From a point (C, 0),three normals are drawn t | vedclass

Name: Exam Paper Generator | JEE NEET Paper Generator | Vedclass
Brand: Vedclass
Rating: 5 (35 reviews)

Question

(C) The equation of the parabola is $y^2 = x$. Comparing this with $y^2 = 4ax$,we get $4a = 1$,so $a = \frac{1}{4}$.Any normal to the parabola $y^2 =

Vedclass · Accepted Answer

$C > \frac{1}{2}$

Vedclass · Answer

(C) The equation of the parabola is $y^2 = x$. Comparing this with $y^2 = 4ax$,we get $4a = 1$,so $a = \frac{1}{4}$.Any normal to the parabola $y^2 = 4ax$ is given by $y = mx - 2am - am^3$.Substituting $a = \frac{1}{4}$,the equation becomes $y = mx - \frac{m}{2} - \frac{m^3}{4}$.If this normal passes through the point $(C, 0)$,then $0 = mC - \frac{m}{2} - \frac{m^3}{4}$.This simplifies to $m(C - \frac{1}{2} - \frac{m^2}{4}) = 0$.One root is $m = 0$,which corresponds to the normal along the $x$-axis.For three distinct normals to be drawn,the quadratic equation $\frac{m^2}{4} = C - \frac{1}{2}$ must have two distinct non-zero real roots.This requires $C - \frac{1}{2} > 0$,which implies $C > \frac{1}{2}$.

From a point $(C, 0)$,three normals are drawn to the parabola $y^2=x$. Then,

Similar Questions

The length of the subnormal to the parabola $y^2 = 4ax$ at any point is equal to

If a point $P$ moves such that its distances from the point $A(1, 1)$ and the line $x+y+2=0$ are equal,then the locus of $P$ is

The equation of the directrix of the parabola whose focus is $(0,0)$ and the tangent at the vertex is $x-y+1=0$ is

The distance of the point $(6, -2 \sqrt{2})$ from the common tangent $y = mx + c$ $(m > 0)$ of the curves $x = 2y^2$ and $x = 1 + y^2$ is

If the line $y = mx + c$ is a tangent to the parabola $y^2 = 4a(x + a)$,then $ma + \frac{a}{m}$ is equal to

Vedclass Test Series

Exam Paper Generator

Online Exam Module