At what point on the parabola y^2 = 4x does t | vedclass

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

Question

(D) The equation of a normal to the parabola $y^2 = 4ax$ at the point $(am^2, -2am)$ is $y = mx - 2am - am^3$.For the parabola $y^2 = 4x$,we have $a =

Vedclass · Accepted Answer

$(1, -2)$

Vedclass · Answer

(D) The equation of a normal to the parabola $y^2 = 4ax$ at the point $(am^2, -2am)$ is $y = mx - 2am - am^3$.For the parabola $y^2 = 4x$,we have $a = 1$. Thus,the normal at $(m^2, -2m)$ is $y = mx - 2m - m^3$.The slope of this normal is $m$.If the normal makes equal angles with the coordinate axes,its inclination $	heta$ must be $45^{\circ}$ or $135^{\circ}$,so the slope $m = 	an(45^{\circ}) = 1$ or $m = 	an(135^{\circ}) = -1$.Case $1$: If $m = 1$,the point is $(1^2, -2(1)) = (1, -2)$.Case $2$: If $m = -1$,the point is $((-1)^2, -2(-1)) = (1, 2)$.Comparing with the given options,the correct point is $(1, -2)$.

At what point on the parabola $y^2 = 4x$ does the normal make equal angles with the coordinate axes?

Similar Questions

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

If the equation of the parabola with vertex $V \left(\frac{3}{2}, 3\right)$ and the directrix $x + 2y = 0$ is $\alpha x^2 + \beta y^2 - \gamma xy - 30x - 60y + 225 = 0$,then $\alpha + \beta + \gamma$ is equal to:

$A$ normal chord $PQ$ drawn at a point $P$ on the parabola $y^2 = 5x$ subtends a right angle at the vertex. If $P$ lies in the first quadrant,then the other end $Q$ of the normal chord is

$A$ point $P(x, y)$ moves such that its distance from the point $(a, 0)$ is always equal to its distance from the line $x + a = 0$. The locus of the point is:

The value of $\lambda$ for which the curve $(7x+5)^{2}+(7y+3)^{2}=\lambda^{2}(4x+3y-24)^{2}$ represents a parabola is

Vedclass Test Series

Exam Paper Generator

Online Exam Module