Show that the point (2,3) lies outside the parabola `y^2=3x`.

Verified by Experts

The correct Answer is:

3x

Similar Questions

Explore conceptually related problems

Find the position of the point (-2,2) with respect to the parabola y^2-4y+9x+13=0 .

Show that the equation of tangent to the parabola y^(2) = 4ax " at " (x_(1), y_(1)) " is " y y_(1)= 2a(x + x_(1))

Find the value of 'a' so that the point (3, a) lies on the line 2x - 3y = 5.

Does the point (-2.5, 3.5) lie inside , outside or on the circle x^(2)+y^(2)=25 ?

The normal to the parabola y^(2)=8 x at the point (2,4) meets the parabola again at the point

The focus of the parabola y=2 x^(2)+x is

The vertex of the parabola y^(2)=x+4y+3 is

Focus of the parabola (y-2)^(2)=20(x+3) is

The point (a,2a) is an interior point of the region bounded by the parabola y^(2)=16x and the double ordinate through the focus the range of 'a' is