NCERT Exemplar Class 11 Maths Solutions Chapter 11 Conic Sections

Question:1

Find the equation of the circle which touches both axes in the first quadrant and whose radius is a.
Answer:

Given that the circle has a radius and touches both axes, the centre is (a, a).
(x-a)²+(y-a)²=a²
x²+y²-2ax-2ay+a²=0

Question:2

Show that the point (x, y) given by $x=\frac{2at}{1+t^2}$ and $x=\frac{a(1-t^2)}{1+t^2}$ lies on a circle for all real values of t such that –1 ≤ t ≤1 where a is any given real numbers.

Answer:

We have variable point as $x=\frac{2at}{1+t^2}$ and $y=\frac{a(1-t^2)}{1+t^2}$
$$\begin{gathered} {x}^2+y^2=\frac{4a^2{t}^2}{{(1+t^2)}^2}+\frac{a^2(1+t^4-2t^2)}{{(1+t^2)}^2} \\ =\frac{a^2+a^2{t}^4+2a^2{t}^2}{{(1+t^2)}^2} \\ =\frac{a^2{(1+t^2)}^2}{{(1+t^2)}^2} \\ =a^2 \end{gathered}$$

Question:3

If a circle passes through the point (0, 0) (a, 0), (0, b) then find the coordinates of its centre.
Answer:

We have a circle through the points A(0,0), B(a,0) and C(0,b). The triangle is right-angled at vertex A. So, the centre of the circle is the mid-point of hypotenuse B, C, which is (a/2,b/2)

Question:4

Find the equation of the circle which touches the x-axis and whose centre is (1, 2).

Answer:

It is given that a circle with a centre (1,2) touches the x-axis
The radius of the circle is r = 2
The o equation of the required circle is
(x-1)²+(y-2)²=(2)²
x²-2x+1+y²-4y+4=4
x²+y²-2x-4y+1=0

Question:5

If the lines 3x – 4y + 4 = 0 and 6x – 8y – 7 = 0 are tangents to a circle, then find the radius of the circle.

Answer:

Given lines are 6x-8y+8=0 and 6x-8y-7=0
These parallel lines are tangent to a circle
Diameter of circle = Distance between the lines
Diameter of circle=$\left|\frac{8-(-7)}{\sqrt{36+64}}\right|=\frac{15}{10}=\frac{3}{2}$
Radius of circle=$\frac{3}{4}$

Question:6

Find the equation of a circle which touches both the axes and the line 3x – 4y + 8 = 0 and lies in the third quadrant.

Answer:

Since the circle touches both axes, its centre is C(-a, a) and its radius is a
Also, the circle touches the line 3x-4y+8=0
The distance from centre C to this line is the radius of the circle
Radius of circle, $a=\left|\frac{-3a+4a+8}{\sqrt{9+16}}\right|=\left|\frac{a+8}{5}\right|$
$\left|\frac{a+8}{5}\right|=\pm a$
a=2 or a=-4/3
a=2
Equation of a required circle
(x+2)²+(y+2)²=2²
x²+y²+4x+4y+4=0

Question:7

If one end of a diameter of the circle x² + y² – 4x – 6y + 11 = 0 is (3, 4), then find the coordinate of the other end of the diameter.

Answer:

Given the equation of the circle is
x²+y²-4x-6y+11=0
2g=-4 and 2f=-6
So centre of circle is C(-g,-f) = C(2,3)
A (3,4) is one end of the diameter. Let the other end of the diameter be B (x₁, y₁)
$2=\frac{3+x_1}{2}$ and $3=\frac{4+y_1}{2}$
x₁=1 and y₁=2

Question:8

Find the equation of the circle having (1, –2) as its centre and passing through 3x + y = 14, 2x + 5y = 18

Answer:

Given lines are
3x+y=14 and 2x+5y=18
Solving these equations, we get the point of intersection of the lines as A (4,2)
Radius= $\sqrt{{(4-1)}^2+{(2-(-2)}^2}=\sqrt{9+16}=5$
So, the equation of the required circle is:
(x-1)²+(y+2)²=5²
x²+y²-2x+4y-20=0

Question:9

If the line $y=\sqrt{3}x+k$ touches the circle x² + y² = 16, then find the value of k.

