B Given,
\(3 x^2+4 y^2=12\)
\(\frac{x^2}{4}+\frac{y^2}{3}=1\)
Here, \(\mathrm{a}^2=4\) and \(\mathrm{b}^2=3\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{3}{4}}\)
\(\mathrm{e}=\frac{1}{2}\)
Focus of ellipse \(( \pm\) ae, 0\()=( \pm 1,0)\)
If required hyperbola is confocal to ellipse then,
Let, \(a^{\prime}, b^{\prime}, e^{\prime}\) are transverse axis and conjugate axis an eccentricity of the hyperbola.
\(a^{\prime} e^{\prime}=1\)
\(\sin \theta e^{\prime}=1\)
\(e=\frac{1}{\sin \theta}\)
Also,
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\left(\mathrm{b}^{\prime}\right)^2=\left(1-\sin ^2 \theta\right)\)
\(\left(\mathrm{b}^{\prime}\right)^2=\cos ^2 \theta\)
\(\therefore\) Required equation of hyperbola is -
\(\frac{x^2}{\left(a^{\prime}\right)^2}-\frac{y^2}{\left(b^{\prime}\right)^2}=1\)
\(\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1\)
\(x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1\)
WB JEE-2018
Hyperbola
120729
Let the eccentricity of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) be reciprocal to that of the ellipse \(x^2+9 y^2=9\), then the ratio \(a^2: b^2\) equals
120730
The equation of the hyperbola with focus \((1,2)\), \(e=\sqrt{3}\) and directrix \(2 x+y=1\) is given by
1 \(2 y^2-12 x y-7 x^2+2 x-14 y+22=0\)
2 \(2 y^2+12 x y+7 x^2-2 x+14 y-22=0\)
3 \(2 y^2-12 x y-7 x^2-2 x-14 y-22=0\)
4 \(2 y^2+12 x y+7 x^2+2 x+14 y+22=0\)
Explanation:
A Given,
Focus \((1,2) \& \mathrm{e}=\sqrt{3}\)
Directrix : \(2 \mathrm{x}+\mathrm{y}=1\)
Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on hyperbola let PM be length of perpendicular from \(\mathrm{P}\) to directrix.
Then, \(\quad \frac{P S}{P M}=e=\sqrt{3}\)
\(\therefore \quad(P S)^2=3(P M)^2\)
\(\text { Or } \quad(x-1)^2+(y-2)^2=3\left|\frac{2 x+y-1}{\sqrt{5}}\right|^2\)
\(5\left[(x-1)^2+(y-2)^2\right]=3\left(4 x^2+y^2+1+4 x y-2 y-4 x\right)\)
\(5\left[x^2+1-2 x+y^2+4-4 y\right]=\left[12 x^2+3 y^2+3+12 x y-\right.\)
\(6 y-12 x]\)
\(\quad -7 x^2+2 y^2-12 x y+2 x-14 y+22=0\)
\(2 y^2-12 x y-7 x^2+2 x-14 y+22=0\)
AP EAMCET-22.09.2020
Hyperbola
120731
If \(e_1\) and \(e_2\) are the eccentricities of the hyperbola \(16 x^2-9 y^2=1\) and its conjugate respectively. Then \(3 \mathrm{e}_1=\)
B Given,
\(3 x^2+4 y^2=12\)
\(\frac{x^2}{4}+\frac{y^2}{3}=1\)
Here, \(\mathrm{a}^2=4\) and \(\mathrm{b}^2=3\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{3}{4}}\)
\(\mathrm{e}=\frac{1}{2}\)
Focus of ellipse \(( \pm\) ae, 0\()=( \pm 1,0)\)
If required hyperbola is confocal to ellipse then,
Let, \(a^{\prime}, b^{\prime}, e^{\prime}\) are transverse axis and conjugate axis an eccentricity of the hyperbola.
