120697
The vertices of the hyperbola are at \((-5,-3)\) and \((-5,-1)\) and the extremities of the conjugate axis are at \((-7,-2)\) and \((-3,-2)\), then the equation of the hyperbola is
1 \(\frac{(y+2)^2}{1}-\frac{(x+5)^2}{4}=1\)
2 \(\frac{(y+2)^2}{4}-\frac{(x+5)^2}{4}=1\)
3 \(\frac{(x+5)^2}{4}-\frac{(y+2)^2}{1}=1\)
4 \(\frac{(x-5)^2}{4}-\frac{(y-2)^2}{1}=1\)
Explanation:
B Given that, the vertices of the hyperbola are \((-5,3),(-5,-1)\) The Extremities of the conjugate axis are at \((-7,-2),(-3,-2)\)
We know that, the distance between the extremities of conjugate axis are \(2 \mathrm{~b}\).
So,
\(2 b =\sqrt{(-7+3)^2+(-2+2)^2}=\sqrt{4^2}=4\)
\(2 b =4 \Rightarrow b=2\)
Given vertices \((-5,-3),(-5,-1)\)
Let's Centre (h, k)
Therefore \((\mathrm{h}, \mathrm{k})=\left(\frac{-5-5}{2}, \frac{-3-1}{2}\right)=(-5,-2)\)
The equation of the hyperbola is given by
\(\frac{(\mathrm{y}-\mathrm{k})^2}{\mathrm{a}^2}-\frac{(\mathrm{x}-\mathrm{h})^2}{\mathrm{~b}^2}=1\)
Substituting the values, we get -
\(\frac{(y+2)^2}{1^2}-\frac{(x+5)^2}{2^2}=1\)
\(\frac{(y+2)^2}{1}-\frac{(x+5)^2}{4}=1\)
JCECE-2017
Hyperbola
120698
The point of contact of the line \(y=x-1\) with \(3 x^2-4 y^2=12\)
1 \((4,3)\)
2 \((3,4)\)
3 \((4,-3)\)
4 None of these
Explanation:
A Given that,
\(y=x-1\)
\(3 x^2-4 y^2=12\)
Putting the value of \(y=x-1\) from equation (i) in equation (ii), we get -
\(3 x^2-4(x-1)^2=12\)
\(3 x^2-4\left(x^2-2 x+1\right)=12\)
\(x^2-8 x+16=0 \Rightarrow x=4\)
Or
Putting the value of \(x\) in equation (i), we get -
\(y=3\)Hence, required point of contact is \((4,3)\).
JCECE-2014
Hyperbola
120699
The eccentricity of the hyperbola, whose length of latus rectum is 8 and conjugate axis is equal to the half of its distance between the foci, is
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{3}}\)
3 \(\frac{2}{\sqrt{3}}\)
4 1
Explanation:
C Given that,
Length of latus rectum \(\frac{2 b^2}{a}=8\) and conjugate axis (2b) \(=\frac{1}{2}(2 \mathrm{ae})\)
\(\therefore \quad \frac{2}{\mathrm{a}}\left(\frac{\mathrm{ae}}{2}\right)^2=8\)
\(\mathrm{ae}^2=16\)
We have, \(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=8\)
\(\mathrm{b}^2=4 \mathrm{a}\)
\(\mathrm{a}^2\left(\mathrm{e}^2-1\right)=4 \mathrm{a}\)
\(\mathrm{e}^2-\mathrm{a}=4\)
\(16-\mathrm{a}=4\)
\(\mathrm{a}=12\)
Substitute \(\mathrm{a}=12\) in equation (i), we get -
\(12 \mathrm{e}^2=16\)
\(\mathrm{e}^2=\frac{4}{3}=\frac{2}{\sqrt{3}}\)
BCECE-2018
Hyperbola
120700
If \(e\) and \(e^{\prime}\) be the eccentricity of a hyperbola and is conjugate, then \(\frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime^2}}\) is equal to
1 0
2 1
3 2
4 3
Explanation:
B We know that,
Equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\because \quad \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2}\)
Equation of conjugate hyperbola is,
\(\begin{cases} \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=-1 \\ \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1 \\ \therefore \quad \mathrm{e}^{\prime 2}=1+\frac{\mathrm{a}^2}{\mathrm{~b}^2} \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{~b}^2}\end{cases}\)
Now, \(\quad \frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime 2}} \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2+\mathrm{b}^2}=1\)
120697
The vertices of the hyperbola are at \((-5,-3)\) and \((-5,-1)\) and the extremities of the conjugate axis are at \((-7,-2)\) and \((-3,-2)\), then the equation of the hyperbola is
1 \(\frac{(y+2)^2}{1}-\frac{(x+5)^2}{4}=1\)
2 \(\frac{(y+2)^2}{4}-\frac{(x+5)^2}{4}=1\)
3 \(\frac{(x+5)^2}{4}-\frac{(y+2)^2}{1}=1\)
4 \(\frac{(x-5)^2}{4}-\frac{(y-2)^2}{1}=1\)
Explanation:
B Given that, the vertices of the hyperbola are \((-5,3),(-5,-1)\) The Extremities of the conjugate axis are at \((-7,-2),(-3,-2)\)
We know that, the distance between the extremities of conjugate axis are \(2 \mathrm{~b}\).
