NEET Test Series from KOTA - 10 Papers In MS WORD
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Hyperbola
120746
If the vertices of a hyperbola be at \((-2,0)\) and \((2,0)\) and one of its foci be at \((-3,0)\), then which one of the following points does not lie on this hyperbola?
1 \((2 \sqrt{6}, 5)\)
2 \((6,5 \sqrt{2})\)
3 \((4, \sqrt{15})\)
4 \((-6,2 \sqrt{10})\)
Explanation:
B
\(\frac{\mathrm{x}^2}{4}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) be the equation
\(\mathrm{ae}=3, \quad \mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{b}^2=(\mathrm{a}, \mathrm{e})^2-\mathrm{a}^2\)
\(b^2=(3)^2-(2)^2\)
\(b^2=9-4\)
\(b^2=5\)
So, equation of hyperbola is \(\frac{x^2}{4}-\frac{y^2}{5}=1\)
From option we can check that the point \((6,5 \sqrt{2})\) does not lie on this hyperbola
JEE Main 12.01.2019
Hyperbola
120747
Let \(S=\left\{(x, y) \in R^2: \frac{y^2}{1+r}-\frac{x^2}{1-r}=1\right\}\), where \(r\) \(\neq \pm 1\). Then, \(S\) represents
1 a hyperbola whose eccentricity is \(\frac{2}{\sqrt{1-\mathrm{r}}}\), when \(0\lt \mathrm{r}\lt 1\).
2 a hyperbola whose eccentricity is \(\frac{2}{\sqrt{\mathrm{r}+1}}\), when \(0\lt \mathrm{r}\lt 1\).
3 an ellipse whose eccentricity is \(\sqrt{\frac{2}{\mathrm{r}+1}}\), when \(r>1\).
4 an ellipse whose eccentricity is \(\frac{1}{\sqrt{\mathrm{r}+1}}\), when \(\mathrm{r}>1\).
Explanation:
C Given,
\(\mathrm{S}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathrm{R}^2: \frac{\mathrm{y}^2}{1+\mathrm{r}}-\frac{\mathrm{x}^2}{1-\mathrm{r}}=1\right\}\)
For, \(\mathrm{r}>1, \frac{\mathrm{y}^2}{1+\mathrm{r}}+\frac{\mathrm{x}^2}{\mathrm{r}-1}=1\)
Represent a vertical ellipse
\([\because\) For, \(\mathrm{r}>1, \mathrm{r}-1\lt \mathrm{r}+1\) and \(\mathrm{r}-1>0\) ]
Now, eccentricity \((e)=\sqrt{1-\frac{r-1}{r+1}}\)
\(\left[\because\right.\) For, \(\left.\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, \mathrm{a}\lt \mathrm{b}, \mathrm{e}=\sqrt{1-\frac{\mathrm{a}^2}{\mathrm{~b}^2}}\right]\)
\(e=\sqrt{\frac{(r+1)-(r-1)}{r+1}}\)
\(e=\sqrt{\frac{2}{r+1}}\)
JEE Main 10.01.2019
Hyperbola
120748
Let \(0\lt \theta\lt \frac{\pi}{2}\). If the eccentricity of the hyperbola \(\frac{x^2}{\cos ^2 \theta}-\frac{y^2}{\sin ^2 \theta}=1\) is greater than 2 , then the length of its latus rectum lies in the interval
1 \(\left(1, \frac{3}{2}\right]\)
2 \((3, \infty)\)
3 \(\left(\frac{3}{2}, 2\right]\)
4 \((2,3)\)
Explanation:
B From question, the hyperbola is given by the equation -
\(\frac{\mathrm{x}^2}{\cos ^2 \theta}-\frac{\mathrm{y}^2}{\sin ^2 \theta}=1\)
On comparing, general equation and given equation -
\(\mathrm{a}^2=\cos ^2 \theta\)
\(\mathrm{b}^2=\sin ^2 \theta\)
Given,
The eccentricity of the hyperbola is greater than 2 .