Answer:

Given line is $y=\sqrt{3}x+k$ and the circle is
x²+y²=16The The The
The Centre of the circle is (0,0) and the radius is 4.
Since the line $y=\sqrt{3}x+k$ touches the circle, the perpendicular distance from (0,0) to the line
is equal to the radius of the circle.
$\left|\frac{0-0+k}{\sqrt{3+1}}\right|=4$
$\pm \frac{k}{2}=4$
k=±8

Question:10

Find the equation of a circle concentric with the circle x² + y² – 6x + 12y + 15 = 0 and that has double its area.

Answer:

The given equation of the circle is :
x²+y²-6x+12y+15=0
(x-3)²+(y+6)²=(30)
Hence, centre is (3, -6) and radius is $\sqrt{30}$
Since the required circle is concentric with the above circle, the centre of the required circle is (3, -6).
Let its radius be r
Area of circle=$\mathrm{\Pi }{r}^2=2\pi {\left(\sqrt{30}\right)}^2$
r²=60
$r=\sqrt{60}$
Equation of required circle (x-3)²+(y+6)²=60
x²+y²-6x+12y-15=0

Question:11

If the latus rectum of an ellipse is equal to half of the minor axis, then find its eccentricity.

Answer:

Consider the equation of the ellipse as $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
It is given that the length of the latus rectum = Half of the minor axis
$\frac{2b^2}{a}=b$
a=2b
b²=a²(1-e²)
b²=4b²(1-e²)

1-e²=1/4
e²=3/4
$e=\frac{\sqrt{3}}{2}$

Question:12

Given the ellipse with equation 9x² + 25y² = 225, find the eccentricity and foci.

Answer:

Given the equation of an ellipse
9x²+25y²=225
$\frac{x^2}{25}+\frac{y^2}{9}=1$
a=5, b=3
b²=a²(1-e²)
9=25(1-e²)
$e=\frac{4}{5}$
$\text{ Foci }\equiv (\pm ae,0)\equiv (\pm 5\times (4/5),0)\equiv (\pm 4,0)$

Question:13

If the eccentricity of an ellipse is 5/8 and the distance between its foci is 10, then find the latus rectum of the ellipse.

Answer:

Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
Distance between foci 2ae=10
It is given that eccentricity $e=\frac{5}{8}$
$a=\frac{10}{2}\ast \frac{8}{5}=8$
$b=\sqrt{a^2(1-e^2)}=\sqrt{39}$
Length of latus rectum of ellipse=$\frac{2b^2}{a}=2\ast \frac{39}{8}=\frac{39}{4}$

Question:14

Find the equation of an ellipse whose eccentricity is 2/3, latus rectum is 5, and the centre is (0, 0).

Answer:

Let the equation of the ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
$e=\frac{2}{3}$
Given that latus rectum $\frac{2b^2}{a}=5$
$b^2=\frac{5a}{2}=a^2(1-e^2)$
$\frac{5}{2}=a(1-{\left(\frac{2}{3}\right)}^2)$
$a=\frac{9}{2}$
$b^2=\frac{45}{4}$
The required equation of the ellipse is$\frac{4x^2}{81}+\frac{4y^2}{45}=1$

Question:15

Find the distance between the directrices of the ellipse $\frac{x^2}{36}+\frac{y^2}{20}=1$

Answer:

The equation of the ellipse is $\frac{x^2}{36}+\frac{y^2}{20}=1$
a=6 and$b=2\sqrt{5}$
We know that $b^2=a^2(1-e^2)$
$\frac{20}{36}=1-e^2$
$e^2=\frac{4}{9}$
$x=\pm \frac{a}{e}$
Distance between directrix=$\frac{2a}{e}=\frac{2\ast 6}{\frac{2}{3}}=18$

Question:16

Find the coordinates of a point on the parabola y² = 8x whose focal distance is 4.

Answer:

Given parabola is y²=8x
On comparing this parabola to the y²=4ax
We get a=2
Focal distance=Distance of any point on parabola from the focus
SP=$\sqrt{{(x_1-2)}^2+{(y_1-0)}^2}$
$\sqrt{x_1^2+y_1^2-4x_1+4}+\sqrt{x_1^2+8x_1-4x_1+4}=|x_1+2|$
|x₁+2|=4
x₁=2,-6
But x≠ -6
So for x=2
$y_1^2=8\ast 2=16$
y ±4 So the points are (2,4) and 2,-4

Question:17

Find the length of the line segment joining the vertex of the parabola y² = 4ax and a point on the parabola where the line segment makes an angle $\theta$ to the x-axis.

