Qus : 1
2
A point P in the first quadrant, lies on y 2 = 4 a x , a > 0, and keeps a distance of 5a units from its focus. Which of the following points lies on the locus of P?
1 (1,0) 2 (1,1) 3 (0,2) 4 (2,0) Go to Discussion
Solution
Locus of Point on Parabola
Given: Point on parabola y 2 = 4 a x is at distance 5 a from focus ( a , 0 ) .
Distance Equation:
( x − a ) 2 + y 2 = 25 a 2 ⇒ ( x − a ) 2 + 4 a x = 25 a 2 ⇒ x 2 + 2 a x − 24 a 2 = 0
Solving gives: x = 4 a , y = 4 a
✅ Final Answer:
( 4 a , 4 a )
Qus : 2
3 A circle touches the x–axis and also touches the circle with centre (0, 3) and radius 2. The locus of the centre of the circle is
1 a circle 2 an ellipse 3 a parabola 4 a hyperbola Go to Discussion
Solution Qus : 3
2 The two parabolas y 2 = 4 a ( x + c ) and y 2 = 4 b x , a > b > 0 cannot
have a common normal unless
1 c > 2 ( a + b ) 2 c > ( a − b ) 3 c < 2 ( a − b ) 4 c < 2 a − b Go to Discussion
Solution Qus : 4
2 Coordinate of the focus of the parabola 4 y 2 + 12 x − 20 y + 67 = 0 is
1 ( − 5 4 , 17 2 ) 2 ( − 17 2 , 5 4 ) 3 ( − 17 4 , 5 2 ) 4 ( − 5 2 , 17 4 ) Go to Discussion
Solution Qus : 5
4 An equilateral triangle is inscribed in the parabola y 2 = 4 a x , such that one of the vertices of the triangle
coincides with the vertex of the parabola. The length of the side of the triangle is:
1 a √ 3 2 2 a √ 3 3 4 a √ 3 4 8 a √ 3 Go to Discussion
Solution Qus : 6
2 The locus of the mid points of all chords of the parabola y 2 = 4 x
which are drawn through its
vertex, is
1 y 2 = 8 x 2 y 2 = 2 x 3 x 2 + 4 y 2 = 16 4 x 2 = 2 y Go to Discussion
Solution Qus : 7
3 If x = 1 is the directrix of the parabola y 2 = k x − 8 , then k is:
1 1 8 2 8 3 4 4 1 4 Go to Discussion
Solution Qus : 8
4 A normal to the curve x 2 = 4 y passes through the point (1, 2). The distance of the origin from the
normal is
1 √ 2 2 2 √ 2 3 1 √ 2 4 3 √ 2 Go to Discussion
Solution Qus : 9
2 The equation of the tangent at any point of curve x = a c o s 2 t , y = 2 √ 2 a s i n t with m as its slope is
1 y = m x + a ( m − 1 m ) 2 y = m x − a ( m + 1 m ) 3 y = m x + a ( a + 1 a ) 4 y = a m x + a ( m − 1 m ) Go to Discussion
Solution Qus : 10
2
The locus of the mid-point of all chords of the parabola y 2 = 4 x which are drawn through its vertex is
1 y 2 = 8 x 2 y 2 = 2 x 3 x 2 + 4 y 2 = 16 4 x 2 = 2 y Go to Discussion
Solution
Locus of Midpoint of Chords
Given Parabola: y 2 = 4 x
Condition: Chords pass through the vertex ( 0 , 0 )
Let the other end of the chord be ( x 1 , y 1 ) , so the midpoint is:
M = ( x 1 2 , y 1 2 ) = ( h , k )
Since the point lies on the parabola: y 2 1 = 4 x 1
⇒ ( 2 k ) 2 = 4 ( 2 h )
⇒ 4 k 2 = 8 h
⇒ k 2 = 2 h
✅ Locus of midpoints:
y 2 = 2 x
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