The eccentricity of the hyperbola is given by the formula -
\(\mathrm{e}=\sqrt{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}\)
Now,
\(\Rightarrow \quad \sqrt{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}>2\)
\(\Rightarrow \quad \sqrt{1+\tan ^2 \theta}>2\)
On squaring both sides -
\(1+\tan ^2 \theta>4\)
\(\tan ^2 \theta>3\)
Taking square root on both sides -
\(\tan \theta=\sqrt{3}\)
Let's take \(\tan \theta\) as \(x\) for easy calculation
\(\Rightarrow \mathrm{x}>\sqrt{3}\)
\(\Rightarrow |\mathrm{x}|>\sqrt{3}\)
\(\Rightarrow \mathrm{x} \in(-\infty,-\sqrt{3}) \cup(\sqrt{3} \infty)\)
From question, \(\quad \theta \in\left(0, \frac{\pi}{2}\right)\)
\(\tan \theta \in(\sqrt{3}, \infty)\)
\(\tan \frac{\pi}{3}=\sqrt{3} \text { and } \tan \frac{\pi}{2}=\infty\)
\(\theta \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\)
The length of the latusrectum of hyperbola is given by
\(\mathrm{LR}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}\)
On substituting the value,
\(\mathrm{LR}=\frac{2\left(\sin ^2 \theta\right)}{\cos \theta}\)
\(\mathrm{LR}=2 \sin \theta \tan \theta\)
The minimum length of the latusrectum \(\left[\theta \in \frac{\pi}{3}\right]\) is
\(\mathrm{LR}_{\min }=2 \sin \left(\frac{\pi}{3}\right) \cdot \tan \left(\frac{\pi}{3}\right)\)
\(\mathrm{LR}_{\min }=2\left(\frac{\sqrt{3}}{2}\right) \sqrt{3}\)
\(\mathrm{LR}_{\min }=3\)
The maximum length of the latusrectum \(\left[\theta \in \frac{\pi}{2}\right]\)
\(\mathrm{LR}_{\max }=2 \sin \left(\frac{\pi}{2}\right) \tan \left(\frac{\pi}{2}\right)\)
\(\mathrm{LR}_{\max }=2(1)(\infty)\)
\(\mathrm{LR}_{\max }=\infty \quad\left[\tan \frac{\pi}{2}=\infty\right]\)
\(\left[\sin \frac{\pi}{2}=1\right]\)So, the interval of the latusrectum is \((3, \infty)\).
JEE Main 09.01.2019
Hyperbola
120749
A hyperbola has its centre at the origin, passes through the point \((4,2)\) and has transverse axis of length 4 along the \(X\)-axis. Then the eccentricity of the hyperbola is
1 2
2 \(\frac{2}{\sqrt{3}}\)
3 \(\frac{3}{2}\)
4 \(\sqrt{3}\)
Explanation:
B Equation of hyperbola \(=\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(2 a=4\)
\(a=2\)
Passes through point \((4,2)\)
\(x=4, \quad y=2\)
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{16}{4}-\frac{4}{b^2}=1\)
\(4-\frac{4}{b^2}=1\)
\(4 b^2-4=b^2\)
\(3 b^2=4\)
\(b=\frac{2}{\sqrt{3}}\)
Eccentricity \((e)=\sqrt{1+\frac{b^2}{a^2}}\)
\(\mathrm{e}=\sqrt{1+\frac{\frac{4}{3}}{4}}\)
\(\mathrm{e}=\sqrt{1+\frac{1}{3}}\)
\(\mathrm{e}=\sqrt{\frac{4}{3}}\)
\(\mathrm{e}=\frac{2}{\sqrt{3}}\)
120746
If the vertices of a hyperbola be at \((-2,0)\) and \((2,0)\) and one of its foci be at \((-3,0)\), then which one of the following points does not lie on this hyperbola?
1 \((2 \sqrt{6}, 5)\)
2 \((6,5 \sqrt{2})\)
3 \((4, \sqrt{15})\)
4 \((-6,2 \sqrt{10})\)
Explanation:
B
\(\frac{\mathrm{x}^2}{4}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) be the equation
\(\mathrm{ae}=3, \quad \mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{b}^2=(\mathrm{a}, \mathrm{e})^2-\mathrm{a}^2\)
\(b^2=(3)^2-(2)^2\)
\(b^2=9-4\)
\(b^2=5\)
So, equation of hyperbola is \(\frac{x^2}{4}-\frac{y^2}{5}=1\)
From option we can check that the point \((6,5 \sqrt{2})\) does not lie on this hyperbola
JEE Main 12.01.2019
Hyperbola
120747
Let \(S=\left\{(x, y) \in R^2: \frac{y^2}{1+r}-\frac{x^2}{1-r}=1\right\}\), where \(r\) \(\neq \pm 1\). Then, \(S\) represents
1 a hyperbola whose eccentricity is \(\frac{2}{\sqrt{1-\mathrm{r}}}\), when \(0\lt \mathrm{r}\lt 1\).
