If the normal to the parabola y^2 = 4ax inter | vedclass

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

Question

(A) The equation of a normal to the parabola $y^2 = 4ax$ is given by $y = mx - 2am - am^3$.If the normal passes through a point on the axis at a dista

Vedclass · Accepted Answer

$A.P.$

Vedclass · Answer

(A) The equation of a normal to the parabola $y^2 = 4ax$ is given by $y = mx - 2am - am^3$.If the normal passes through a point on the axis at a distance of $4a$ from the vertex $(0, 0)$,the point is $(4a, 0)$.Substituting $(4a, 0)$ into the equation: $0 = m(4a) - 2am - am^3$.$0 = 2am - am^3$.Since $a 
eq 0$,we have $2m - m^3 = 0$,which implies $m(2 - m^2) = 0$.Thus,the possible slopes are $m = 0$ and $m^2 = 2$,so $m = 0, \sqrt{2}, -\sqrt{2}$.The slopes are $-\sqrt{2}, 0, \sqrt{2}$,which form an $A.P.$ with a common difference of $\sqrt{2}$.

If the normal to the parabola $y^2 = 4ax$ intersects the axis of the parabola at a distance of $4a$ from the vertex,then the slopes of the normal are in which progression?

Similar Questions

For the parabola $y^2+6y-2x+5=0$:
$(I)$ The vertex is $(-2,-3)$
$(II)$ The directrix is $y+3=0$
Which of the following is correct?

What is the locus of the vertices $(x, y)$ of the parabolas $y = \frac{a^3x^2}{3} + \frac{a^2x}{2} - 2a$?

The angle of intersection between the curves $x^2 = 4(y + 1)$ and $x^2 = -4(y + 1)$ is

$A$ tangent and a normal are drawn at the point $P(2, -4)$ on the parabola $y^{2} = 8x$,which meet the directrix of the parabola at the points $A$ and $B$ respectively. If $Q(a, b)$ is a point such that $AQBP$ is a square,then $2a + b$ is equal to:

The parabola with directrix $x+2y-1=0$ and focus $(1,0)$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module