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

Question

y=3 x-2 is a straight line touching the parabola (y-3)2=12(x-2). If a line drawn perpendicular to this line at P on it, touches the given parabola, then the point P is (A) (-1,-5) (B) (-1,5) (C) (-2,-8) (D) (2,4)

Tardigrade Admin · Accepted Answer

According to the given information, Angle between the tangents is 90° . ⇒ The point on the directrix Here, x-2=-3 x=-1 is directrix Given tangent is y=3 x-2 ∴ y=3(-1)-2=-5 ∴ Point P is (-1,-5)

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

$(- 1, - 5)$

$(- 1, 5)$

$(- 2, - 8)$

$(2, 4)$

According to the given information, Angle between the tangents is $9 0^{\circ} .$
$\Rightarrow$ The point on the directrix
Here, $x - 2 = - 3$
$x = - 1$ is directrix
Given tangent is $y = 3 x - 2$
$∴ y = 3 (- 1) - 2 = - 5$
$∴$ Point $P$ is $(- 1, - 5)$

Q. 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 P on it, touches the given parabola, then the point P is

(−1,−5)

(−1,5)

(−2,−8)

(2,4)