2 a hyperbola whose eccentricity is \(\frac{2}{\sqrt{\mathrm{r}+1}}\), when \(0\lt \mathrm{r}\lt 1\).
3 an ellipse whose eccentricity is \(\sqrt{\frac{2}{\mathrm{r}+1}}\), when \(r>1\).
4 an ellipse whose eccentricity is \(\frac{1}{\sqrt{\mathrm{r}+1}}\), when \(\mathrm{r}>1\).
Explanation:
C Given,
\(\mathrm{S}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathrm{R}^2: \frac{\mathrm{y}^2}{1+\mathrm{r}}-\frac{\mathrm{x}^2}{1-\mathrm{r}}=1\right\}\)
For, \(\mathrm{r}>1, \frac{\mathrm{y}^2}{1+\mathrm{r}}+\frac{\mathrm{x}^2}{\mathrm{r}-1}=1\)
Represent a vertical ellipse
\([\because\) For, \(\mathrm{r}>1, \mathrm{r}-1\lt \mathrm{r}+1\) and \(\mathrm{r}-1>0\) ]
Now, eccentricity \((e)=\sqrt{1-\frac{r-1}{r+1}}\)
\(\left[\because\right.\) For, \(\left.\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, \mathrm{a}\lt \mathrm{b}, \mathrm{e}=\sqrt{1-\frac{\mathrm{a}^2}{\mathrm{~b}^2}}\right]\)
\(e=\sqrt{\frac{(r+1)-(r-1)}{r+1}}\)
\(e=\sqrt{\frac{2}{r+1}}\)
JEE Main 10.01.2019
Hyperbola
120748
Let \(0\lt \theta\lt \frac{\pi}{2}\). If the eccentricity of the hyperbola \(\frac{x^2}{\cos ^2 \theta}-\frac{y^2}{\sin ^2 \theta}=1\) is greater than 2 , then the length of its latus rectum lies in the interval
1 \(\left(1, \frac{3}{2}\right]\)
2 \((3, \infty)\)
3 \(\left(\frac{3}{2}, 2\right]\)
4 \((2,3)\)
Explanation:
B From question, the hyperbola is given by the equation -
\(\frac{\mathrm{x}^2}{\cos ^2 \theta}-\frac{\mathrm{y}^2}{\sin ^2 \theta}=1\)
On comparing, general equation and given equation -
\(\mathrm{a}^2=\cos ^2 \theta\)
\(\mathrm{b}^2=\sin ^2 \theta\)
Given,
The eccentricity of the hyperbola is greater than 2 .
The eccentricity of the hyperbola is given by the formula -
\(\mathrm{e}=\sqrt{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}\)
Now,
\(\Rightarrow \quad \sqrt{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}>2\)
\(\Rightarrow \quad \sqrt{1+\tan ^2 \theta}>2\)
On squaring both sides -
\(1+\tan ^2 \theta>4\)
\(\tan ^2 \theta>3\)
Taking square root on both sides -
\(\tan \theta=\sqrt{3}\)
Let's take \(\tan \theta\) as \(x\) for easy calculation
\(\Rightarrow \mathrm{x}>\sqrt{3}\)
\(\Rightarrow |\mathrm{x}|>\sqrt{3}\)
\(\Rightarrow \mathrm{x} \in(-\infty,-\sqrt{3}) \cup(\sqrt{3} \infty)\)
From question, \(\quad \theta \in\left(0, \frac{\pi}{2}\right)\)
\(\tan \theta \in(\sqrt{3}, \infty)\)
\(\tan \frac{\pi}{3}=\sqrt{3} \text { and } \tan \frac{\pi}{2}=\infty\)
\(\theta \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\)
The length of the latusrectum of hyperbola is given by
\(\mathrm{LR}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}\)
On substituting the value,
\(\mathrm{LR}=\frac{2\left(\sin ^2 \theta\right)}{\cos \theta}\)
\(\mathrm{LR}=2 \sin \theta \tan \theta\)
The minimum length of the latusrectum \(\left[\theta \in \frac{\pi}{3}\right]\) is
\(\mathrm{LR}_{\min }=2 \sin \left(\frac{\pi}{3}\right) \cdot \tan \left(\frac{\pi}{3}\right)\)
\(\mathrm{LR}_{\min }=2\left(\frac{\sqrt{3}}{2}\right) \sqrt{3}\)
\(\mathrm{LR}_{\min }=3\)
The maximum length of the latusrectum \(\left[\theta \in \frac{\pi}{2}\right]\)
\(\mathrm{LR}_{\max }=2 \sin \left(\frac{\pi}{2}\right) \tan \left(\frac{\pi}{2}\right)\)
\(\mathrm{LR}_{\max }=2(1)(\infty)\)
\(\mathrm{LR}_{\max }=\infty \quad\left[\tan \frac{\pi}{2}=\infty\right]\)
\(\left[\sin \frac{\pi}{2}=1\right]\)So, the interval of the latusrectum is \((3, \infty)\).
