Conic Sections Test

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Conic Sections Test - 66

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Question 1
1 / -0

Determine the equation for the ellipse if center is at (0, 0), the major axis on the y-axis and ellipse passes through the points (4, 3) and (2, 5).

A
\(4 y ^{2}+3 x ^{2}=1\)

B
\(3 y ^{2}+4 x ^{2}=1\)

C
\(4 y ^{2}+3 x ^{2}=91\)

D
\(3 y ^{2}+4 x ^{2}=91\)

Solution

As we know,
The standard equation of an ellipse,
\(\frac{ x ^{2}}{ a ^{2}}+\frac{ y ^{2}}{ b ^{2}}=1\)
Where \(2 a\) and \(2 b\) are the length of the major axis and minor axis respectively and center \((0,0)\).
Given,
Center is \((0,0)\).
So standard equation \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Since \((4,3)\) lies on ellipse satisfy the equation of ellipse
\(\frac{16}{ a ^{2}}+\frac{9}{ b ^{2}}=1\)...(i)
Similarly for \((2,5)\)
\(\frac{4}{a^{2}}+\frac{25}{b^{2}}=1\)...(ii)
Substracting the equations (i) from \(4 \times\) (ii), we get
\(\frac{4 \times 25}{ b ^{2}}-\frac{9}{ b ^{2}}=4-1\)
\(\Rightarrow \frac{91}{ b ^{2}}=3 \)
\(\Rightarrow b ^{2}=\frac{ 9 1 }{ 3 }\)
Putting the value of \(b^2\) in equation (i), we get
\(\frac{16}{a^{2}}+\frac{9 \times 3}{91}=1 \)
\(\Rightarrow \frac{16}{a^{2}}=1-\frac{27}{91} \)
\(\Rightarrow \frac{16}{a^{2}}=\frac{64}{91} \)
\(\Rightarrow a ^{2}=\frac{91}{4}\)
Equation of ellipse is:
\(\frac{4 x^{2}}{91}+\frac{3 y^{2}}{91}=1 \)
\(\Rightarrow 4 x^{2}+3 y^{2}=91\)
Question 2
1 / -0

Find the length of the major axis of the ellipse \(\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\).

A
6

B
8

C
10

D
None of these

Solution

As we know,
The properties of a horizontal ellipse \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) where \(0< b < a\) are as follows:
Centre of ellipse is \((0,0)\).
Vertices of ellipse are \((-a, 0)\) and \((a, 0)\).
Foci of ellipse are \((-a e, 0)\) and \((a e, 0)\).
Length of major axis is \(2 a\).
Length of minor axis is \(2 b\).
Eccentricity of ellipse is given by,
\(e=\frac{\sqrt{a^{2}-b^{2}}}{a}\)
Given,
Equation of ellipse \(=\frac{x^{2}}{16}+\frac{y^{2}}{9}=1\)
As we can see that, the given ellipse is an horizontal ellipse.
By comparing the given equation of ellipse with \(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\) where \(0< b < a\), we get
\( a =4\) and \(b =3\)
As we know that, the length of the major axis is given by \(2 a\).
So, the length of the major axis for the given ellipse \(=2 \times 4=8\) units
Question 3
1 / -0

The length of latus rectum of the hyperbola \(\frac{x^{2}}{100}-\frac{y^{2}}{75}=1\) is:

A
10

B
12

C
14

D
15

Solution

As we know,
Standard equation of an hyperbola \(=\frac{ x ^{2}}{ a ^{2}}-\frac{ y ^{2}}{ b ^{2}}=1\)
Coordinates of foci \(=(\pm a e, 0)\)
Eccentricity \((e)=\sqrt{1+\frac{b^{2}}{a^{2}}}= a^{2} e^{2}=a^{2}+b^{2}\)
Length of Latus rectum \(=\frac{2 b ^{2}}{ a }\)
Given,
Equation of hyperbola is \(\frac{x^{2}}{100}-\frac{y^{2}}{75}=1\).
Compare with the standard equation of a hyperbola, we get
\(a^{2}=100\) and \(b^{2}=75\)
\(\therefore a =10\)
Length of latus rectum \(=\frac{2 b ^{2}}{ a }=\frac{2 \times 75}{10}=15\)
Question 4
1 / -0

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

A
\(\frac{1}{2}\)

B
\(\frac{1}{\sqrt{2}}\)

C
\(\frac{\sqrt{3}}{2}\)

D
\(\frac{\sqrt{3}}{4}\)