So,
\(2 b =\sqrt{(-7+3)^2+(-2+2)^2}=\sqrt{4^2}=4\)
\(2 b =4 \Rightarrow b=2\)
Given vertices \((-5,-3),(-5,-1)\)
Let's Centre (h, k)
Therefore \((\mathrm{h}, \mathrm{k})=\left(\frac{-5-5}{2}, \frac{-3-1}{2}\right)=(-5,-2)\)
The equation of the hyperbola is given by
\(\frac{(\mathrm{y}-\mathrm{k})^2}{\mathrm{a}^2}-\frac{(\mathrm{x}-\mathrm{h})^2}{\mathrm{~b}^2}=1\)
Substituting the values, we get -
\(\frac{(y+2)^2}{1^2}-\frac{(x+5)^2}{2^2}=1\)
\(\frac{(y+2)^2}{1}-\frac{(x+5)^2}{4}=1\)
JCECE-2017
Hyperbola
120698
The point of contact of the line \(y=x-1\) with \(3 x^2-4 y^2=12\)
1 \((4,3)\)
2 \((3,4)\)
3 \((4,-3)\)
4 None of these
Explanation:
A Given that,
\(y=x-1\)
\(3 x^2-4 y^2=12\)
Putting the value of \(y=x-1\) from equation (i) in equation (ii), we get -
\(3 x^2-4(x-1)^2=12\)
\(3 x^2-4\left(x^2-2 x+1\right)=12\)
\(x^2-8 x+16=0 \Rightarrow x=4\)
Or
Putting the value of \(x\) in equation (i), we get -
\(y=3\)Hence, required point of contact is \((4,3)\).
JCECE-2014
Hyperbola
120699
The eccentricity of the hyperbola, whose length of latus rectum is 8 and conjugate axis is equal to the half of its distance between the foci, is
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{3}}\)
3 \(\frac{2}{\sqrt{3}}\)
4 1
Explanation:
C Given that,
Length of latus rectum \(\frac{2 b^2}{a}=8\) and conjugate axis (2b) \(=\frac{1}{2}(2 \mathrm{ae})\)
\(\therefore \quad \frac{2}{\mathrm{a}}\left(\frac{\mathrm{ae}}{2}\right)^2=8\)
\(\mathrm{ae}^2=16\)
We have, \(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=8\)
\(\mathrm{b}^2=4 \mathrm{a}\)
\(\mathrm{a}^2\left(\mathrm{e}^2-1\right)=4 \mathrm{a}\)
\(\mathrm{e}^2-\mathrm{a}=4\)
\(16-\mathrm{a}=4\)
\(\mathrm{a}=12\)
Substitute \(\mathrm{a}=12\) in equation (i), we get -
\(12 \mathrm{e}^2=16\)
\(\mathrm{e}^2=\frac{4}{3}=\frac{2}{\sqrt{3}}\)
BCECE-2018
Hyperbola
120700
If \(e\) and \(e^{\prime}\) be the eccentricity of a hyperbola and is conjugate, then \(\frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime^2}}\) is equal to
1 0
2 1
3 2
4 3
Explanation:
B We know that,
Equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\because \quad \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2}\)
Equation of conjugate hyperbola is,
\(\begin{cases} \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=-1 \\ \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1 \\ \therefore \quad \mathrm{e}^{\prime 2}=1+\frac{\mathrm{a}^2}{\mathrm{~b}^2} \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{~b}^2}\end{cases}\)
Now, \(\quad \frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime 2}} \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2+\mathrm{b}^2}=1\)
120697
The vertices of the hyperbola are at \((-5,-3)\) and \((-5,-1)\) and the extremities of the conjugate axis are at \((-7,-2)\) and \((-3,-2)\), then the equation of the hyperbola is
1 \(\frac{(y+2)^2}{1}-\frac{(x+5)^2}{4}=1\)
2 \(\frac{(y+2)^2}{4}-\frac{(x+5)^2}{4}=1\)
3 \(\frac{(x+5)^2}{4}-\frac{(y+2)^2}{1}=1\)
4 \(\frac{(x-5)^2}{4}-\frac{(y-2)^2}{1}=1\)
Explanation:
B Given that, the vertices of the hyperbola are \((-5,3),(-5,-1)\) The Extremities of the conjugate axis are at \((-7,-2),(-3,-2)\)
We know that, the distance between the extremities of conjugate axis are \(2 \mathrm{~b}\).
