120711
Let \(\mathrm{T}\) and \(\mathrm{C}\) respectively be the transverse and conjugate axes of the hyperbola \(16 x^2-y^2+64 x\) \(+4 y+44=0\). Then the area of the region above the parabola \(x^2=y+4\), below the transverse axis \(T\) and on the right of the conjugate axis \(C\) is :
1 \(4 \sqrt{6}+\frac{44}{3}\)
2 \(4 \sqrt{6}-\frac{28}{3}\)
3 \(4 \sqrt{6}+\frac{28}{3}\)
4 \(4 \sqrt{6}-\frac{44}{3}\)
Explanation:
C Given, \(16 x^2-y^2+64 x+4 y+44=0\)
\(\Rightarrow \quad 16\left(\mathrm{x}^2+4 \mathrm{x}\right)-\left(\mathrm{y}^2-4 \mathrm{y}\right)+44=0\)
\(16(\mathrm{x}+2)^2-(\mathrm{y}-2)^2=16\)
\(\Rightarrow \quad \frac{(\mathrm{x}+2)^2}{(1)^2}-\frac{(\mathrm{y}-2)^2}{(4)^2}=1\)
Centre \((-2,2) \& \mathrm{a}^2=1\) and \(\mathrm{b}^2=(4)^2\) Given, parabola : \(\mathrm{x}^2=4+\mathrm{y}\)
Required area,
\(A=\int_{-2}^{\sqrt{6}}\left[2-\left(x^2-4\right)\right] d x\)
\(A=\int_{-2}^{\sqrt{6}}\left(6-x^2\right) d x=\left(6 x-\frac{x^3}{3}\right)_{-2}^{\sqrt{6}}\)
\(A=\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right)\)
\(A=\frac{12 \sqrt{6}}{3}+\frac{28}{3}\)
\(A=4 \sqrt{6}+\frac{28}{3}\)
JEE Main-25.01.2023
Hyperbola
120712
If the line \(x-1=0\) is a directrix of the hyperbola \(\mathrm{kx}^2-\mathrm{y}^2=6\), then the hyperbola passes through the point
1 \((-2 \sqrt{5}, 6)\)
2 \((-\sqrt{5}, 3)\)
3 \((\sqrt{5},-2)\)
4 \((2 \sqrt{5}, 3 \sqrt{6})\)
Explanation:
C \(\mathrm{H}: \mathrm{kx}^2-\mathrm{y}^2=6\)
\(\frac{x^2}{(6 / \mathrm{k})}-\frac{\mathrm{y}^2}{6}=1\)
Now, \(\quad \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=1+\frac{6}{6 / \mathrm{k}}\)
\(\mathrm{e}=\sqrt{1+\mathrm{k}}\)
And, \(\quad \mathrm{a}=\sqrt{\frac{6}{\mathrm{k}}}\)
Equation of directrix, \(x=\frac{a}{e}=\sqrt{\frac{6}{k(k+1)}}\)
\(\Rightarrow \quad \frac{6}{\substack{\mathrm{k}(\mathrm{k}+1) \\ \mathrm{k}=2}}=1\)
Now, equation (i) is,
\(\frac{x^2}{3}-\frac{y^2}{6}=1\)
\(\Rightarrow \quad 2 x^2-y^2 =6\)From options, \((\sqrt{5},-2)\) lies on this hyperbola.