Solution

As we know,
In an ellipse \(\frac{ x ^{2}}{ a ^{2}}+\frac{ y ^{2}}{ b ^{2}}=1, a >b\)
The length of the latus rectum \(=\frac{2 b ^{2}}{ a }\)
Its eccentricity is given by,
\(e=\sqrt{1-\frac{b^{2}}{a^{2}}}\)
Let the equation of the ellipse be \(\frac{ x ^{2}}{ a ^{2}}+\frac{ y ^{2}}{ b ^{2}}=1\) where \(a >b\).
According to the question,
Length of the latus rectum \(=\) Half of its minor axis
\(\frac{2 b ^{2}}{ a }=b\)
\(\Rightarrow \frac{2 b ^{2}}{b}=a\)
\(\therefore a =2 b\)
Now, the eccentricity of the ellipse \((e)=\sqrt{1-\frac{b^{2}}{a^{2}}}\)
\(=\sqrt{1-\frac{b^{2}}{ (2b)^{2}}}\)
\(=\sqrt{1-\frac{b^{2}}{ 4b^{2}}}\)
\(=\sqrt{1-\frac{1}{4}}\)
\(=\sqrt{\frac{3}{4}}\)
\(=\frac{\sqrt{3}}{2}\)
Question 5
1 / -0

Find the foci of the ellipse \(4 x^{2}+9 y^{2}+16 x+18 y-11=0\).

A
\((-3 \pm \sqrt{5},-2)\)

B
\((-2 \pm \sqrt{5},-1)\)

C
\((-2 \pm \sqrt{5}, 1)\)

D
\((2 \pm \sqrt{5},-1)\)

Solution

As we know,
The properties of a horizontal ellipse \(\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\) where \(0< b< a\) are as follows:
Centre of ellipse is \(( h , k )\).
Vertices of ellipse are: \(( h - a , k )\) and \(( h + a , k )\)
Foci of ellipse are: \(( h - ae , k )\) and \(( h + ae , k )\)
Eccentricity of ellipse is given by,
\(e=\frac{\sqrt{a^{2}-b^{2}}}{a}\)
Given,
The equation of ellipse is \(4 x^{2}+9 y^{2}+16 x+18 y-11=0\).
The given equation can be re-written as \(\frac{(x+2)^{2}}{9}+\frac{(y+1)^{2}}{4}=1\)...(1)
As we can see that, the given ellipse is a horizontal ellipse.
So, by comparing equation (1) with respect to \(\frac{(x-h)^{2}}{a^{2}}+\frac{(y-k)^{2}}{b^{2}}=1\), we get
\( h =-2, k =-1, a =3\) and \(b =2\)
As we know that,
Eccentricity of an ellipse is given by,
\(e=\frac{\sqrt{a^{2}-b^{2}}}{a}\)
\(\Rightarrow e=\frac{\sqrt{3^{2}-2^{2}}}{3}=\frac{\sqrt{5}}{3}\)
\(\Rightarrow 3e =\sqrt{5}\)
\(\Rightarrow a \cdot e =\sqrt{5}\)
As we know,
Foci of a horizontal ellipse are given by \(( h \pm ae, k )\).
So, the required foci of the given ellipse are \((-2 \pm \sqrt{5},-1)\).
Question 6
1 / -0

Find the length of latus rectum of hyperbola \(25 y^{2}-24 x^{2}=600\).

A
\(\frac{25}{\sqrt{6}}\) units

B
\(\frac{5}{\sqrt{6}}\) units

C
\(\frac{48}{5}\) units

D
\(\frac{5}{3}\) units

Solution

As we know,
If Equation of hyperbola is \(\frac{ x ^{2}}{ a ^{2}}-\frac{ y ^{2}}{ b ^{2}}=1\), then
Length of latus rectum \(=\frac{2 b^{2}}{a}\)
If Equation of hyperbola is \(-\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\), then
Length of latus rectum \(=\frac{2 a^{2}}{b}\)
Given,
Equation is \(25 y^{2}-24 x^{2}=600\)
The given equation of hyperbola can be re-written as,
\(-\frac{x^{2}}{25}+\frac{y^{2}}{24}=1\)
On comparing with standard equation, we get
\(a=5\) and \(b =2 \sqrt{6}\)
So, Length of latus rectum \(=\frac{2 a^{2}}{b}=\frac{2 \times 25}{2 \sqrt{6}} \)
\(=\frac{25}{\sqrt{6}}\) units
Question 7
1 / -0

Find the equation of the directrix of the parabola x² = 64y.

A
y = 16

B
x = 16

C
y = - 16

D
x = -16

Solution

Given,
\(x^{2}=64 y\)
The above equation can be written as,
\(x^{2}=4 \cdot 16 \cdot y\)
By comparing the above equation with the standard equation of the parabola (\(x^{2}=4 ay\)), we get
\( a=16\)
As we know that,
The equation of directrix of the parabola \(x^{2}=4 a y\) is given by,
\(y=-a\)
\(\Rightarrow y=-16\)
Question 8
1 / -0

The equation of the ellipse whose vertices are at \((\pm 5,0)\) and foci at \((\pm 4,0)\) is:

A
\(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)

B
\(\frac{x^{2}}{9}+\frac{y^{2}}{25}=1\)

C
\(\frac{x^{2}}{16}+\frac{y^{2}}{25}=1\)

D
\(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)