So,
\(2 b =\sqrt{(-7+3)^2+(-2+2)^2}=\sqrt{4^2}=4\)
\(2 b =4 \Rightarrow b=2\)
Given vertices \((-5,-3),(-5,-1)\)
Let's Centre (h, k)
Therefore \((\mathrm{h}, \mathrm{k})=\left(\frac{-5-5}{2}, \frac{-3-1}{2}\right)=(-5,-2)\)
The equation of the hyperbola is given by
\(\frac{(\mathrm{y}-\mathrm{k})^2}{\mathrm{a}^2}-\frac{(\mathrm{x}-\mathrm{h})^2}{\mathrm{~b}^2}=1\)
Substituting the values, we get -
\(\frac{(y+2)^2}{1^2}-\frac{(x+5)^2}{2^2}=1\)
\(\frac{(y+2)^2}{1}-\frac{(x+5)^2}{4}=1\)
JCECE-2017
Hyperbola
120698
The point of contact of the line \(y=x-1\) with \(3 x^2-4 y^2=12\)
1 \((4,3)\)
2 \((3,4)\)
3 \((4,-3)\)
4 None of these
Explanation:
A Given that,
\(y=x-1\)
\(3 x^2-4 y^2=12\)
Putting the value of \(y=x-1\) from equation (i) in equation (ii), we get -
\(3 x^2-4(x-1)^2=12\)
\(3 x^2-4\left(x^2-2 x+1\right)=12\)
\(x^2-8 x+16=0 \Rightarrow x=4\)
Or
Putting the value of \(x\) in equation (i), we get -
\(y=3\)Hence, required point of contact is \((4,3)\).
JCECE-2014
Hyperbola
120699
The eccentricity of the hyperbola, whose length of latus rectum is 8 and conjugate axis is equal to the half of its distance between the foci, is
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{3}}\)
3 \(\frac{2}{\sqrt{3}}\)
4 1
Explanation:
C Given that,
Length of latus rectum \(\frac{2 b^2}{a}=8\) and conjugate axis (2b) \(=\frac{1}{2}(2 \mathrm{ae})\)
\(\therefore \quad \frac{2}{\mathrm{a}}\left(\frac{\mathrm{ae}}{2}\right)^2=8\)
\(\mathrm{ae}^2=16\)
We have, \(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=8\)
\(\mathrm{b}^2=4 \mathrm{a}\)
\(\mathrm{a}^2\left(\mathrm{e}^2-1\right)=4 \mathrm{a}\)
\(\mathrm{e}^2-\mathrm{a}=4\)
\(16-\mathrm{a}=4\)
\(\mathrm{a}=12\)
Substitute \(\mathrm{a}=12\) in equation (i), we get -
\(12 \mathrm{e}^2=16\)
\(\mathrm{e}^2=\frac{4}{3}=\frac{2}{\sqrt{3}}\)
BCECE-2018
Hyperbola
120700
If \(e\) and \(e^{\prime}\) be the eccentricity of a hyperbola and is conjugate, then \(\frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime^2}}\) is equal to
1 0
2 1
3 2
4 3
Explanation:
B We know that,
Equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\because \quad \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2}\)
Equation of conjugate hyperbola is,
\(\begin{cases} \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=-1 \\ \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1 \\ \therefore \quad \mathrm{e}^{\prime 2}=1+\frac{\mathrm{a}^2}{\mathrm{~b}^2} \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{~b}^2}\end{cases}\)
Now, \(\quad \frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime 2}} \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2+\mathrm{b}^2}=1\)
120697
The vertices of the hyperbola are at \((-5,-3)\) and \((-5,-1)\) and the extremities of the conjugate axis are at \((-7,-2)\) and \((-3,-2)\), then the equation of the hyperbola is
1 \(\frac{(y+2)^2}{1}-\frac{(x+5)^2}{4}=1\)
2 \(\frac{(y+2)^2}{4}-\frac{(x+5)^2}{4}=1\)
3 \(\frac{(x+5)^2}{4}-\frac{(y+2)^2}{1}=1\)
4 \(\frac{(x-5)^2}{4}-\frac{(y-2)^2}{1}=1\)
Explanation:
B Given that, the vertices of the hyperbola are \((-5,3),(-5,-1)\) The Extremities of the conjugate axis are at \((-7,-2),(-3,-2)\)
We know that, the distance between the extremities of conjugate axis are \(2 \mathrm{~b}\).
