (x - 2)^2 + (y - 4)^2 = 65$
$\Rightarrow (x - 2)^2 + (y - 4)^2 = {(\sqrt {65} )^2}$
Comparing it with $(x - h)^2 + (y - k)^2 = r^2,$ we have
$h = 2 , k = 4$ 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
$h = -5, k = 3$ and $r = 6$
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
$(x - h)^2 + (y - k)^2 = r^2$
$\therefore (x + a)^2 + (y + b)^2 = {\left( {\sqrt {{a^2} - {b^2}} } \right)^2}$
$\Rightarrow x^2 + a^2 + 2ax + y^2 + b^2 +2by = a^2 - b^2$
$\Rightarrow x^2 + y^2 + 2ax + 2by + 2b^2 = 0$
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
$(x - h)^2 + (y - k)^2 = r^2$
$\therefore (x - 1)^2 + (y - 1)^2 = {(\sqrt 2 )^2}$
$\Rightarrow x^2 + 1 - 2x + y^2 + 1 - 2y = 2$
$\Rightarrow x^2 + y^2 - 2x - 2y = 0$
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, $ we get
$\left( x - \frac { 1 } { 2 } \right) ^ { 2 } + \left( y - \frac { 1 } { 4 } \right) ^ { 2 } = \left( \frac { 1 } { 12 } \right) ^ { 2 }$
$\Rightarrow x^2 + \frac { 1 } { 4 } - x + y^2 + \frac { 1 } { 16 } - \frac { y } { 2 } = \frac { 1 } { 144 }$
$\Rightarrow x^2 + y^2 - x - \frac { y } { 2 } + \frac { 1 } { 4 } + \frac { 1 } { 16 } - \frac { 1 } { 144 } = 0$
$\Rightarrow x^2 + y^2 - x - \frac { y } { 2 } + \frac { 11 } { 36 } = 0$
$\Rightarrow 36x^2 + 36y^2- 36x - 18y + 11 = 0$
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
$(x - h)^2 + (y - k)^2 = r^2$
$\therefore (x + 2)^2 + (y - 3)^2 = (4)^2$
$\Rightarrow x^2 + 4 + 4x + y^2 + 9 - 6y = 16$
$\Rightarrow x^2 + y^2 + 4x - 6y - 3 = 0$
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 $OA = a$ and $OB = b$
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.
$x^2 + y^2 + 2gx + 2fy + c = 0. . . (i)$
$c = 0$
$a^2 + 2ga = 0 \Rightarrow a(a + 2 g) = 0 \Rightarrow g = \frac{{ - 1}}{2}a$
$b^2 + 2fb = 0 \Rightarrow b (b + 2 f ) = 0 \Rightarrow f = \frac{{ - 1}}{2}b$
Putting these values of $g, f$ and c in $(i)$ we have
${x^2} + {y^2} + 2 \times \frac{{ - 1}}{2}ax + 2 \times \frac{{ - 1}}{2}by + 0 = 0$
$\Rightarrow x^2 + y^2 - ax - by = 0$
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
$(x - h)^2 + (y - k)^2 = r^2$
$\therefore (x - 0)^2 + (y - 2)^2 = (2)^2$
$\Rightarrow x^2 + y^2 + 4 - 4y = 4$
$x^2 + y^2 - 4y = 0$
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
$x^2 = 4ay$
$x^2 = 4(2)y$ i.e., $x^2 = 8y$
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 $y^2 = 4ax$ or $y^2 = -4ax.$
Since the directrix is $x = – 2$ and the focus is $(2, 0),$
the parabola is to be of the form $y^2 = 4ax$ with $a = 2.$
Hence the required equation is $y^2= 4(2)x = 8x$
View full question & answer→Question 572 Marks
Find the coordinates of focus, axis, the equation of directrix and latus rectum of parabola $y^2 = 8x.$
AnswerWe have, equation of parabola is $y^2 = 8x.$
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 $y^2 = 4ax,$ we get,
$a = 2$
Thus, focus $= (2, 0)$
Equation of directrix, $x = - 2$
Length of latus rectum is $4a = 4 \times 2 = 8.$
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 $(x - h)^2 + (y - k)^2 = r^2 ...(i)$
Since, circle passes through the points $(2, - 2)$ and $(3, 4),$ so the points $(2, - 2)$ and $(3, 4)$ will lie on Eq. $(i).$
$\therefore (2 - h)^2 + (- 2 - k)^2 = r^2 ...(ii)$
and $(3 - h)^2 + (4 - k)^2 = r^2...(iii)$
Now, from Eqs. $(ii)$ and $(iii),$ we get
$(2 - h)^2 + (- 2 - k)^2 = (3 - h)^2 + (4 - k)^2$
$\Rightarrow 4 + h^2 - 4h + 4 + k^2 + 4k = 9 + h^2 - 6h + 16 + k^2- 8k$
$\Rightarrow 2h + 12k = 17 ...(iv)$
Also, given that centre $(h, k)$ lies on $x + y = 2.$ So, it will satisfy it.
$\therefore h + k = 2 ...(v)$
On solving Eqs. $(iv)$ and $(v)$, we get
$h = 0.7, k = 1.3$
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: $x^2 + y^2 + 8x + 10y – 8 = 0$
AnswerThe given equation is $(x^2 + 8x) + (y^2 + 10y) = 8$
Now, completing the squares within the parenthesis, we get
$(x^2 + 8x + 16) + (y^2 + 10y + 25) = 8 + 16 + 25$
i.e. $(x + 4)^2 + (y + 5)^2 = 49$
i.e. $\{x – (– 4)\}^2 + \{y – (–5)\}^2 = 49 [$comparing with $\{x – (h)\}^2 + \{y – (k)\}^2 = r^2,$ centre$(-h, -k)$ 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 $\theta $ 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→