1
IIT-JEE 2011 Paper 2 Offline
+3
-0.75
Let $$P(6, 3)$$ be a point on the hyperbola $${{{x^2}} \over {{a^2}}} - {{{y^2}} \over {{b^2}}} = 1$$. If the normal at the point $$P$$ intersects the $$x$$-axis at $$(9, 0)$$, then the eccentricity of the hyperbola is
A
$$\sqrt {{5 \over 2}}$$
B
$$\sqrt {{3 \over 2}}$$
C
$${\sqrt 2 }$$
D
$${\sqrt 3 }$$
2
IIT-JEE 2010 Paper 1 Offline
+4
-1
The circle $${x^2} + {y^2} - 8x = 0$$ and hyperbola $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ intersect at the points $$A$$ and $$B$$.

Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

A
$$2x - \sqrt {5y} - 20 = 0$$
B
$$2x - \sqrt {5y} + 4 = 0$$
C
$$3x - 4y + 8 = 0$$
D
$$4x - 3y + 4 = 0$$
3
IIT-JEE 2010 Paper 1 Offline
+4
-1
The circle $${x^2} + {y^2} - 8x = 0$$ and hyperbola $${{{x^2}} \over 9} - {{{y^2}} \over 4} = 1$$ intersect at the points $$A$$ and $$B$$.

Equation of the circle with $$AB$$ as its diameter is

A
$${x^2} + {y^2} - 12x + 24 = 0$$
B
$${x^2} + {y^2} + 12x + 24 = 0$$
C
$${x^2} + {y^2} + 24x - 12 = 0$$
D
$${x^2} + {y^2} - 24x - 12 = 0$$
4
IIT-JEE 2008 Paper 2 Offline
+3
-1
Consider a branch of the hyperbola $${x^2} - 2{y^2} - 2\sqrt 2 x - 4\sqrt 2 y - 6 = 0$$\$

with vertex at the point $$A$$. Let $$B$$ be one of the end points of its latus rectum. If $$C$$ is the focus of the hyperbola nearest to the point $$A$$, then the area of the triangle $$ABC$$ is

A
$$1 - \sqrt {{2 \over 3}}$$
B
$$\sqrt {{3 \over 2}} - 1$$
C
$$1 + \sqrt {{2 \over 3}}$$
D
$$\sqrt {{3 \over 2}} + 1$$
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