Let E denote the parabola y^2=8x. Let P=(-2,4 | vedclass

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

Question

(D) The parabola is $y^2=8x$,so $4a=8$,which gives $a=2$. The focus $F$ is $(2,0)$ and the directrix is $x=-2$.Point $P=(-2,4)$ lies on the directrix

Vedclass · Accepted Answer

$A, B, D$

Vedclass · Answer

(D) The parabola is $y^2=8x$,so $4a=8$,which gives $a=2$. The focus $F$ is $(2,0)$ and the directrix is $x=-2$.Point $P=(-2,4)$ lies on the directrix $x=-2$.It is a known property that the tangents drawn from a point on the directrix to a parabola are perpendicular to each other,and the chord of contact $QQ^{\prime}$ passes through the focus $F$.Since $PQ$ and $PQ^{\prime}$ are tangents from $P$ to the parabola,$\angle QPQ^{\prime} = 90^{\circ}$,so $(B)$ is $TRUE$.The chord of contact $QQ^{\prime}$ passes through the focus $F$,so $(D)$ is $TRUE$.The distance $PF = \sqrt{(2 - (-2))^2 + (0 - 4)^2} = \sqrt{4^2 + (-4)^2} = \sqrt{16+16} = \sqrt{32} = 4\sqrt{2}$. Thus,$(C)$ is $FALSE$.For a parabola,the angle subtended by the chord of contact at the focus is $180^{\circ}$ if the tangents are perpendicular,but the triangle $PFQ$ is right-angled at $F$ because the tangent at $Q$ makes an angle with the focal radius $FQ$ such that $\angle FQP = 90^{\circ}$ is not generally true,but here the geometry confirms $\angle PFQ = 90^{\circ}$ is not the case,however,$\angle FQP = 90^{\circ}$ is a property of tangents. Specifically,the triangle $PFQ$ is right-angled at $Q$ because the tangent at $Q$ is perpendicular to the line joining the focus to the point of contact. Thus,$(A)$ is $TRUE$.Therefore,the correct statements are $(A), (B),$ and $(D)$.

$A. \ 2x-5=0$	$I. \ \text{Vertex}$
$B. \ (\frac{3}{2}, -1)$	$II. \ \text{Focus}$
$C. \ y+1=0$	$III. \ \text{Equation of directrix}$
$D. \ (2, -1)$	$IV. \ \text{Equation of the axis}$
	$V. \ \text{Equation of the Latus rectum}$

Similar Questions

The locus of the vertices of the family of parabolas $y = \frac{a^3 x^2}{3} + \frac{a^2 x}{2} - 2a$ is

If the point on the curve $y^{2}=6x$,nearest to the point $\left(3, \frac{3}{2}\right)$ is $(\alpha, \beta)$,then $2(\alpha+\beta)$ is equal to $.....$

At which point does the line $x + y = 1$ intersect the parabola $y = x - x^2$?

The length of the latus rectum of the parabola $4y^2 + 2x - 20y + 17 = 0$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module