JEE Main-26.07.2022
Hyperbola
120713
Let \(H\) be the hyperbola, whose foci are \((1 \pm \sqrt{2}, 0)\) and eccentricity is \(\sqrt{2}\). Then the length of its latus rectum is :
1 3
2 \(\frac{5}{2}\)
3 2
4 \(\frac{3}{2}\)
Explanation:
C Given,
Foci: \((1 \pm \sqrt{2}, 0)\) and \(\mathrm{e}=\sqrt{2}\)
\(2 \mathrm{ae}=|(1+\sqrt{2})-(1-\sqrt{2})|=2 \sqrt{2}\)
ae \(=\sqrt{2}\)
\(\mathrm{a}=\frac{\sqrt{2}}{\mathrm{e}}=\frac{\sqrt{2}}{\sqrt{2}}\)
\(\mathrm{a}=1\)
and \(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(b^2=(1)^2(2-1)\)
\(\mathrm{b}=1\)
Letus rectum, \(l=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=2 \times \frac{(1)^2}{1}\)
\(l=2\)
JEE Main-31.01.2023
Hyperbola
120714
If \(e\) and \(e^{\prime}\) are the eccentricities of the ellipse \(5 x^2+9 y^2=45\) and the hyperbola \(5 x^2-4 y^2=45\) respectively, the ee' is equal to
120711
Let \(\mathrm{T}\) and \(\mathrm{C}\) respectively be the transverse and conjugate axes of the hyperbola \(16 x^2-y^2+64 x\) \(+4 y+44=0\). Then the area of the region above the parabola \(x^2=y+4\), below the transverse axis \(T\) and on the right of the conjugate axis \(C\) is :
1 \(4 \sqrt{6}+\frac{44}{3}\)
2 \(4 \sqrt{6}-\frac{28}{3}\)
3 \(4 \sqrt{6}+\frac{28}{3}\)
4 \(4 \sqrt{6}-\frac{44}{3}\)
Explanation:
C Given, \(16 x^2-y^2+64 x+4 y+44=0\)
\(\Rightarrow \quad 16\left(\mathrm{x}^2+4 \mathrm{x}\right)-\left(\mathrm{y}^2-4 \mathrm{y}\right)+44=0\)
\(16(\mathrm{x}+2)^2-(\mathrm{y}-2)^2=16\)
\(\Rightarrow \quad \frac{(\mathrm{x}+2)^2}{(1)^2}-\frac{(\mathrm{y}-2)^2}{(4)^2}=1\)
Centre \((-2,2) \& \mathrm{a}^2=1\) and \(\mathrm{b}^2=(4)^2\) Given, parabola : \(\mathrm{x}^2=4+\mathrm{y}\)
Required area,
\(A=\int_{-2}^{\sqrt{6}}\left[2-\left(x^2-4\right)\right] d x\)
\(A=\int_{-2}^{\sqrt{6}}\left(6-x^2\right) d x=\left(6 x-\frac{x^3}{3}\right)_{-2}^{\sqrt{6}}\)
\(A=\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right)\)
\(A=\frac{12 \sqrt{6}}{3}+\frac{28}{3}\)
\(A=4 \sqrt{6}+\frac{28}{3}\)
JEE Main-25.01.2023
Hyperbola
120712
If the line \(x-1=0\) is a directrix of the hyperbola \(\mathrm{kx}^2-\mathrm{y}^2=6\), then the hyperbola passes through the point
1 \((-2 \sqrt{5}, 6)\)
2 \((-\sqrt{5}, 3)\)
3 \((\sqrt{5},-2)\)
4 \((2 \sqrt{5}, 3 \sqrt{6})\)
Explanation:
C \(\mathrm{H}: \mathrm{kx}^2-\mathrm{y}^2=6\)
\(\frac{x^2}{(6 / \mathrm{k})}-\frac{\mathrm{y}^2}{6}=1\)
Now, \(\quad \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=1+\frac{6}{6 / \mathrm{k}}\)
\(\mathrm{e}=\sqrt{1+\mathrm{k}}\)
And, \(\quad \mathrm{a}=\sqrt{\frac{6}{\mathrm{k}}}\)
Equation of directrix, \(x=\frac{a}{e}=\sqrt{\frac{6}{k(k+1)}}\)
\(\Rightarrow \quad \frac{6}{\substack{\mathrm{k}(\mathrm{k}+1) \\ \mathrm{k}=2}}=1\)
Now, equation (i) is,
\(\frac{x^2}{3}-\frac{y^2}{6}=1\)
\(\Rightarrow \quad 2 x^2-y^2 =6\)From options, \((\sqrt{5},-2)\) lies on this hyperbola.
JEE Main-26.07.2022
Hyperbola
120713
Let \(H\) be the hyperbola, whose foci are \((1 \pm \sqrt{2}, 0)\) and eccentricity is \(\sqrt{2}\). Then the length of its latus rectum is :
1 3
2 \(\frac{5}{2}\)
3 2
4 \(\frac{3}{2}\)
Explanation:
C Given,
Foci: \((1 \pm \sqrt{2}, 0)\) and \(\mathrm{e}=\sqrt{2}\)
