y = 3x - 2 is a straight line touching the pa | vedclass

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

Question

(A) The given parabola is $(y - 3)^2 = 12(x - 2)$.Comparing this with $(y - k)^2 = 4a(x - h)$,we get $h = 2, k = 3$,and $4a = 12$,so $a = 3$.The direc

Vedclass · Accepted Answer

$(-1, -5)$

Vedclass · Answer

(A) The given parabola is $(y - 3)^2 = 12(x - 2)$.Comparing this with $(y - k)^2 = 4a(x - h)$,we get $h = 2, k = 3$,and $4a = 12$,so $a = 3$.The directrix of the parabola is $x = h - a = 2 - 3 = -1$.It is a known property of parabolas that the locus of the intersection of two perpendicular tangents is the directrix.Since the two tangents are perpendicular and meet at point $P$,point $P$ must lie on the directrix $x = -1$.Substituting $x = -1$ into the equation of the first tangent $y = 3x - 2$:$y = 3(-1) - 2 = -3 - 2 = -5$.Therefore,the point $P$ is $(-1, -5)$.

List-$I$	List-$II$
$I$. Vertex	$A$. $8$
$II$. Length of latus rectum	$B$. $(\frac{29}{10}, \frac{-38}{10})$
$III$. Directrix	$C$. $3x+4y-1=0$
$IV$. One end of the latus rectum	$D$. $(\frac{-2}{5}, \frac{-16}{5})$
	$E$. $6$

$y = 3x - 2$ is a straight line touching the parabola $(y - 3)^2 = 12(x - 2)$. If a line drawn perpendicular to this line at point $P$ on it touches the given parabola,then the point $P$ is:

Similar Questions

If $P$ is a point on the parabola $y=x^{2}+4$ which is closest to the straight line $y =4 x -1,$ then the coordinates of $P$ are:

If the normal to the parabola $y^2 = 4ax$ at the point with parameter $t_1$ cuts the parabola again at the point with parameter $t_2$,then:

The equation of the parabola with its vertex at $(1, 1)$ and focus at $(3, 1)$ is

Find the equation of the common tangent to the parabolas $y^2 = 2x$ and $x^2 = 16y$.

Vedclass Test Series

Exam Paper Generator

Online Exam Module