\(a^{\prime} e^{\prime}=1\)
\(\sin \theta e^{\prime}=1\)
\(e=\frac{1}{\sin \theta}\)
Also,
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\left(\mathrm{b}^{\prime}\right)^2=\left(1-\sin ^2 \theta\right)\)
\(\left(\mathrm{b}^{\prime}\right)^2=\cos ^2 \theta\)
\(\therefore\) Required equation of hyperbola is -
\(\frac{x^2}{\left(a^{\prime}\right)^2}-\frac{y^2}{\left(b^{\prime}\right)^2}=1\)
\(\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1\)
\(x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1\)
WB JEE-2018
Hyperbola
120729
Let the eccentricity of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) be reciprocal to that of the ellipse \(x^2+9 y^2=9\), then the ratio \(a^2: b^2\) equals
120730
The equation of the hyperbola with focus \((1,2)\), \(e=\sqrt{3}\) and directrix \(2 x+y=1\) is given by
1 \(2 y^2-12 x y-7 x^2+2 x-14 y+22=0\)
2 \(2 y^2+12 x y+7 x^2-2 x+14 y-22=0\)
3 \(2 y^2-12 x y-7 x^2-2 x-14 y-22=0\)
4 \(2 y^2+12 x y+7 x^2+2 x+14 y+22=0\)
Explanation:
A Given,
Focus \((1,2) \& \mathrm{e}=\sqrt{3}\)
Directrix : \(2 \mathrm{x}+\mathrm{y}=1\)
Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on hyperbola let PM be length of perpendicular from \(\mathrm{P}\) to directrix.
Then, \(\quad \frac{P S}{P M}=e=\sqrt{3}\)
\(\therefore \quad(P S)^2=3(P M)^2\)
\(\text { Or } \quad(x-1)^2+(y-2)^2=3\left|\frac{2 x+y-1}{\sqrt{5}}\right|^2\)
\(5\left[(x-1)^2+(y-2)^2\right]=3\left(4 x^2+y^2+1+4 x y-2 y-4 x\right)\)
\(5\left[x^2+1-2 x+y^2+4-4 y\right]=\left[12 x^2+3 y^2+3+12 x y-\right.\)
\(6 y-12 x]\)
\(\quad -7 x^2+2 y^2-12 x y+2 x-14 y+22=0\)
\(2 y^2-12 x y-7 x^2+2 x-14 y+22=0\)
AP EAMCET-22.09.2020
Hyperbola
120731
If \(e_1\) and \(e_2\) are the eccentricities of the hyperbola \(16 x^2-9 y^2=1\) and its conjugate respectively. Then \(3 \mathrm{e}_1=\)
B Given,
\(3 x^2+4 y^2=12\)
\(\frac{x^2}{4}+\frac{y^2}{3}=1\)
Here, \(\mathrm{a}^2=4\) and \(\mathrm{b}^2=3\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{3}{4}}\)
\(\mathrm{e}=\frac{1}{2}\)
Focus of ellipse \(( \pm\) ae, 0\()=( \pm 1,0)\)
If required hyperbola is confocal to ellipse then,
Let, \(a^{\prime}, b^{\prime}, e^{\prime}\) are transverse axis and conjugate axis an eccentricity of the hyperbola.
\(a^{\prime} e^{\prime}=1\)
\(\sin \theta e^{\prime}=1\)
\(e=\frac{1}{\sin \theta}\)
Also,
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\left(\mathrm{b}^{\prime}\right)^2=\left(1-\sin ^2 \theta\right)\)
\(\left(\mathrm{b}^{\prime}\right)^2=\cos ^2 \theta\)
\(\therefore\) Required equation of hyperbola is -
\(\frac{x^2}{\left(a^{\prime}\right)^2}-\frac{y^2}{\left(b^{\prime}\right)^2}=1\)
\(\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1\)
\(x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1\)
WB JEE-2018
Hyperbola
120729
Let the eccentricity of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) be reciprocal to that of the ellipse \(x^2+9 y^2=9\), then the ratio \(a^2: b^2\) equals
120730
The equation of the hyperbola with focus \((1,2)\), \(e=\sqrt{3}\) and directrix \(2 x+y=1\) is given by
1 \(2 y^2-12 x y-7 x^2+2 x-14 y+22=0\)
2 \(2 y^2+12 x y+7 x^2-2 x+14 y-22=0\)
3 \(2 y^2-12 x y-7 x^2-2 x-14 y-22=0\)
4 \(2 y^2+12 x y+7 x^2+2 x+14 y+22=0\)
Explanation:
A Given,
Focus \((1,2) \& \mathrm{e}=\sqrt{3}\)
Directrix : \(2 \mathrm{x}+\mathrm{y}=1\)
Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on hyperbola let PM be length of perpendicular from \(\mathrm{P}\) to directrix.