Answer:

Let the point on the parabola be P (x₁,y₁)
Slope of OP=$\tan \theta =\frac{y_1}{x_1}$
$y_1^2=4a{x}_1$
$x_1=\frac{4a}{{\tan }^2\theta }$
$OP=\frac{4a\sec \theta }{{\tan }^2\theta }=\frac{4a\cos \theta }{{\sin }^2\theta }$

Question:18

If the points (0, 4) and (0, 2) are respectively the vertex and focus of a parabola, then find the equation of the parabola.

Answer:

Given that the vertex of the parabola is A(0,4) and its focus is S(0,2)
So, directrix of the parabola is y=6
Now, by definition of the parabola, for any point P (x,y) on the parabola, a SP = PM
$\sqrt{{(x-0)}^2+{(y-2)}^2}=\left|\frac{0+y-6}{\sqrt{(}0+1)}\right|$
x²+y²-4y+4=y²-12y+36
x²+8y=32

Question:19

If the line y = mx + 1 is tangent to the parabola y² = 4x, then find the value of m.

Answer:

Given that line, y=mx+1 is tangent to the parabola y²=4x
Solving the line with a parabola, we have
(mx+1)²=4x
m²x²+2mx+1=4x
m²x²+x(2m-4)+1=0
Since the line touches the parabola, the above equation must have equal roots.
Discriminant D=0
(2m-4)²-4m²=0
4m²-16m+16-4m²=0
16m=16
m=1

Question:20

If the distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$, then obtain the equation of the hyperbola
Answer:

Let the equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Distance between foci=2ae=16
$e=\sqrt{2}$
$a=4\sqrt{2}$
We know that $b=\sqrt{a^2(1-e^2)}=\sqrt{32}$
$\frac{x^2}{32}-\frac{y^2}{32}=1$
x²-y²=32

Question:21

Find the eccentricity of the hyperbola 9y² – 4x² = 36

Answer:

We have the hyperbola 9y²-4x²=36
$\frac{x^2}{9}-\frac{y^2}{4}=-1$
a²=9
b²=4
a²=b²(e²-1)
$e=\sqrt{1+\frac{9}{4}}=\frac{\sqrt{13}}{2}$

Question:22

Find the equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0).

Answer:

Let the equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
ae=2
$a\ast \frac{3}{2}=2$
$a=\frac{4}{3}$
b²=a²(e²-1)
$b^2=\frac{16}{9}\ast (\frac{9}{4}-1)=\frac{20}{9}$
$\frac{x^2}{\frac{16}{9}}-\frac{y^2}{\frac{20}{9}}=1$
$\frac{x^2}{16}-\frac{y^2}{20}=\frac{1}{9}$

Question:23

If the lines 2x – 3y = 5 and 3x – 4y = 7 are the diameters of a circle of area 154 square units, then obtain the equation of the circle.

Answer:

Given that the lines 2x-3y-5=0 and 3x-4y-7=0 are diameters of the circle.
2x-3y-5=0.....(i)
3x-4y-7=0....(ii)
3 (Equation (i)) - 2 (Equation (ii) )
6x-9y-15-6x+8y+14=0.
y = -1
Put the value of y in equation (i)
x = 1
Solving these lines, we will get the intersection as (1,-1), which is the centre of the circle.
Also, given that the area of the circle is 154 sq units
πr²=154
r²=154*7/22=49
r=7
(x-1)²+(y+1)²=49
x²+y²-2x+2y-47=0

Question:24

Find the equation of the circle which passes through the points (2, 3) and (4, 5) and the centre lies on the straight line y – 4x + 3 = 0.

Answer:

Let the centre of the circle be C(h,k)
Given that the centre lies on the line y-4x+3=0
y-4x+3=0
k-4h+3=0
k=4h-3
AC²=BC²
(h-2)²+(4h-3-3)²=(h-4)²+(4h-3-5)²
h²-4h+4+16h²-48h+36=h²-8h+16+16h²-64h+64
20h=40
h=2
k=4h-3=5
radius=$\sqrt{{(2-2)}^2+{(3-5)}^2}=2$
Therefore, the equation of the circle is : (x-2)²+(y-5)²=4
x²+y²-4x-10y+25=0

Question:25

Find the equation of a circle whose centre is (3, –1) and which cuts off a chord of length 6 units on the line 2x – 5y + 18 = 0.