Solution

As we know,
Equation of ellipse \(=\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Eccentricity \(( e )=\sqrt{1-\frac{ b ^{2}}{ a ^{2}}}\)
Where, vertices \(=(\pm a, 0)\) and focus \(=(\pm a e, 0)\)
Given,
Vertices of ellipse \((\pm 5,0)\) and foci \((\pm 4,0)\)
So, \(a=\pm 5\)
\( \Rightarrow a^{2}=25\)
\(a e=4\)
\( \Rightarrow e=\frac{4 }{ a}\)
\( \Rightarrow e=\frac{4 }{ 5}\)
Now, \(\frac{4 }{ 5}=\sqrt{1-\frac{ b ^{2}}{5^{2}}}\)
\(\Rightarrow \frac{16}{25}=\frac{25- b ^{2}}{25}\)
\(\Rightarrow 16=25- b ^{2}\)
\(\Rightarrow b ^{2}=9\)
\(\therefore\) Equation of ellipse is \(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\).
Question 9
1 / -0

The curve represented by the equations
\(x=3(\cos t+\sin t)\)
\(y=4(\cos t-\sin t)\) is:

A
A straight line

B
A circle

C
A hyperbola

D
An ellipse

Solution

Equation of an ellipse is:
\(\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1\)
Given, \(x=3(\cos t+\sin t) \)
\(y=4(\cos t-\sin t) \)
\(\left(\frac{x}{3}\right)^{2}=1+\sin 2 t \ldots(1)\)
\(\left(\frac{y}{4}\right)^{2}=1-\sin 2 t \ldots(2)\)
Adding equation (1) and (2), we get
\(\frac{x^{2}}{9}+\frac{y^{2}}{16}=2\)
Thus, the given curve represents an ellipse.
Question 10
1 / -0

The equation of parabola with the focus (2, 0) and directrix as x + 2 = 0 is:

A
\(x^{2}=-8 y\)

B
\(x^{2}=8 y\)

C
\(y^{2}=-8 x\)

D
\(y^{2}=8 x\)

Solution

Let a point \((x, y)\) lie on the parabola, then its distance from focus is always equal to its distance from directrix.
i.e., Distance between \((x, y)\) and focus \((2,0)=\) Distance between \((x, y)\) and directrix i.e., \(x+2=0\)
\(\therefore \sqrt{( x -2)^{2}+( y -0)^{2}}=\frac{ x +2}{\sqrt{1^{2}+0^{2}}} \)
\(\Rightarrow \sqrt{( x -2)^{2}+ y ^{2}}=\frac{ x +2}{\sqrt{1}}\)
By squaring both sides we get,
\(( x -2)^{2}+ y ^{2}=\frac{( x +2)^{2}}{1} \)
\(\Rightarrow( x -2)^{2}+ y ^{2}=( x +2)^{2} \)
\(\Rightarrow x ^{2}-4 x +4+ y ^{2}= x ^{2}+4 x +4 \)
\(\Rightarrow-4 x + y ^{2}=4 x \)
\(\Rightarrow y ^{2}=8 x\)

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Re-Attempt Weekly Quiz Competition

Conic Sections Test - 66

Result

Conic Sections Test - 66

Score

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Rank

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SHARING IS CARING

Question 1

\(4 y ^{2}+3 x ^{2}=1\)

\(3 y ^{2}+4 x ^{2}=1\)

\(4 y ^{2}+3 x ^{2}=91\)

\(3 y ^{2}+4 x ^{2}=91\)

Solution

Question 2

6

8

10

None of these

Solution

Question 3

10

12

14

15

Solution

Question 4

\(\frac{1}{2}\)

\(\frac{1}{\sqrt{2}}\)

\(\frac{\sqrt{3}}{2}\)

\(\frac{\sqrt{3}}{4}\)

Solution

Question 5

\((-3 \pm \sqrt{5},-2)\)

\((-2 \pm \sqrt{5},-1)\)

\((-2 \pm \sqrt{5}, 1)\)

\((2 \pm \sqrt{5},-1)\)

Solution

Question 6

\(\frac{25}{\sqrt{6}}\) units

\(\frac{5}{\sqrt{6}}\) units

\(\frac{48}{5}\) units

\(\frac{5}{3}\) units

Solution

Question 7

y = 16

x = 16

y = - 16

x = -16

Solution

Question 8

\(\frac{x^{2}}{25}+\frac{y^{2}}{9}=1\)

\(\frac{x^{2}}{9}+\frac{y^{2}}{25}=1\)

\(\frac{x^{2}}{16}+\frac{y^{2}}{25}=1\)

\(\frac{x^{2}}{25}+\frac{y^{2}}{16}=1\)

Solution

Question 9

A straight line

A circle

A hyperbola

An ellipse

Solution

Question 10

\(x^{2}=-8 y\)

\(x^{2}=8 y\)

\(y^{2}=-8 x\)

\(y^{2}=8 x\)

Solution

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