The equation of the parabola with x+2y=1 as d | vedclass

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

Question

(A) The directrix is $x+2y-1=0$ and the focus is $S(1,0)$.Let $P(x, y)$ be any point on the parabola.By definition,the distance from $P$ to the focus

Vedclass · Accepted Answer

$4x^2-4xy+y^2-8x+4y+4=0$

Vedclass · Answer

(A) The directrix is $x+2y-1=0$ and the focus is $S(1,0)$.Let $P(x, y)$ be any point on the parabola.By definition,the distance from $P$ to the focus equals the perpendicular distance from $P$ to the directrix $(PS = PM)$.$PS^2 = PM^2$$(x-1)^2 + (y-0)^2 = \frac{(x+2y-1)^2}{1^2+2^2}$$5((x-1)^2 + y^2) = (x+2y-1)^2$$5(x^2 - 2x + 1 + y^2) = x^2 + 4y^2 + 1 + 4xy - 2x - 4y$$5x^2 - 10x + 5 + 5y^2 = x^2 + 4y^2 + 4xy - 2x - 4y + 1$$4x^2 - 4xy + y^2 - 8x + 4y + 4 = 0$

The equation of the parabola with $x+2y=1$ as directrix and $(1,0)$ as focus is

Similar Questions

Let $P : y^{2} = 4ax, a > 0$ be a parabola with focus $S$. Let the tangents to the parabola $P$ that make an angle of $\frac{\pi}{4}$ with the line $y = 3x + 5$ touch the parabola $P$ at $A$ and $B$. Then the value of $a$ for which $A, B$ and $S$ are collinear is:

In the parabola $y^2 = 6x$,the equation of the chord passing through the vertex and the negative end of the latus rectum is

What are the equations of the tangents at the endpoints of the latus rectum of the parabola $y^2 = 4ax$?

The length of the normal chord of the parabola $y^2 = 4x$,which makes an angle of $\frac{\pi}{4}$ with the $x$-axis,is:

The acute angle between the tangents drawn at the point of intersection (other than the origin) of the curves $x^2=4y$ and $y^2=4x$ is

Vedclass Test Series

Exam Paper Generator

Online Exam Module