Question

Find the equation of the parabola whose focus is (1, 1) and the directrix is x + y + 1 = 0.

Solution

Correct option is

x2 – 2xy + y2 – 6x – 6y + 3 = 0

Let P(x, y) be any point on the parabola.

Then the distance of (x, y) from the focus (1, 1).

         = distance of P(x, y) from the directrix (x + y + 1 = 0)

Squaring (1), we get

             

or    2[x2 + 1 – 2x + y2 + 1 – 2y] = x2 + y2 + 2xy + 2y + 2x + 1

or     x2 – 2xy + y2 – 6x – 6y + 3 = 0

This is the required equation of the parabola.

SIMILAR QUESTIONS

Q1

The equation of the parabola whose axis is vertical and passes through the points (0, 0), (3, 0) and (–1, 4), is

Q2

The points on the parabola y2 = 36x whose ordinate is three times the abscissa are

Q3

The points on the parabola y2 = 12x whose focal distance is 4, are

Q4

Axis of the parabola x2 – 4x – 3+ 10 = 0 is  

Q5

The equation of the latus rectum of the parabola x2 + 4x + 2= 0 is 

Q6

x – 2 = t2y = 2t are the parameter equations of the parabola 

Q7

The equation  represents a parabola if  is

Q8

t1’ and ‘t2’ are two points on the parabola y2 = 4x. If the chord joining them is a normal to the parabola at ‘t1’ then

Q9

The vertex of the parabola y2 = 8x is at the center of a circle and the parabola cuts the circle at the ends of its latus rectum. Then the equation of the circle is

Q10

If the line 2x + 3y = 1 touch the parabola y2 = 4ax at the pointP. Find the focal distance of the point P.