The condition that the parabolas y^2 = 4ax an | vedclass

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

Question

(B) The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$  $(1)$.The equation of a normal to the parabola $y^2

Vedclass · Accepted Answer

$\frac{b}{a - c} > 2$

Vedclass · Answer

(B) The equation of a normal to the parabola $y^2 = 4ax$ with slope $m$ is $y = mx - 2am - am^3$  $(1)$.The equation of a normal to the parabola $y^2 = 4c(x - b)$ with slope $m$ is $y = m(x - b) - 2cm - cm^3$,which simplifies to $y = mx - (b + 2c)m - cm^3$  $(2)$.For a common normal,the equations $(1)$ and $(2)$ must represent the same line. Equating the constant terms:$-2am - am^3 = -(b + 2c)m - cm^3$$(c - a)m^3 + (b + 2c - 2a)m = 0$Since the common normal is not the $x$-axis,$m 
eq 0$. Dividing by $m$:$(c - a)m^2 + (b + 2c - 2a) = 0$$m^2 = \frac{2a - 2c - b}{c - a} = \frac{2(a - c) - b}{c - a} = -2 + \frac{b}{a - c}$.Since $m^2 > 0$ for a real normal,we have:$-2 + \frac{b}{a - c} > 0$$\frac{b}{a - c} > 2$.

The condition that the parabolas $y^2 = 4ax$ and $y^2 = 4c(x - b)$ have a common normal other than the $x$-axis ($a, b, c$ being distinct positive real numbers) is

Similar Questions

If the tangent to the curve $y^2 = 4x$ at point $(1, 2)$ cuts the coordinate axes at points $A$ and $B$,then the area of $\Delta AOB$ is (where $'O'$ is the origin).

Let $PQ$ be a chord of the parabola $y^2=12x$ and the midpoint of $PQ$ be at $(4,1)$. Then,which of the following points lies on the line passing through the points $P$ and $Q$?

The directrix of the parabola $2 y^2+25 x=0$ is $........$

The length of the chord of the parabola $y^2 = x$ which is bisected at the point $(2, 1)$ is

The coordinates of the focus of the parabola $5x^2 = -12y$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module