JEE Main 09.01.2019
Hyperbola
120749
A hyperbola has its centre at the origin, passes through the point \((4,2)\) and has transverse axis of length 4 along the \(X\)-axis. Then the eccentricity of the hyperbola is
1 2
2 \(\frac{2}{\sqrt{3}}\)
3 \(\frac{3}{2}\)
4 \(\sqrt{3}\)
Explanation:
B Equation of hyperbola \(=\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(2 a=4\)
\(a=2\)
Passes through point \((4,2)\)
\(x=4, \quad y=2\)
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{16}{4}-\frac{4}{b^2}=1\)
\(4-\frac{4}{b^2}=1\)
\(4 b^2-4=b^2\)
\(3 b^2=4\)
\(b=\frac{2}{\sqrt{3}}\)
Eccentricity \((e)=\sqrt{1+\frac{b^2}{a^2}}\)
\(\mathrm{e}=\sqrt{1+\frac{\frac{4}{3}}{4}}\)
\(\mathrm{e}=\sqrt{1+\frac{1}{3}}\)
\(\mathrm{e}=\sqrt{\frac{4}{3}}\)
\(\mathrm{e}=\frac{2}{\sqrt{3}}\)
120746
If the vertices of a hyperbola be at \((-2,0)\) and \((2,0)\) and one of its foci be at \((-3,0)\), then which one of the following points does not lie on this hyperbola?
1 \((2 \sqrt{6}, 5)\)
2 \((6,5 \sqrt{2})\)
3 \((4, \sqrt{15})\)
4 \((-6,2 \sqrt{10})\)
Explanation:
B
\(\frac{\mathrm{x}^2}{4}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) be the equation
\(\mathrm{ae}=3, \quad \mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{b}^2=(\mathrm{a}, \mathrm{e})^2-\mathrm{a}^2\)
\(b^2=(3)^2-(2)^2\)
\(b^2=9-4\)
\(b^2=5\)
So, equation of hyperbola is \(\frac{x^2}{4}-\frac{y^2}{5}=1\)
From option we can check that the point \((6,5 \sqrt{2})\) does not lie on this hyperbola
JEE Main 12.01.2019
Hyperbola
120747
Let \(S=\left\{(x, y) \in R^2: \frac{y^2}{1+r}-\frac{x^2}{1-r}=1\right\}\), where \(r\) \(\neq \pm 1\). Then, \(S\) represents
1 a hyperbola whose eccentricity is \(\frac{2}{\sqrt{1-\mathrm{r}}}\), when \(0\lt \mathrm{r}\lt 1\).
2 a hyperbola whose eccentricity is \(\frac{2}{\sqrt{\mathrm{r}+1}}\), when \(0\lt \mathrm{r}\lt 1\).
3 an ellipse whose eccentricity is \(\sqrt{\frac{2}{\mathrm{r}+1}}\), when \(r>1\).
4 an ellipse whose eccentricity is \(\frac{1}{\sqrt{\mathrm{r}+1}}\), when \(\mathrm{r}>1\).