Answer:

Given centre of the circle O(3, -1)
The chord of the circle is AB
Given that the equation of AB is 2x-5y+18=0
Perpendicular distance from O to AB is OP= $\left|\frac{2(3)-5(-1)+18}{\sqrt{4+25}}\right|=\frac{29}{\sqrt{29}}=\sqrt{29}$
OB²=OP²+PB²
OB²=29+9=38
OB=$\sqrt{38}$
Equation of the circle is (x-3)²+(y+1)²=38
x²+y²-6x+2y=28

Question:26

Find the equation of a circle of radius 5 which is touching another circle x² + y² – 2x – 4y – 20 = 0 at (5, 5).

Answer:

The given circle is x²+y²-2x-4y=20
(x-1)²+(y-2)²=5²
Centre of this circle is C₁(1,2).
Now, the required circle of radius 5 touches the above circle at P(5,5).
Let the centre of the required circle be C₂(h,k).
Since the radius of the given circle and the required circle are the same, point P is the midpoint of C₁C₂.
$5=\frac{1+h}{2}$ and $5=\frac{2+k}{2}$
h=9 and k=8
So the equation of the required circle is (x-9)²+(y-8)²=25
x²+y²-18x-16y+120=0

Question:27

Find the equation of a circle passing through the point (7, 3), having radius 3 units and whose centre lies on the line y = x – 1.

Answer:

Given that the circle passes through the point A(7,3) and its radius is 3
Therefore, the centre of the circle is C(h,h-1)
Now, the radius of the circle is AC=3
(h-7)²+(h-1-3)²=3²
2h²-22h+56=0
h²-11h+28=0
(h-4)(h-7)=0
h=4 or 7
If the centre of the circle is C(4,3)⇒(x-4)²+(y-3)²=9⇒x²+y²-8x-6y+16=0
If the centre is C(7,6)⇒(x-7)²+(y-6)²=9⇒x²+y²-14x-12y+76=0

Question:28

Find the equation of each of the following parabolas
(a) Directrix x = 0, focus at (6, 0)
(b) Vertex at (0, 4), focus at (0, 2)
(c) Focus at (–1, –2), directrix x – 2y + 3 = 0

Answer:

We know that the distance of any point on the parabola from its focus and its directrix is the same.
i) Given that the directrix x=0 and the focus (6,0)
So, for any point P(x,y) on the parabola
Distance of P from directrix=Distance of P from focus x²=(x-6)²+y²
y²-12x+36=0
ii) Given that vertex=(0,4) and focus (0,2)
Now the distance between the vertex and the directrix is the same as the distance between the vertex and the focus.
Distance of P from directrix = Distance of P from focus
$|y-6|=\sqrt{{(x-0)}^2+{(y-2)}^2}$
y²-12y+36=x²+y²-4y+4
x²=32-8y
iii) Given that the focus is at (-1,-2)
and directrix x-2y+3=0
$\sqrt{{(x+1)}^2+{(y+2)}^2}=\left|\frac{x-2y+3}{\sqrt{1+4}}\right|$
x²+2x+1+y²+4y+4=1/5[x²+4y²+9+6x-4xy-12y]
4x²+4xy+y²+4x+32y+16=0

Question:29

Find the equation of the set of all points the sum of whose distances from the points (3, 0) and (9, 0) is 12.

Answer:

Let the coordinates of the variable point be (x,y)
Then according to the question
$\sqrt{{(x-3)}^2+y^2}=12-\sqrt{{(x-9)}^2+y^2}$
x²-6x+9+y²=$144+x^2-18x+81+y^2-24\sqrt{{(x-9)}^2+y^2}$
$(x-18)=-2\sqrt{{(x-9)}^2+y^2}$
Again, squaring both sides, we get
x²-36x+324=4(x²-18x+81+y²)
3x²+4y²-36x=0, which is an ellipse.

Question:30

Find the equation of the set of all points whose distance from (0, 4) is 2/3 of their distance from the line y = 9.

