Two parabolas y^2 = 4a(x - l_1) and x^2 = 4a( | vedclass

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

Question

(C) Let the point of contact be $(x_1, y_1)$.For the first parabola $y^2 = 4a(x - l_1)$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} =

Vedclass · Accepted Answer

$xy = 4a^2$

Vedclass · Answer

(C) Let the point of contact be $(x_1, y_1)$.For the first parabola $y^2 = 4a(x - l_1)$,differentiating with respect to $x$ gives $2y \frac{dy}{dx} = 4a$,so $\frac{dy}{dx} = \frac{2a}{y_1}$.For the second parabola $x^2 = 4a(y - l_2)$,differentiating with respect to $x$ gives $2x = 4a \frac{dy}{dx}$,so $\frac{dy}{dx} = \frac{x_1}{2a}$.Since the parabolas touch at $(x_1, y_1)$,their slopes must be equal at this point.Therefore,$\frac{2a}{y_1} = \frac{x_1}{2a}$.This implies $x_1 y_1 = 4a^2$.Thus,the locus of the point of contact is $xy = 4a^2$.

Two parabolas $y^2 = 4a(x - l_1)$ and $x^2 = 4a(y - l_2)$ always touch one another,where $l_1$ and $l_2$ are variable. The locus of their point of contact has the equation:

Similar Questions

Let $P$ be the parabola,whose focus is $(-2, 1)$ and directrix is $2x + y + 2 = 0$. Then the sum of the ordinates of the points on $P$,whose abscissa is $-2$,is

Let the tangent and normal at any point $P(at^2, 2at)$, $(a > 0)$, on the parabola $y^2 = 4ax$ meet the axis of the parabola at $T$ and $G$ respectively. Then the radius of the circle through $P, T$ and $G$ is

Let chord $PQ$ of length $3\sqrt{13}$ of the parabola $y^2 = 12x$ be such that the ordinates of points $P$ and $Q$ are in the ratio $1:2$. If the chord $PQ$ subtends an angle $\alpha$ at the focus of the parabola, then $\sin \alpha$ is equal to:

Find the equation of the tangent at the vertex of the parabola $4y^2 + 6x = 8y + 7$.

If the line $y = 2x + k$ is a normal to the parabola $y^2 = 4x$ at the point $(t^2, 2t)$,then:

Vedclass Test Series

Exam Paper Generator

Online Exam Module