The number of normals drawn to the parabola y | vedclass

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

Question

(B) The equation of the parabola is $y^2=4ax$,where $a=1$. The point $(1,0)$ is the focus of the parabola.Any normal to the parabola $y^2=4ax$ at poin

Vedclass · Accepted Answer

$1$

Vedclass · Answer

(B) The equation of the parabola is $y^2=4ax$,where $a=1$. The point $(1,0)$ is the focus of the parabola.Any normal to the parabola $y^2=4ax$ at point $(at^2, 2at)$ is given by the equation $y = -tx + 2at + at^3$.If this normal passes through the point $(h, k) = (1, 0)$,then $0 = -t(1) + 2(1)t + (1)t^3$.This simplifies to $0 = t + t^3$,or $t(1 + t^2) = 0$.The real solution for $t$ is $t=0$.For $t=0$,the normal is $y = 0$,which is the $x$-axis.Thus,only $1$ normal can be drawn from the point $(1,0)$ to the parabola $y^2=4x$.

The number of normals drawn to the parabola $y^2=4x$ from the point $(1,0)$ is

Similar Questions

The focus of the parabola $4y^2 - 6x - 4y = 5$ is

Let $y^{2}=12x$ be the parabola with its vertex at $O(0,0)$. Let $P$ be a point on the parabola and $A$ be a point on the $x$-axis such that $\angle OPA=90^{\circ}$. Then the locus of the centroid of such triangles $OPA$ is:

The radius of the smallest circle which touches the parabolas $y = x^2 + 2$ and $x = y^2 + 2$ is

The coordinates of a point on the parabola $y^2 = 8x$ whose focal distance is $4$ are:

Let $A$ and $B$ be two distinct points on the parabola $y^2 = 4x$. If the axis of the parabola touches a circle of radius $r$ having $AB$ as its diameter,then the slope of the line joining $A$ and $B$ can be

Vedclass Test Series

Exam Paper Generator

Online Exam Module