1
IIT-JEE 1989
MCQ (Single Correct Answer)
+2
-0.5
The lines 2x - 3y = 5 and 3x - 4y = 7 are diameters of a circle of area 154 sq. units. Then the equation of this circle is
A
$${x^2} + {y^2} + 2x - 2y = 62$$
B
$${x^2} + {y^2} + 2x - 2y = 47$$
C
$${x^2} + {y^2} - 2x + 2y = 47$$
D
$${x^2} + {y^2} - 2x + 2y = 62$$c
2
IIT-JEE 1989
Subjective
+2
-0
If $$\left( {{m_i},{1 \over {{m_i}}}} \right),\,{m_i}\, > \,0,\,i\, = 1,\,2,\,3,\,4$$ are four distinct points on a circle, then show that $${m_1}\,{m_2}\,{m_3}\,{m_4}\, = 1$$
3
IIT-JEE 1989
Subjective
+2
-0
If $$x = \sec \theta - \cos \theta $$ and $$y = {\sec ^n}\theta - {\cos ^n}\theta $$, then show
that $$\left( {{x^2} + 4} \right){\left( {{{dy} \over {dx}}} \right)^2} = {n^2}\left( {{y^2} + 4} \right)$$
4
IIT-JEE 1989
Subjective
+5
-0
$$ABC$$ is a triangular park with $$AB=AC=100$$ $$m$$. A television tower stands at the midpoint of $$BC$$. The angles of elevetion of the top of the tower at $$A, B, C$$ are 45$$^ \circ $$, 60$$^ \circ $$, 60$$^ \circ $$, respectively. Find the height of the tower.
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