\(2 \mathrm{ae}=|(1+\sqrt{2})-(1-\sqrt{2})|=2 \sqrt{2}\)
ae \(=\sqrt{2}\)
\(\mathrm{a}=\frac{\sqrt{2}}{\mathrm{e}}=\frac{\sqrt{2}}{\sqrt{2}}\)
\(\mathrm{a}=1\)
and \(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(b^2=(1)^2(2-1)\)
\(\mathrm{b}=1\)
Letus rectum, \(l=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=2 \times \frac{(1)^2}{1}\)
\(l=2\)
JEE Main-31.01.2023
Hyperbola
120714
If \(e\) and \(e^{\prime}\) are the eccentricities of the ellipse \(5 x^2+9 y^2=45\) and the hyperbola \(5 x^2-4 y^2=45\) respectively, the ee' is equal to
120711
Let \(\mathrm{T}\) and \(\mathrm{C}\) respectively be the transverse and conjugate axes of the hyperbola \(16 x^2-y^2+64 x\) \(+4 y+44=0\). Then the area of the region above the parabola \(x^2=y+4\), below the transverse axis \(T\) and on the right of the conjugate axis \(C\) is :
1 \(4 \sqrt{6}+\frac{44}{3}\)
2 \(4 \sqrt{6}-\frac{28}{3}\)
3 \(4 \sqrt{6}+\frac{28}{3}\)
4 \(4 \sqrt{6}-\frac{44}{3}\)
Explanation:
C Given, \(16 x^2-y^2+64 x+4 y+44=0\)
\(\Rightarrow \quad 16\left(\mathrm{x}^2+4 \mathrm{x}\right)-\left(\mathrm{y}^2-4 \mathrm{y}\right)+44=0\)
\(16(\mathrm{x}+2)^2-(\mathrm{y}-2)^2=16\)
\(\Rightarrow \quad \frac{(\mathrm{x}+2)^2}{(1)^2}-\frac{(\mathrm{y}-2)^2}{(4)^2}=1\)
Centre \((-2,2) \& \mathrm{a}^2=1\) and \(\mathrm{b}^2=(4)^2\) Given, parabola : \(\mathrm{x}^2=4+\mathrm{y}\)
Required area,
\(A=\int_{-2}^{\sqrt{6}}\left[2-\left(x^2-4\right)\right] d x\)
\(A=\int_{-2}^{\sqrt{6}}\left(6-x^2\right) d x=\left(6 x-\frac{x^3}{3}\right)_{-2}^{\sqrt{6}}\)
\(A=\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right)\)
\(A=\frac{12 \sqrt{6}}{3}+\frac{28}{3}\)
\(A=4 \sqrt{6}+\frac{28}{3}\)
JEE Main-25.01.2023
Hyperbola
120712
If the line \(x-1=0\) is a directrix of the hyperbola \(\mathrm{kx}^2-\mathrm{y}^2=6\), then the hyperbola passes through the point
1 \((-2 \sqrt{5}, 6)\)
2 \((-\sqrt{5}, 3)\)
3 \((\sqrt{5},-2)\)
4 \((2 \sqrt{5}, 3 \sqrt{6})\)
Explanation:
C \(\mathrm{H}: \mathrm{kx}^2-\mathrm{y}^2=6\)
\(\frac{x^2}{(6 / \mathrm{k})}-\frac{\mathrm{y}^2}{6}=1\)
Now, \(\quad \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=1+\frac{6}{6 / \mathrm{k}}\)
\(\mathrm{e}=\sqrt{1+\mathrm{k}}\)
And, \(\quad \mathrm{a}=\sqrt{\frac{6}{\mathrm{k}}}\)
Equation of directrix, \(x=\frac{a}{e}=\sqrt{\frac{6}{k(k+1)}}\)
\(\Rightarrow \quad \frac{6}{\substack{\mathrm{k}(\mathrm{k}+1) \\ \mathrm{k}=2}}=1\)
Now, equation (i) is,
\(\frac{x^2}{3}-\frac{y^2}{6}=1\)
\(\Rightarrow \quad 2 x^2-y^2 =6\)From options, \((\sqrt{5},-2)\) lies on this hyperbola.
JEE Main-26.07.2022
Hyperbola
120713
Let \(H\) be the hyperbola, whose foci are \((1 \pm \sqrt{2}, 0)\) and eccentricity is \(\sqrt{2}\). Then the length of its latus rectum is :
1 3
2 \(\frac{5}{2}\)
3 2
4 \(\frac{3}{2}\)
Explanation:
C Given,
Foci: \((1 \pm \sqrt{2}, 0)\) and \(\mathrm{e}=\sqrt{2}\)
\(2 \mathrm{ae}=|(1+\sqrt{2})-(1-\sqrt{2})|=2 \sqrt{2}\)
ae \(=\sqrt{2}\)
\(\mathrm{a}=\frac{\sqrt{2}}{\mathrm{e}}=\frac{\sqrt{2}}{\sqrt{2}}\)
\(\mathrm{a}=1\)
and \(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(b^2=(1)^2(2-1)\)
\(\mathrm{b}=1\)
