Question

Find the equation of the parabola whose focus is (- 3,2) and the directrix is

x + y = 4.


​

Answer 1

Answer:

Let A(x,y) be the Pt. on the parabola

focus = (2,3) & eq

n

of directrix =x−4y+3=0

Now let AM be a perpendicular from A on the directrix.

then, we have (SA)

2

=(AM)

2

(x−2)

2

+(y−3)

2

=(

1

2

+(−4)

2

x−4y+3

)

2

x

2

−4x+4+y

2

−6y+9=

17

(x+4y+3)

2

17x

2

+17y

2

−68x−102y+221=x

2

+16y

2

+9−8xy+6x−24y

⇒16x

2

+y

2

+8xy−74x−78y+212=0

Which is the required eq

n

of parabola.

Latus rectum = 2× (length of ⊥ from focus in th directrix)

=2×

∣

∣

∣

∣

∣

1

2

+4

2

(2−2×3+3)

∣

∣

∣

∣

∣

⇒2×

∣

∣

∣

∣

∣

1+16

−7

∣

∣

∣

∣

∣

⇒

17

+14

Latus Rectum =

17

14

Answer 2

Step-by-step explanation:

Let the focus, S=( -3,2)

Given equation of the directrix is x-y+5=0

Let P(x,y) be a point on the parabola. Draw PM perpendicular to the directrix x-y+5=0

From the definition of the parabola SP/PM =1

SP= 1×PM

SP=PM

SP^2=PM^2

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

x^2+(3)^2+2×x×3+y^2+(2)^2-2×y×2=

x^2 + 9 + 6x + y^2 + 4 - 4y=