Let a parabola P be such that its vertex and | vedclass

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

Question

(C) The vertex of the parabola is $V(2,0)$ and the focus is $F(4,0)$.Thus,the distance $VF = a = 4 - 2 = 2$.The equation of the parabola is $(y - 0)^2

Vedclass · Accepted Answer

$16$

Vedclass · Answer

(C) The vertex of the parabola is $V(2,0)$ and the focus is $F(4,0)$.Thus,the distance $VF = a = 4 - 2 = 2$.The equation of the parabola is $(y - 0)^2 = 4a(x - 2)$,which simplifies to $y^2 = 8(x - 2)$.Let the tangent from the origin $O(0,0)$ to the parabola be $y = mx + c$. Since it passes through $(0,0)$,$c = 0$,so $y = mx$.Substituting $y = mx$ into $y^2 = 8x - 16$,we get $(mx)^2 = 8x - 16$,or $m^2x^2 - 8x + 16 = 0$.For tangency,the discriminant $D = (-8)^2 - 4(m^2)(16) = 0$.$64 - 64m^2 = 0$ $\Rightarrow m^2 = 1$ $\Rightarrow m = \pm 1$.The points of contact $S$ and $R$ are found by solving $x^2 - 8x + 16 = 0$ (for $m=1$) and $(-x)^2 - 8x + 16 = 0$ (for $m=-1$).Both give $(x-4)^2 = 0$,so $x = 4$.For $x = 4$,$y = \pm 4$. Thus,the points are $R(4, 4)$ and $S(4, -4)$.The base of $	riangle SOR$ is the segment $RS$,which has length $4 - (-4) = 8$.The height of $	riangle SOR$ with respect to base $RS$ is the distance from $O(0,0)$ to the line $x = 4$,which is $4$.Area of $	riangle SOR = \frac{1}{2} 	imes 	ext{base} 	imes 	ext{height} = \frac{1}{2} 	imes 8 	imes 4 = 16 	ext{ sq. units}$.

Let a parabola $P$ be such that its vertex and focus lie on the positive $x$-axis at a distance $2$ and $4$ units from the origin,respectively. If tangents are drawn from $O(0,0)$ to the parabola $P$ which meet $P$ at $S$ and $R$,then the area (in $sq. \text{ units}$) of $\triangle SOR$ is equal to:

Similar Questions

If $y = 2x - 3$ is a tangent to the parabola $y^2 = 4a(x - \frac{1}{3})$,then $a$ is equal to:

Find the equation of the parabola with focus $(2, 0)$ and directrix $x = -2$.

From the point $(-1, 2)$, tangents are drawn to the parabola $y^2 = 4x$. The area of the triangle formed by the chord of contact and the tangents is: (in $\sqrt{2}$)

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:

The normal meets the parabola $y^2 = 4ax$ at a point where the abscissa is equal to the ordinate. Find this point.

Vedclass Test Series

Exam Paper Generator

Online Exam Module