Explanation:
C Given,
\(\mathrm{S}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathrm{R}^2: \frac{\mathrm{y}^2}{1+\mathrm{r}}-\frac{\mathrm{x}^2}{1-\mathrm{r}}=1\right\}\)
For, \(\mathrm{r}>1, \frac{\mathrm{y}^2}{1+\mathrm{r}}+\frac{\mathrm{x}^2}{\mathrm{r}-1}=1\)
Represent a vertical ellipse
\([\because\) For, \(\mathrm{r}>1, \mathrm{r}-1\lt \mathrm{r}+1\) and \(\mathrm{r}-1>0\) ]
Now, eccentricity \((e)=\sqrt{1-\frac{r-1}{r+1}}\)
\(\left[\because\right.\) For, \(\left.\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, \mathrm{a}\lt \mathrm{b}, \mathrm{e}=\sqrt{1-\frac{\mathrm{a}^2}{\mathrm{~b}^2}}\right]\)
\(e=\sqrt{\frac{(r+1)-(r-1)}{r+1}}\)
\(e=\sqrt{\frac{2}{r+1}}\)
JEE Main 10.01.2019
Hyperbola
120748
Let \(0\lt \theta\lt \frac{\pi}{2}\). If the eccentricity of the hyperbola \(\frac{x^2}{\cos ^2 \theta}-\frac{y^2}{\sin ^2 \theta}=1\) is greater than 2 , then the length of its latus rectum lies in the interval
1 \(\left(1, \frac{3}{2}\right]\)
2 \((3, \infty)\)
3 \(\left(\frac{3}{2}, 2\right]\)
4 \((2,3)\)
Explanation:
B From question, the hyperbola is given by the equation -
\(\frac{\mathrm{x}^2}{\cos ^2 \theta}-\frac{\mathrm{y}^2}{\sin ^2 \theta}=1\)
On comparing, general equation and given equation -
\(\mathrm{a}^2=\cos ^2 \theta\)
\(\mathrm{b}^2=\sin ^2 \theta\)
Given,
The eccentricity of the hyperbola is greater than 2 .
The eccentricity of the hyperbola is given by the formula -
\(\mathrm{e}=\sqrt{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}\)
Now,
\(\Rightarrow \quad \sqrt{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}>2\)
\(\Rightarrow \quad \sqrt{1+\tan ^2 \theta}>2\)
On squaring both sides -
\(1+\tan ^2 \theta>4\)
\(\tan ^2 \theta>3\)
Taking square root on both sides -
\(\tan \theta=\sqrt{3}\)
Let's take \(\tan \theta\) as \(x\) for easy calculation
\(\Rightarrow \mathrm{x}>\sqrt{3}\)
\(\Rightarrow |\mathrm{x}|>\sqrt{3}\)
\(\Rightarrow \mathrm{x} \in(-\infty,-\sqrt{3}) \cup(\sqrt{3} \infty)\)
From question, \(\quad \theta \in\left(0, \frac{\pi}{2}\right)\)
\(\tan \theta \in(\sqrt{3}, \infty)\)
\(\tan \frac{\pi}{3}=\sqrt{3} \text { and } \tan \frac{\pi}{2}=\infty\)
\(\theta \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\)
The length of the latusrectum of hyperbola is given by
\(\mathrm{LR}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}\)
On substituting the value,
\(\mathrm{LR}=\frac{2\left(\sin ^2 \theta\right)}{\cos \theta}\)
\(\mathrm{LR}=2 \sin \theta \tan \theta\)
The minimum length of the latusrectum \(\left[\theta \in \frac{\pi}{3}\right]\) is
\(\mathrm{LR}_{\min }=2 \sin \left(\frac{\pi}{3}\right) \cdot \tan \left(\frac{\pi}{3}\right)\)
\(\mathrm{LR}_{\min }=2\left(\frac{\sqrt{3}}{2}\right) \sqrt{3}\)
\(\mathrm{LR}_{\min }=3\)
The maximum length of the latusrectum \(\left[\theta \in \frac{\pi}{2}\right]\)
\(\mathrm{LR}_{\max }=2 \sin \left(\frac{\pi}{2}\right) \tan \left(\frac{\pi}{2}\right)\)
\(\mathrm{LR}_{\max }=2(1)(\infty)\)
\(\mathrm{LR}_{\max }=\infty \quad\left[\tan \frac{\pi}{2}=\infty\right]\)
\(\left[\sin \frac{\pi}{2}=1\right]\)So, the interval of the latusrectum is \((3, \infty)\).