Answer:

Let the point be P (x,y)
According to the question
$\sqrt{x^2+{(y-4)}^2}=\frac{2}{3}\left|\frac{y-9}{1}\right|$
9(x²+y²-8y+16)=4(y²-18y+81)
9x²+5y²=180

Question:31

Show that the set of all points such that the difference of their distances from (4, 0) and (– 4, 0) is always equal to 2 represents a hyperbola.

Answer:

Let the point be P (x,y)
According to the question
Distance of P from (-4,0) -Distance of P from (4,0)=2
$\sqrt{{(x+4)}^2+y^2}=2+\sqrt{{(x-4)}^2+y^2}$
Squaring both sides
$x^2+8x+16+y^2=4+x^2-8x+16+y^2+4\sqrt{{(x-4)}^2+y^2}$
$4x-1=\sqrt{{(x-4)}^2+y^2}$
16x²-8x+1=x²+16-8x+y²
15x²-y²=15
Which is an equation of a hyperbola.

Question:32

Find the equation of the hyperbola with

(i) Vertices (± 5, 0), foci (± 7, 0)
(ii) Vertices (0, ± 7), e = $\frac{4}{3}$
(iii) Foci $(0,\pm \sqrt{10})$, passing through (2, 3)

Answer:

a) Given that a=5 and ae = 7
e=7/5
b²=a²⁽e²-1)=25(49/25-1)=24
So, the equation of the hyperbola is
$\frac{x^2}{25}-\frac{y^2}{24}=1$
b) b=7,e=4/3
$a^2=b^2(e^2-1)=\frac{343}{9}$
$\frac{x^2}{\frac{343}{9}}-\frac{y^2}{49}=-1$
$9x^2-7y^2+343=0$
c) Given that foci=(0,±10 ) $be=\sqrt{10}$
a²=b²(e²-1)
a²=b²e²-b²
a²=10-b²

$\frac{x^2}{10-b^2}-\frac{y^2}{b^2}=-1$

Since a hyperbola passes through the point (2,3)
4b²-9(10-b²)=-b²(10-b²)
b⁴-23b²+90=0
b⁴-18b²-5b²+90=0
(b²-18)(b²-5)=0
b²=5
a²+b²=10⇒a²=5
$\frac{x^2}{5}-\frac{y^2}{5}=-1$
y²-x²=5

Question:33

The line x + 3y = 0 is athediameter of the circle x² + y² + 6x + 2y = 0.

Answer:

False
Given equation of the circle is x²+y²+6x+2y=0
Centre=(-3,-1)
It does not lie on the line x+3y=0 as-3+3(-1)=-6
So this line is not the diameter of the circle

Question:34

The shortest distance from the point (2, –7) to the circle x² + y² – 14x – 10y – 151 = 0 is equal to 5.

False

$\begin{array}{rl}& \text{ Centre }\equiv C(7,5)\\ & \text{ Radius }=\sqrt{49+25+151}=\sqrt{225}=15\end{array}$
Distance between the point $P(2,-7)$ and centre

$=\sqrt{(2-7{)}^2+(-7-5{)}^2}=\sqrt{169}=13$
The shortest distance of point P from the circle $=|13-15|=2$

Question:35

If the line lx + my = 1 is tangent to the circle x² + y² = a², then the point (l, m) lies on a circle.

Answer:

True
Given circle is x²+y²=a²
The given line lx+my-1=0 is tangent to the circle
Distance of (0,0) from the line lx+my-1=0 is equal to the radius a
$\left|\frac{0+0+1}{\sqrt{l^{2+m^2}}}\right|=a$
$l^2+m^2=\frac{1}{a^2}$
The locus of (l,m) is $x^2+y^2=\frac{1}{a^2}$. which is an equation of a circle.

Question:36
The point (1, 2) lies inside the circle x² + y² – 2x + 6y + 1 = 0.

Answer:

False
Given circle is x²+y²-2x+6y+1=0 or (x-1)²+(y+3)²=3²
Centre is C(1,-3) and the radius is 3.
The distance of point P (1,2) from the centre is 5.
Thus CP>radius
So, point P lies outside the circle.

Question:37

The line lx + my + n = 0 will touch the parabola y² = 4ax if ln = am².