So,
\(2 b =\sqrt{(-7+3)^2+(-2+2)^2}=\sqrt{4^2}=4\)
\(2 b =4 \Rightarrow b=2\)
Given vertices \((-5,-3),(-5,-1)\)
Let's Centre (h, k)
Therefore \((\mathrm{h}, \mathrm{k})=\left(\frac{-5-5}{2}, \frac{-3-1}{2}\right)=(-5,-2)\)
The equation of the hyperbola is given by
\(\frac{(\mathrm{y}-\mathrm{k})^2}{\mathrm{a}^2}-\frac{(\mathrm{x}-\mathrm{h})^2}{\mathrm{~b}^2}=1\)
Substituting the values, we get -
\(\frac{(y+2)^2}{1^2}-\frac{(x+5)^2}{2^2}=1\)
\(\frac{(y+2)^2}{1}-\frac{(x+5)^2}{4}=1\)
JCECE-2017
Hyperbola
120698
The point of contact of the line \(y=x-1\) with \(3 x^2-4 y^2=12\)
1 \((4,3)\)
2 \((3,4)\)
3 \((4,-3)\)
4 None of these
Explanation:
A Given that,
\(y=x-1\)
\(3 x^2-4 y^2=12\)
Putting the value of \(y=x-1\) from equation (i) in equation (ii), we get -
\(3 x^2-4(x-1)^2=12\)
\(3 x^2-4\left(x^2-2 x+1\right)=12\)
\(x^2-8 x+16=0 \Rightarrow x=4\)
Or
Putting the value of \(x\) in equation (i), we get -
\(y=3\)Hence, required point of contact is \((4,3)\).
JCECE-2014
Hyperbola
120699
The eccentricity of the hyperbola, whose length of latus rectum is 8 and conjugate axis is equal to the half of its distance between the foci, is
1 \(\frac{1}{2}\)
2 \(\frac{1}{\sqrt{3}}\)
3 \(\frac{2}{\sqrt{3}}\)
4 1
Explanation:
C Given that,
Length of latus rectum \(\frac{2 b^2}{a}=8\) and conjugate axis (2b) \(=\frac{1}{2}(2 \mathrm{ae})\)
\(\therefore \quad \frac{2}{\mathrm{a}}\left(\frac{\mathrm{ae}}{2}\right)^2=8\)
\(\mathrm{ae}^2=16\)
We have, \(\frac{2 \mathrm{~b}^2}{\mathrm{a}}=8\)
\(\mathrm{b}^2=4 \mathrm{a}\)
\(\mathrm{a}^2\left(\mathrm{e}^2-1\right)=4 \mathrm{a}\)
\(\mathrm{e}^2-\mathrm{a}=4\)
\(16-\mathrm{a}=4\)
\(\mathrm{a}=12\)
Substitute \(\mathrm{a}=12\) in equation (i), we get -
\(12 \mathrm{e}^2=16\)
\(\mathrm{e}^2=\frac{4}{3}=\frac{2}{\sqrt{3}}\)
BCECE-2018
Hyperbola
120700
If \(e\) and \(e^{\prime}\) be the eccentricity of a hyperbola and is conjugate, then \(\frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime^2}}\) is equal to
1 0
2 1
3 2
4 3
Explanation:
B We know that,
Equation of hyperbola is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\because \quad \mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=\frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2}\)
Equation of conjugate hyperbola is,
\(\begin{cases} \frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=-1 \\ \frac{\mathrm{y}^2}{\mathrm{~b}^2}-\frac{\mathrm{x}^2}{\mathrm{a}^2}=1 \\ \therefore \quad \mathrm{e}^{\prime 2}=1+\frac{\mathrm{a}^2}{\mathrm{~b}^2} \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{~b}^2}\end{cases}\)
Now, \(\quad \frac{1}{\mathrm{e}^2}+\frac{1}{\mathrm{e}^{\prime 2}} \Rightarrow \frac{\mathrm{a}^2+\mathrm{b}^2}{\mathrm{a}^2+\mathrm{b}^2}=1\)