JEE Main 09.01.2019
Hyperbola
120749
A hyperbola has its centre at the origin, passes through the point \((4,2)\) and has transverse axis of length 4 along the \(X\)-axis. Then the eccentricity of the hyperbola is
1 2
2 \(\frac{2}{\sqrt{3}}\)
3 \(\frac{3}{2}\)
4 \(\sqrt{3}\)
Explanation:
B Equation of hyperbola \(=\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(2 a=4\)
\(a=2\)
Passes through point \((4,2)\)
\(x=4, \quad y=2\)
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{16}{4}-\frac{4}{b^2}=1\)
\(4-\frac{4}{b^2}=1\)
\(4 b^2-4=b^2\)
\(3 b^2=4\)
\(b=\frac{2}{\sqrt{3}}\)
Eccentricity \((e)=\sqrt{1+\frac{b^2}{a^2}}\)
\(\mathrm{e}=\sqrt{1+\frac{\frac{4}{3}}{4}}\)
\(\mathrm{e}=\sqrt{1+\frac{1}{3}}\)
\(\mathrm{e}=\sqrt{\frac{4}{3}}\)
\(\mathrm{e}=\frac{2}{\sqrt{3}}\)
120746
If the vertices of a hyperbola be at \((-2,0)\) and \((2,0)\) and one of its foci be at \((-3,0)\), then which one of the following points does not lie on this hyperbola?
1 \((2 \sqrt{6}, 5)\)
2 \((6,5 \sqrt{2})\)
3 \((4, \sqrt{15})\)
4 \((-6,2 \sqrt{10})\)
Explanation:
B
\(\frac{\mathrm{x}^2}{4}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1\) be the equation
\(\mathrm{ae}=3, \quad \mathrm{~b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\mathrm{b}^2=(\mathrm{a}, \mathrm{e})^2-\mathrm{a}^2\)
\(b^2=(3)^2-(2)^2\)
\(b^2=9-4\)
\(b^2=5\)
So, equation of hyperbola is \(\frac{x^2}{4}-\frac{y^2}{5}=1\)
From option we can check that the point \((6,5 \sqrt{2})\) does not lie on this hyperbola
JEE Main 12.01.2019
Hyperbola
120747
Let \(S=\left\{(x, y) \in R^2: \frac{y^2}{1+r}-\frac{x^2}{1-r}=1\right\}\), where \(r\) \(\neq \pm 1\). Then, \(S\) represents
1 a hyperbola whose eccentricity is \(\frac{2}{\sqrt{1-\mathrm{r}}}\), when \(0\lt \mathrm{r}\lt 1\).
2 a hyperbola whose eccentricity is \(\frac{2}{\sqrt{\mathrm{r}+1}}\), when \(0\lt \mathrm{r}\lt 1\).
3 an ellipse whose eccentricity is \(\sqrt{\frac{2}{\mathrm{r}+1}}\), when \(r>1\).
4 an ellipse whose eccentricity is \(\frac{1}{\sqrt{\mathrm{r}+1}}\), when \(\mathrm{r}>1\).
Explanation:
C Given,
\(\mathrm{S}=\left\{(\mathrm{x}, \mathrm{y}) \in \mathrm{R}^2: \frac{\mathrm{y}^2}{1+\mathrm{r}}-\frac{\mathrm{x}^2}{1-\mathrm{r}}=1\right\}\)
For, \(\mathrm{r}>1, \frac{\mathrm{y}^2}{1+\mathrm{r}}+\frac{\mathrm{x}^2}{\mathrm{r}-1}=1\)
Represent a vertical ellipse
\([\because\) For, \(\mathrm{r}>1, \mathrm{r}-1\lt \mathrm{r}+1\) and \(\mathrm{r}-1>0\) ]
Now, eccentricity \((e)=\sqrt{1-\frac{r-1}{r+1}}\)
\(\left[\because\right.\) For, \(\left.\frac{\mathrm{x}^2}{\mathrm{a}^2}-\frac{\mathrm{y}^2}{\mathrm{~b}^2}=1, \mathrm{a}\lt \mathrm{b}, \mathrm{e}=\sqrt{1-\frac{\mathrm{a}^2}{\mathrm{~b}^2}}\right]\)
\(e=\sqrt{\frac{(r+1)-(r-1)}{r+1}}\)
\(e=\sqrt{\frac{2}{r+1}}\)
JEE Main 10.01.2019
Hyperbola
120748
Let \(0\lt \theta\lt \frac{\pi}{2}\). If the eccentricity of the hyperbola \(\frac{x^2}{\cos ^2 \theta}-\frac{y^2}{\sin ^2 \theta}=1\) is greater than 2 , then the length of its latus rectum lies in the interval
1 \(\left(1, \frac{3}{2}\right]\)
2 \((3, \infty)\)
3 \(\left(\frac{3}{2}, 2\right]\)
4 \((2,3)\)
Explanation:
B From question, the hyperbola is given by the equation -
\(\frac{\mathrm{x}^2}{\cos ^2 \theta}-\frac{\mathrm{y}^2}{\sin ^2 \theta}=1\)
On comparing, general equation and given equation -
\(\mathrm{a}^2=\cos ^2 \theta\)
\(\mathrm{b}^2=\sin ^2 \theta\)
Given,
The eccentricity of the hyperbola is greater than 2 .