Answer:

True
Give line lx+my+n=0
and parabola y²=4ax
Solving the line and parabola for their point of intersection, we get
$l\left(\frac{y^2}{4a}\right)+my+n=0$
Since the line touches the parabola, the above equation must have equal roots
D=0
$m^2-4\ast \left(\frac{1}{4a}\right)\ast n=0$
am²=nl

Question:38

If P is a point on the ellipse $\frac{x^2}{16}+\frac{x^2}{25}=1$ whose foci are S and S′, then PS + PS′ = 8.
Answer:

False
We have definition of the ellipse is $\frac{x^2}{16}+\frac{y^2}{25}=1$
From the definition of the ellipse, we know that the sum of the distances of any point P
on the ellipse from the two foci is equal to the length of the major axis.
Here major axis=2b=2*5=10
S and S'are foci, then SP+S'P=10

Question:39

The line 2x + 3y = 12 touches the ellipse$\frac{x^2}{9}+\frac{y^2}{4}=2$ at the point (3, 2).

Answer:

True
Given line is 2x+3y=12
And the ellipse 4x²+9y²=72
Solving the line and the ellipse, we get (12-3y)²+9y²=72
(4-y)²+y²=8
2y²-8y+8=0
y²-4y+4=0
(y-2)²=0
y=2;x=3

Question:40

The locus of the point of intersection of lines $\sqrt{3}x-y-4\sqrt{3}k=0$ and $\sqrt{3}kx+ky-4\sqrt{3}=0$ for different values of k is a hyperbola whose eccentricity is 2.

Answer:

True
Given equation of lines are $\sqrt{3}x-y-4\sqrt{3}k=0$
$\sqrt{3}kx+ky-4\sqrt{3}=0$
$k=\frac{\sqrt{3}x-y}{4\sqrt{3}}$ and$k=\frac{4\sqrt{3}}{\sqrt{3}x+y}$
$\frac{\sqrt{3}x-y}{4\sqrt{3}}$ $=\frac{4\sqrt{3}}{\sqrt{3}x+y}$
3x²-y²=48
$\frac{x^2}{16}-\frac{y^2}{48}=1$ which is the equation of hyperbola
a²=16 and b²=48
e²=1+48/16=4
e=2

Question:41

The equation of the circle having a centre at (3, – 4) and touching the line 5x + 12y – 12 = 0 is ______.

Answer:

The perpendicular distance from centre (3,-4) to the given line is,$r=\left|\frac{5(3)+12(-4)-12}{\sqrt{25-144}}\right|$
=$\frac{45}{13}$ which is radius of the circle
Required equation of the circle is ${(x-3)}^2+{(y+4)}^2={\left(\frac{45}{13}\right)}^2$

Question:42

The equation of the circle circumscribing the triangle whose sides are the lines y = x + 2, 3y = 4x, and 2y = 3x is __________ .

Answer:

Given equation of the line is y=x+2
3y=4x
2y=3x
Solving these lines, we get points of intersection A(6,8), B (4,6) and C(0,0).
Let the equation of the circle circumscribing the given triangle be
x²+y²+2gx+2fy+c=0
36+64+12g+16f+c=0⇒12g+16f+c=-100
16+36+8g+12f+c=0⇒8g+12f+c=-52
0+0+0+0+c=0⇒c=0
3g+4f=-25
2g+3f=-13
On solving, we get g=-23 and f=11
The equation of a circle is x²+y²-46x+22y=0

Question:43

An ellipse is described by using an endless string which is passed over two pins. If the axes are 6 cm and 4 cm, the length of the string and distance between the pins are ____________.

Answer:

Let equation of ellipse be $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
a=3 and b=2
b²=a²(1-e²)
e²=1-4/9=5/9
$e=\frac{\sqrt{5}}{3}$
From the definition of the ellipse, for any point P on the ellipse, we have
SP+S'P=2a
Length of endless string=SP+S'P+SS'=2a+2ae=$2(3)+2(3)\ast \frac{\sqrt{5}}{3}=6+2\sqrt{5}$

Question:44

The equation of the ellipse having foci (0, 1), (0, –1) and minor axis of length 1 is_________

Answer:

be=1
Length of minor axis 2a=1
a=1/2
a²=b²(1-e²)
1/4=b²-b²e²
b²=1/4+1=5/4
$\frac{x^2}{\frac{1}{4}}+\frac{y^2}{\frac{5}{4}}=1$
So, the equation of ellipse is $4x^{+}\frac{4y^2}{5}=1$

Question:45

The equation of the parabola having focus at (–1, –2) and the directrix x – 2y + 3 = 0 is ____________ .

