1
JEE Main 2022 (Online) 26th July Evening Shift
+4
-1
Out of Syllabus

The acute angle between the pair of tangents drawn to the ellipse $$2 x^{2}+3 y^{2}=5$$ from the point $$(1,3)$$ is :

A
$$\tan ^{-1}\left(\frac{16}{7 \sqrt{5}}\right)$$
B
$$\tan ^{-1}\left(\frac{24}{7 \sqrt{5}}\right)$$
C
$$\tan ^{-1}\left(\frac{32}{7 \sqrt{5}}\right)$$
D
$$\tan ^{-1}\left(\frac{3+8 \sqrt{5}}{35}\right)$$
2
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
$$\frac{2 \sqrt{5}}{7}$$
3
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
4
JEE Main 2022 (Online) 27th June Morning Shift
+4
-1

Let the eccentricity of an ellipse $${{{x^2}} \over {{a^2}}} + {{{y^2}} \over {{b^2}}} = 1$$, $$a > b$$, be $${1 \over 4}$$. If this ellipse passes through the point $$\left( { - 4\sqrt {{2 \over 5}} ,3} \right)$$, then $${a^2} + {b^2}$$ is equal to :

A
29
B
31
C
32
D
34
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