Question

The equation of the parabola where focus is
(-3,2) and directrix is x+y=4 is
a) x2 + y2 + 3xy - 2y + 1 = 0
b) x² + y2 – 2x + 4y +10=0
c) x2 + y2 - 3xy + 10x = 0
d) x2 + y2 – 2xy + 20x + 10 = 0
osagarc​

Answer 1

Answer:

Option(D)

Step-by-step explanation:

Given, focus : (-3,2) .

directrix : x + y = 4.

Let (x ,y) is the point on the parabola.

∴ Distance of point from focus = distance of point from directrix

=> √{(x + 3)² + (y - 2)² = |x + y - 4|/√2

Multiply both sides by √2

=> √2 * √{(x + 3)² + (y - 2)² = |x + y - 4|

Square on both sides

=> 2{(x + 3)² + (y - 2)² = (x + y - 4)²

=> 2[x² + 9 + 6x + y² + 4 - 4y] = x² + y² + 16 + 2xy - 8x - 8y

=> 2x² + 18 + 12x + 2y² + 8 - 8y - x² - y² - 16 - 2xy + 8x + 8y = 0

=> x² + y² - 2xy + 20x + 10 = 0

Equation is x² + y² - 2xy + 20x + 10 = 0.

Hope this help!

Answer 2

Answer:

D) is the answer.

$\huge\mathfrak\purple{Solution..}$

Given focus is:(-3,2) and directrix is: x+y =4

Let (x,y) be the parabola.

Distance of point from focus = distance of point from directrix.

Or, √{(x+3)^2+(y-2)^2}={x+y-4}/√2

Multiplying √2 in both sides,

√2 × √{(x+3)^2+(y-2)^2=x+y-4

Then by squaring both sides we get,

2{(x+3)^2+(y-2)^2=(x+y-4)^2

or, 2{x^2+6x+9+y^2-4y+4}=x^2+2xy+y^2-8x-8y+16

or, 2x^2+12x+18+2y^2-8y+8=x^2+2xy+y2-8x-8y+16

or, 2x^2+12x+18+2y^2-8y+8-x^2-2xy-y2+8x+8y-16=0

or, x^2+20x+10+y^2-2xy=0

So, x^2+y^2-2xy+20x+10=0✔✔✔

So, Answer is: option d.

$x {}^{2} + y {}^{2} - 2xy + 20x + 10 = 0$

That's the correct one✔✔✔

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