Answer:

Let any point on parabola be P(x,y)
Length of the perpendicular from S to the directrix
$\frac{{(x-2y+3)}^2}{5}={(x+1)}^2+{(y+2)}^2$
4x²+y²+4x+32y+16=0

Question:46

The equation of the hyperbola with vertices at (0, ±6) and eccentricity 5/3 is ________ and its foci are __________ .

Answer:

Let the equation of parabola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=-1$
b=6
e=5/3
a²=b²(e²-1)
a²=64
$\frac{x^2}{64}-\frac{y^2}{36}=-1$
$\text{ foci }=(0,\pm be)\equiv (0,\pm \frac{5}{3}\times 6)=(0,\pm 10)$

Question:47

The area of the circle centred at (1, 2) and passing through (4, 6) is
A. 5π
B. 10π
C. 25π
D. none of these

Answer:

Centre of the circle is C(1,2)
Radius=$\sqrt{{(4-1)}^2+{(6-2)}^2}=5$
Area of circle= π(5)²=25π square units

Question:48

The equation of a circle which passes through (3, 6) and touches the axes is
A. x² + y² + 6x + 6y + 3 = 0
B. x² + y² – 6x – 6y – 9 = 0
C. x² + y² – 6x – 6y + 9 = 0
D. none of these

Answer:

Given that the circle touches both axes
Therefore, the equation of the circle is (x-a)²+(y-a) 2 = a2
circle passes through (3,6)
(3-a)²+(6-a)²=a²
a²-18a+45=0
(a-3)(a-15)=0
a = 3 or a =15
For an =3 equation of a circle
(x-3)²+(y-3)²=9
x²+y²-6x-6y+9=0

Question:49

The equation of the circle with the centre on the y-axis and passing through the origin and the point (2, 3) is
A. x² + y² + 13y = 0
B. 3x² + 3y² + 13x + 3 = 0
C. 6x² + 6y² – 13x = 0
D. x² + y² +13x +3 = 0

Solution:

The centre of the circle lies on the y-axis
So, let the centre be C(0,k)
Circle passes through O(0,0)and A(2,3)
OC²=AC²
k²=(2-0)²+(3-k)²
k=13/6
(x-0)²+(y-13/6)²=(13/6)²
3x²+3y²-13y=0
Answer- None

Question:50

The equation of a circle with origin as centre and passing through the vertices of an equilateral triangle whose median is of length 3a is
A. x² + y² = 9a²
B. x² + y² = 16a²
C.x² + y² = 4a²
D. x² + y² = a²
Answer:

Option (C)
median length=3a
Radius of circle=2/3*median length=2a
Equation of a circle
x²+y²=4a²

Question:51

If the focus of a parabola is (0, –3) and its directrix is y = 3, then its equation is
A. x² = –12y
B. x² = 12y
C. y² = –12x
D. y² = 12x

Answer:

a Given that, focus of parabola is at S(0,-3)and equation of directrix is y=3
For any point P(x,y)on the parabola, we have
SP=PM
$\sqrt{(x-0{)}^2+(y+3{)}^2}=|y-3|$
x²+y²+6y+9=y²-6y+9
x²=-12y

Question:52

If the parabola y² = 4ax passes through the point (3, 2), then the length of its latus rectum is
A. 2/3
B. 4/3
C. 1/3
D. 4
Answer:

Option (b) is correct.
Parabola y²=4ax passes through the point (3,2)
4=4a(3)
a=1/3
Length of latus rectum 4a=4/3

Question:53

If the vertex of the parabola is the point (–3, 0) and the directrix is the line x + 5 = 0, then its equation is
A. y² = 8 (x + 3)
B. x² = 8 (y + 3)
C. y² = – 8 (x + 3)
D. y² = 8 ( x + 5)

Answer:

Given: Vertex =(–3, 0) and the directrix x = -5(M)
So focus S (-1,0)
For any point of the parabola P (x,y), we have
SP=PM
$\sqrt{(x+1{)}^2+y^2}=|x+5|$
x²+y²+2x+1=x²+10x+25
y²=8x+24