The eccentricity of the hyperbola is given by the formula -
\(\mathrm{e}=\sqrt{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}\)
Now,
\(\Rightarrow \quad \sqrt{1+\frac{\sin ^2 \theta}{\cos ^2 \theta}}>2\)
\(\Rightarrow \quad \sqrt{1+\tan ^2 \theta}>2\)
On squaring both sides -
\(1+\tan ^2 \theta>4\)
\(\tan ^2 \theta>3\)
Taking square root on both sides -
\(\tan \theta=\sqrt{3}\)
Let's take \(\tan \theta\) as \(x\) for easy calculation
\(\Rightarrow \mathrm{x}>\sqrt{3}\)
\(\Rightarrow |\mathrm{x}|>\sqrt{3}\)
\(\Rightarrow \mathrm{x} \in(-\infty,-\sqrt{3}) \cup(\sqrt{3} \infty)\)
From question, \(\quad \theta \in\left(0, \frac{\pi}{2}\right)\)
\(\tan \theta \in(\sqrt{3}, \infty)\)
\(\tan \frac{\pi}{3}=\sqrt{3} \text { and } \tan \frac{\pi}{2}=\infty\)
\(\theta \in\left(\frac{\pi}{3}, \frac{\pi}{2}\right)\)
The length of the latusrectum of hyperbola is given by
\(\mathrm{LR}=\frac{2 \mathrm{~b}^2}{\mathrm{a}}\)
On substituting the value,
\(\mathrm{LR}=\frac{2\left(\sin ^2 \theta\right)}{\cos \theta}\)
\(\mathrm{LR}=2 \sin \theta \tan \theta\)
The minimum length of the latusrectum \(\left[\theta \in \frac{\pi}{3}\right]\) is
\(\mathrm{LR}_{\min }=2 \sin \left(\frac{\pi}{3}\right) \cdot \tan \left(\frac{\pi}{3}\right)\)
\(\mathrm{LR}_{\min }=2\left(\frac{\sqrt{3}}{2}\right) \sqrt{3}\)
\(\mathrm{LR}_{\min }=3\)
The maximum length of the latusrectum \(\left[\theta \in \frac{\pi}{2}\right]\)
\(\mathrm{LR}_{\max }=2 \sin \left(\frac{\pi}{2}\right) \tan \left(\frac{\pi}{2}\right)\)
\(\mathrm{LR}_{\max }=2(1)(\infty)\)
\(\mathrm{LR}_{\max }=\infty \quad\left[\tan \frac{\pi}{2}=\infty\right]\)
\(\left[\sin \frac{\pi}{2}=1\right]\)So, the interval of the latusrectum is \((3, \infty)\).
JEE Main 09.01.2019
Hyperbola
120749
A hyperbola has its centre at the origin, passes through the point \((4,2)\) and has transverse axis of length 4 along the \(X\)-axis. Then the eccentricity of the hyperbola is
1 2
2 \(\frac{2}{\sqrt{3}}\)
3 \(\frac{3}{2}\)
4 \(\sqrt{3}\)
Explanation:
B Equation of hyperbola \(=\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(2 a=4\)
\(a=2\)
Passes through point \((4,2)\)
\(x=4, \quad y=2\)
\(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\)
\(\frac{16}{4}-\frac{4}{b^2}=1\)
\(4-\frac{4}{b^2}=1\)
\(4 b^2-4=b^2\)
\(3 b^2=4\)
\(b=\frac{2}{\sqrt{3}}\)
Eccentricity \((e)=\sqrt{1+\frac{b^2}{a^2}}\)
\(\mathrm{e}=\sqrt{1+\frac{\frac{4}{3}}{4}}\)
\(\mathrm{e}=\sqrt{1+\frac{1}{3}}\)
\(\mathrm{e}=\sqrt{\frac{4}{3}}\)
\(\mathrm{e}=\frac{2}{\sqrt{3}}\)