Then, \(\quad \frac{P S}{P M}=e=\sqrt{3}\)
\(\therefore \quad(P S)^2=3(P M)^2\)
\(\text { Or } \quad(x-1)^2+(y-2)^2=3\left|\frac{2 x+y-1}{\sqrt{5}}\right|^2\)
\(5\left[(x-1)^2+(y-2)^2\right]=3\left(4 x^2+y^2+1+4 x y-2 y-4 x\right)\)
\(5\left[x^2+1-2 x+y^2+4-4 y\right]=\left[12 x^2+3 y^2+3+12 x y-\right.\)
\(6 y-12 x]\)
\(\quad -7 x^2+2 y^2-12 x y+2 x-14 y+22=0\)
\(2 y^2-12 x y-7 x^2+2 x-14 y+22=0\)
AP EAMCET-22.09.2020
Hyperbola
120731
If \(e_1\) and \(e_2\) are the eccentricities of the hyperbola \(16 x^2-9 y^2=1\) and its conjugate respectively. Then \(3 \mathrm{e}_1=\)
B Given,
\(3 x^2+4 y^2=12\)
\(\frac{x^2}{4}+\frac{y^2}{3}=1\)
Here, \(\mathrm{a}^2=4\) and \(\mathrm{b}^2=3\)
\(\mathrm{e}=\sqrt{1-\frac{\mathrm{b}^2}{\mathrm{a}^2}}=\sqrt{1-\frac{3}{4}}\)
\(\mathrm{e}=\frac{1}{2}\)
Focus of ellipse \(( \pm\) ae, 0\()=( \pm 1,0)\)
If required hyperbola is confocal to ellipse then,
Let, \(a^{\prime}, b^{\prime}, e^{\prime}\) are transverse axis and conjugate axis an eccentricity of the hyperbola.
\(a^{\prime} e^{\prime}=1\)
\(\sin \theta e^{\prime}=1\)
\(e=\frac{1}{\sin \theta}\)
Also,
\(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(\left(\mathrm{b}^{\prime}\right)^2=\left(1-\sin ^2 \theta\right)\)
\(\left(\mathrm{b}^{\prime}\right)^2=\cos ^2 \theta\)
\(\therefore\) Required equation of hyperbola is -
\(\frac{x^2}{\left(a^{\prime}\right)^2}-\frac{y^2}{\left(b^{\prime}\right)^2}=1\)
\(\frac{x^2}{\sin ^2 \theta}-\frac{y^2}{\cos ^2 \theta}=1\)
\(x^2 \operatorname{cosec}^2 \theta-y^2 \sec ^2 \theta=1\)
WB JEE-2018
Hyperbola
120729
Let the eccentricity of the hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) be reciprocal to that of the ellipse \(x^2+9 y^2=9\), then the ratio \(a^2: b^2\) equals
120730
The equation of the hyperbola with focus \((1,2)\), \(e=\sqrt{3}\) and directrix \(2 x+y=1\) is given by
1 \(2 y^2-12 x y-7 x^2+2 x-14 y+22=0\)
2 \(2 y^2+12 x y+7 x^2-2 x+14 y-22=0\)
3 \(2 y^2-12 x y-7 x^2-2 x-14 y-22=0\)
4 \(2 y^2+12 x y+7 x^2+2 x+14 y+22=0\)
Explanation:
A Given,
Focus \((1,2) \& \mathrm{e}=\sqrt{3}\)
Directrix : \(2 \mathrm{x}+\mathrm{y}=1\)
Let \(\mathrm{P}(\mathrm{x}, \mathrm{y})\) be any point on hyperbola let PM be length of perpendicular from \(\mathrm{P}\) to directrix.
Then, \(\quad \frac{P S}{P M}=e=\sqrt{3}\)
\(\therefore \quad(P S)^2=3(P M)^2\)
\(\text { Or } \quad(x-1)^2+(y-2)^2=3\left|\frac{2 x+y-1}{\sqrt{5}}\right|^2\)
\(5\left[(x-1)^2+(y-2)^2\right]=3\left(4 x^2+y^2+1+4 x y-2 y-4 x\right)\)
\(5\left[x^2+1-2 x+y^2+4-4 y\right]=\left[12 x^2+3 y^2+3+12 x y-\right.\)
\(6 y-12 x]\)
\(\quad -7 x^2+2 y^2-12 x y+2 x-14 y+22=0\)
\(2 y^2-12 x y-7 x^2+2 x-14 y+22=0\)
AP EAMCET-22.09.2020
Hyperbola
120731
If \(e_1\) and \(e_2\) are the eccentricities of the hyperbola \(16 x^2-9 y^2=1\) and its conjugate respectively. Then \(3 \mathrm{e}_1=\)