1
JEE Main 2022 (Online) 25th July Evening Shift
+4
-1

If the ellipse $$\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$$ meets the line $$\frac{x}{7}+\frac{y}{2 \sqrt{6}}=1$$ on the $$x$$-axis and the line $$\frac{x}{7}-\frac{y}{2 \sqrt{6}}=1$$ on the $$y$$-axis, then the eccentricity of the ellipse is

A
$$\frac{5}{7}$$
B
$$\frac{2 \sqrt{6}}{7}$$
C
$$\frac{3}{7}$$
D
$$\frac{2 \sqrt{5}}{7}$$
2
JEE Main 2022 (Online) 25th July Evening Shift
+4
-1

The tangents at the points $$A(1,3)$$ and $$B(1,-1)$$ on the parabola $$y^{2}-2 x-2 y=1$$ meet at the point $$P$$. Then the area (in unit $${ }^{2}$$ ) of the triangle $$P A B$$ is:

A
4
B
6
C
7
D
8
3
JEE Main 2022 (Online) 25th July Evening Shift
+4
-1

Let the foci of the ellipse $$\frac{x^{2}}{16}+\frac{y^{2}}{7}=1$$ and the hyperbola $$\frac{x^{2}}{144}-\frac{y^{2}}{\alpha}=\frac{1}{25}$$ coincide. Then the length of the latus rectum of the hyperbola is:

A
$$\frac{32}{9}$$
B
$$\frac{18}{5}$$
C
$$\frac{27}{4}$$
D
$$\frac{27}{10}$$
4
JEE Main 2022 (Online) 30th June Morning Shift
+4
-1

Let the eccentricity of the ellipse $${x^2} + {a^2}{y^2} = 25{a^2}$$ be b times the eccentricity of the hyperbola $${x^2} - {a^2}{y^2} = 5$$, where a is the minimum distance between the curves y = ex and y = logex. Then $${a^2} + {1 \over {{b^2}}}$$ is equal to :

A
$${3 \over 2}$$
B
$${5 \over 2}$$
C
3
D
5
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