Let P(4, -4) and Q(9, 6) be two points on the | vedclass

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

Question

(B) The area of $\Delta PXQ$ is maximum when the tangent at $X$ is parallel to the chord $PQ$.The slope of the chord $PQ$ is $m = \frac{6 - (-4)}{9 -

Vedclass · Accepted Answer

$\frac{125}{4}$

Vedclass · Answer

(B) The area of $\Delta PXQ$ is maximum when the tangent at $X$ is parallel to the chord $PQ$.The slope of the chord $PQ$ is $m = \frac{6 - (-4)}{9 - 4} = \frac{10}{5} = 2$.The equation of the parabola is $y^2 = 4x$,so $2y \frac{dy}{dx} = 4$,which gives $\frac{dy}{dx} = \frac{2}{y}$.Setting the slope of the tangent equal to the slope of the chord: $\frac{2}{y} = 2 \Rightarrow y = 1$.Since $X$ lies on $y^2 = 4x$,for $y = 1$,we have $1^2 = 4x \Rightarrow x = \frac{1}{4}$. Thus,$X = (\frac{1}{4}, 1)$.The area of $\Delta PXQ$ with vertices $P(4, -4)$,$Q(9, 6)$,and $X(\frac{1}{4}, 1)$ is given by the determinant formula:Area $= \frac{1}{2} |x_P(y_X - y_Q) + x_X(y_Q - y_P) + x_Q(y_P - y_X)|$Area $= \frac{1}{2} |4(1 - 6) + \frac{1}{4}(6 - (-4)) + 9(-4 - 1)|$Area $= \frac{1}{2} |4(-5) + \frac{1}{4}(10) + 9(-5)|$Area $= \frac{1}{2} |-20 + 2.5 - 45| = \frac{1}{2} |-62.5| = 31.25 = \frac{125}{4}$ sq. units.

Let $P(4, -4)$ and $Q(9, 6)$ be two points on the parabola $y^2 = 4x$. Let $X$ be any point on the arc $POQ$ of this parabola,where $O$ is the vertex,such that the area of $\Delta PXQ$ is maximum. Then this maximum area (in sq. units) is

Similar Questions

If a normal drawn to the parabola $y^2 = 4ax$ at the point $(a, 2a)$ meets the parabola again at $(at^2, 2at)$,then the value of $t$ is:

Let the line $y - \sqrt{3}x + 3 = 0$ cut the parabola $2y^2 = 2x + 3$ at $A$ and $B$. If $P(\sqrt{3}, 0)$,then the value of $|PA - PB|$ is [where $PA$ denotes the distance between points $P$ and $A$].

The equations of two sides of a variable triangle are $x = 0$ and $y = 3$,and its third side is a tangent to the parabola $y^2 = 6x$. The locus of its circumcentre is:

Find the angle $\alpha$ between the tangents drawn from the point $(-2, -1)$ to the parabola $y^2 = 4x$. What is the value of $\tan \alpha$?

If $mx - y + c = 0$ is a normal at a point $P$ on the parabola $y^2 = 16x$ and the focal distance of $P$ is $40$ units,then $|c| =$

Vedclass Test Series

Exam Paper Generator

Online Exam Module