Question:54

The equation of the ellipse whose focus is (1, –1), the directrix the line x – y – 3 = 0 and eccentricity 1/2 is
A. 7x² + 2xy + 7y² – 10x + 10y + 7 = 0
B. 7x² + 2xy + 7y² + 7 = 0
C. 7x² + 2xy + 7y² + 10x – 10y – 7 = 0D. none

Answer:

Option (a) is correct.
Given that focus of the ellipse is S(1,-1)and the equation of directrix is x-y-3=0
Also, e=1/2
From definition of ellipse, for any point P (x,y) on the ellipse, we have SP=ePM, where
M is the root of the perpendicular from point P to the directrix.
$\sqrt{(x-1{)}^2+(y+1{)}^2}=\frac{1}{2}\frac{|x-y-3|}{\sqrt{2}}$
7x²+7y²+2xy-10x+10y+7=0

Question:55

The length of the latus rectum of the ellipse 3x² + y² = 12 is
A. 4
B. 3
C. 8
D. 12
Answer:

Option (d) is correct.
Given ellipse is 3x²+y²=12
$\frac{x^2}{4}+\frac{y^2}{12}=1$
a²=4
a=2
b²=12

b=$2\sqrt{3}$
length of latus rectum=$\frac{2a^2}{b}=\frac{4}{\sqrt{3}}$

Question:56

If e is the eccentricity of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a<b)$, then
A. b² = a² (1 – e²)
B. a² = b² (1 – e²)
C. a² = b² (e² – 1)
D. b² = a² (e² – 1)

Answer:

Option (b) is correct.
Given that $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$
We know that a²=b²(1-e²)

Question:57

The eccentricity of the hyperbola whose latus rectum is 8 and whose conjugate axis is equal to half of the distance between the foci is
A. 4/3
B. $\frac{4}{\sqrt{3}}$
C. $\frac{2}{\sqrt{3}}$
D. none of these

Answer:

Let the equation of the hyperbola be x²/a²-y²/b²=1
length of latus rectum=8
2b²/a=8
b²=4a
Conjugate axis = half of the distance between the foci
2b=ae
b²=a²(e²-1)
$\frac{a^2{e}^2}{4}$=a²(e²-1)
e²=4/3
e=$\frac{2}{\sqrt{3}}$

Question:58

The distance between the foci of a hyperbola is 16 and its eccentricity is $\sqrt{2}$. Its equation is
A. x² – y² = 32
B. $\frac{x^2}{4}-\frac{y^2}{9}=1$
C. 2x – 3y² = 7
D. none of these
Answer:

Option (A) is correct.
Let the equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
Given that Foci=(±ae,0)
Distance between foci=2ae=16
$2\ast a\ast \sqrt{2}=16$
$a=4\sqrt{2}$
$b^2=a^2(e^2-1)=(4\sqrt{2}{)}^2((\sqrt{2}{)}^2-1)$
=32(2-1)=32
Hence, equation is $\frac{x^2}{32}-\frac{y^2}{32}=1$

Question:59

The equation of the hyperbola with eccentricity 3/2 and foci at (± 2, 0) is
A. $\frac{x^2}{4}-\frac{y^2}{5}=\frac{4}{9}$
B. $\frac{x^2}{9}-\frac{y^2}{9}=\frac{4}{9}$
C. $\frac{x^2}{4}-\frac{y^2}{9}=1$
D. None of these

Answer:

Option (A) is correct.
Let the equation of the hyperbola be $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
e=3/2
Foci=(±ae),0=(±2,0)
So, after comparing the equations, ae=2
a*3/2=2
a=4/3
b²=a²(e²-1)
b²=(4/3)²((3/2)²-1)
=(16/9)(9/4-1)
=16/9*5/4=20/9
Equation of hyperbola is $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$
=$\frac{x^2}{{\left(\frac{4}{3}\right)}^2}-\frac{y^2}{\frac{20}{9}}=1$
$\frac{x^2}{\frac{16}{9}}-\frac{y^2}{\frac{20}{9}}=1$
Hence, $\frac{x^2}{16}-\frac{y^2}{20}=\frac{1}{9}$ is the required equation.