JEE MATHEMATICS MCQ: 3D GEOMETRY SYSTEM

1 Easy

1 of 1

Application +4 | -1

Let

P

be the point (1, 0) and

{\cal Q}

be the point on

{y^2} = 8x

. The locus of mid point of

P{\cal Q}

View answer and explanation

Correct option: D

{y^2} - 4x + 2 = 0

Option D

{y^2} - 4x + 2 = 0

is correct because it is the locus of the midpoint derived from the given parabola equation.

More Questions from 3D GEOMETRY SYSTEM

Practice All Chapter MCQs →

\left( { - 10,\;8} \right)

\left( { - 6,\;12} \right)

\left( {4,\; - 2} \right)

\left( { - 2,\;4} \right)

y z

(1,-2,4)

(3,-4

5)

p

2 x+y-3=0

3 x-y-2=0

x-p y+5=0

O\left( {0,0} \right),A\left( {a,0} \right)

B\left( {0,a} \right).

A(1,0,3)

B(4,7,1)

C(3,5,3)

\frac{x}{1}=\frac{y}{2}=\frac{z}{3}, \frac{x}{2}=\frac{y}{1}=\frac{z}{3}

\frac{x-1}{1}=\frac{2-y}{2}, z=3