The equation of the parabola with focus (0,0) | vedclass

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

Question

(A) Given focus of parabola is $S(0,0)$.Equation of directrix is $x+y-4=0$.Let $P(x, y)$ be any point on the parabola.By definition,the distance from

Vedclass · Accepted Answer

$x^2+y^2-2xy+8x+8y-16=0$

Vedclass · Answer

(A) Given focus of parabola is $S(0,0)$.Equation of directrix is $x+y-4=0$.Let $P(x, y)$ be any point on the parabola.By definition,the distance from $P$ to the focus equals the distance from $P$ to the directrix,so $SP^2 = PM^2$.$SP^2 = (x-0)^2 + (y-0)^2 = x^2 + y^2$.The perpendicular distance $PM$ from $P(x, y)$ to the line $x+y-4=0$ is $\frac{|x+y-4|}{\sqrt{1^2+1^2}} = \frac{|x+y-4|}{\sqrt{2}}$.Thus,$x^2 + y^2 = \left(\frac{x+y-4}{\sqrt{2}}\right)^2$.$x^2 + y^2 = \frac{(x+y-4)^2}{2}$.$2(x^2 + y^2) = x^2 + y^2 + 16 + 2xy - 8x - 8y$.$x^2 + y^2 - 2xy + 8x + 8y - 16 = 0$.

The equation of the parabola with focus $(0,0)$ and directrix $x+y=4$ is

Similar Questions

If the line $y=2x+k$ is a normal to the parabola $y^2=4x$,then $k=$

What is the length of the latus rectum of the parabola $x^2 - 4x - 8y + 12 = 0$?

From the focus of the parabola $y^2 = 12x$, a ray of light is directed in a direction making an angle $\tan^{-1} \frac{3}{4}$ with the $x$-axis. Then the equation of the line along which the reflected ray leaves the parabola is

The shortest distance between the line $y = x$ and the curve $y^2 = x - 2$ is

If $x-2y+k=0$ is a tangent to the parabola $y^2-4x-4y+8=0$,then the slope of the tangent drawn at $(1, k)$ on the given parabola is

Vedclass Test Series

Exam Paper Generator

Online Exam Module