The minimum distance between the parabola y^2 | vedclass

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

Question

(C) The parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$. The general point on the parabola is $P(2t^2, 4t)$.The slope of the tangent at $P$ is

Vedclass · Accepted Answer

$4\sqrt{2}$

Vedclass · Answer

(C) The parabola is $y^2 = 8x$,so $4a = 8$,which gives $a = 2$. The general point on the parabola is $P(2t^2, 4t)$.The slope of the tangent at $P$ is given by $\frac{dy}{dx} = \frac{4}{2t} = \frac{2}{t}$.For the minimum distance,the tangent at $P$ must be parallel to the line $x + y + 4 = 0$,which has a slope of $-1$.Setting $\frac{2}{t} = -1$,we get $t = -2$.The point $P$ is $(2(-2)^2, 4(-2)) = (8, -8)$.The distance from point $P(8, -8)$ to the line $x + y + 4 = 0$ is $d = \frac{|8 - 8 + 4|}{\sqrt{1^2 + 1^2}} = \frac{4}{\sqrt{2}} = 2\sqrt{2}$.The minimum distance between the parabola and its image is twice the distance from the point $P$ to the line of reflection,which is $2 	imes (2\sqrt{2}) = 4\sqrt{2}$.

List-$A$	List-$B$
$(A)$. The vertex of the parabola $y^2+4x-2y+3=0$ is	$(I)$. $\left(\frac{5}{4}, 1\right)$
$(B)$. The vertex of the parabola $x^2+8x+12y+4=0$ is	$(II)$. $\left(1, \frac{5}{4}\right)$
$(C)$. The focus of the parabola $y^2-x-2y+2=0$ is	$(III)$. $\left(-\frac{1}{2}, 1\right)$
$(D)$. The focus of the parabola $x^2-2x-8y-23=0$ is	$(IV)$. $(1, -1)$
	$(V)$. $(-4, 1)$

The minimum distance between the parabola $y^2 = 8x$ and its image with respect to the line $x + y + 4 = 0$ is:

Similar Questions

If the vertices of a triangle are $(am_1^2, 2am_1), (am_2^2, 2am_2),$ and $(am_3^2, 2am_3),$ then the area of the triangle is

Let $P$ be the point $(1, 0)$ and $Q$ be a point on the parabola $y^2 = 8x$. Find the locus of the midpoint of $PQ$.

If $ax^2+2hxy+by^2-82x+98y+144=0$ is the equation of a parabola with focus $(2,-3)$ and directrix $3x-2y+5=0$,then $ax^2+2hxy+by^2=0$ represents

The parametric equations of the parabola $x^2-8 x+12 y+15=0$ are

Vedclass Test Series

Exam Paper Generator

Online Exam Module