Letus rectum, \(l=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=2 \times \frac{(1)^2}{1}\)
\(l=2\)
JEE Main-31.01.2023
Hyperbola
120714
If \(e\) and \(e^{\prime}\) are the eccentricities of the ellipse \(5 x^2+9 y^2=45\) and the hyperbola \(5 x^2-4 y^2=45\) respectively, the ee' is equal to
120711
Let \(\mathrm{T}\) and \(\mathrm{C}\) respectively be the transverse and conjugate axes of the hyperbola \(16 x^2-y^2+64 x\) \(+4 y+44=0\). Then the area of the region above the parabola \(x^2=y+4\), below the transverse axis \(T\) and on the right of the conjugate axis \(C\) is :
1 \(4 \sqrt{6}+\frac{44}{3}\)
2 \(4 \sqrt{6}-\frac{28}{3}\)
3 \(4 \sqrt{6}+\frac{28}{3}\)
4 \(4 \sqrt{6}-\frac{44}{3}\)
Explanation:
C Given, \(16 x^2-y^2+64 x+4 y+44=0\)
\(\Rightarrow \quad 16\left(\mathrm{x}^2+4 \mathrm{x}\right)-\left(\mathrm{y}^2-4 \mathrm{y}\right)+44=0\)
\(16(\mathrm{x}+2)^2-(\mathrm{y}-2)^2=16\)
\(\Rightarrow \quad \frac{(\mathrm{x}+2)^2}{(1)^2}-\frac{(\mathrm{y}-2)^2}{(4)^2}=1\)
Centre \((-2,2) \& \mathrm{a}^2=1\) and \(\mathrm{b}^2=(4)^2\) Given, parabola : \(\mathrm{x}^2=4+\mathrm{y}\)
Required area,
\(A=\int_{-2}^{\sqrt{6}}\left[2-\left(x^2-4\right)\right] d x\)
\(A=\int_{-2}^{\sqrt{6}}\left(6-x^2\right) d x=\left(6 x-\frac{x^3}{3}\right)_{-2}^{\sqrt{6}}\)
\(A=\left(6 \sqrt{6}-\frac{6 \sqrt{6}}{3}\right)-\left(-12+\frac{8}{3}\right)\)
\(A=\frac{12 \sqrt{6}}{3}+\frac{28}{3}\)
\(A=4 \sqrt{6}+\frac{28}{3}\)
JEE Main-25.01.2023
Hyperbola
120712
If the line \(x-1=0\) is a directrix of the hyperbola \(\mathrm{kx}^2-\mathrm{y}^2=6\), then the hyperbola passes through the point
1 \((-2 \sqrt{5}, 6)\)
2 \((-\sqrt{5}, 3)\)
3 \((\sqrt{5},-2)\)
4 \((2 \sqrt{5}, 3 \sqrt{6})\)
Explanation:
C \(\mathrm{H}: \mathrm{kx}^2-\mathrm{y}^2=6\)
\(\frac{x^2}{(6 / \mathrm{k})}-\frac{\mathrm{y}^2}{6}=1\)
Now, \(\quad \mathrm{e}^2=1+\frac{\mathrm{b}^2}{\mathrm{a}^2}=1+\frac{6}{6 / \mathrm{k}}\)
\(\mathrm{e}=\sqrt{1+\mathrm{k}}\)
And, \(\quad \mathrm{a}=\sqrt{\frac{6}{\mathrm{k}}}\)
Equation of directrix, \(x=\frac{a}{e}=\sqrt{\frac{6}{k(k+1)}}\)
\(\Rightarrow \quad \frac{6}{\substack{\mathrm{k}(\mathrm{k}+1) \\ \mathrm{k}=2}}=1\)
Now, equation (i) is,
\(\frac{x^2}{3}-\frac{y^2}{6}=1\)
\(\Rightarrow \quad 2 x^2-y^2 =6\)From options, \((\sqrt{5},-2)\) lies on this hyperbola.
JEE Main-26.07.2022
Hyperbola
120713
Let \(H\) be the hyperbola, whose foci are \((1 \pm \sqrt{2}, 0)\) and eccentricity is \(\sqrt{2}\). Then the length of its latus rectum is :
1 3
2 \(\frac{5}{2}\)
3 2
4 \(\frac{3}{2}\)
Explanation:
C Given,
Foci: \((1 \pm \sqrt{2}, 0)\) and \(\mathrm{e}=\sqrt{2}\)
\(2 \mathrm{ae}=|(1+\sqrt{2})-(1-\sqrt{2})|=2 \sqrt{2}\)
ae \(=\sqrt{2}\)
\(\mathrm{a}=\frac{\sqrt{2}}{\mathrm{e}}=\frac{\sqrt{2}}{\sqrt{2}}\)
\(\mathrm{a}=1\)
and \(\mathrm{b}^2=\mathrm{a}^2\left(\mathrm{e}^2-1\right)\)
\(b^2=(1)^2(2-1)\)
\(\mathrm{b}=1\)
Letus rectum, \(l=\frac{2 \mathrm{~b}^2}{\mathrm{a}}=2 \times \frac{(1)^2}{1}\)
\(l=2\)
JEE Main-31.01.2023
Hyperbola
120714
If \(e\) and \(e^{\prime}\) are the eccentricities of the ellipse \(5 x^2+9 y^2=45\) and the hyperbola \(5 x^2-4 y^2=45\) respectively, the ee' is equal to
