Answer on Question #80400 – Math – Analytic Geometry
Question
Obtain the equation of the parabola with focus (2,3) and directrix 3 x − 4 y + 9 = 0 3x - 4y + 9 = 0 3 x − 4 y + 9 = 0
Solution
Let ( x , y ) (x,y) ( x , y )
The distance between the point ( x , y ) (x,y) ( x , y ) ( x , y ) (x,y) ( x , y )
( x − 2 ) 2 + ( y − 3 ) 2 = ∣ 3 ⋅ x − 4 ⋅ y + 9 ∣ 3 2 + ( − 4 ) 2 , \sqrt{(x - 2)^2 + (y - 3)^2} = \frac{|3 \cdot x - 4 \cdot y + 9|}{\sqrt{3^2 + (-4)^2}}, ( x − 2 ) 2 + ( y − 3 ) 2 = 3 2 + ( − 4 ) 2 ∣3 ⋅ x − 4 ⋅ y + 9∣ ,
now by squaring both sides
( ( x − 2 ) 2 + ( y − 3 ) 2 ) 2 = ( ∣ 3 ⋅ x − 4 ⋅ y + 9 ∣ 3 2 + ( − 4 ) 2 ) 2 ; \left(\sqrt{(x - 2)^2 + (y - 3)^2}\right)^2 = \left(\frac{|3 \cdot x - 4 \cdot y + 9|}{\sqrt{3^2 + (-4)^2}}\right)^2; ( ( x − 2 ) 2 + ( y − 3 ) 2 ) 2 = ( 3 2 + ( − 4 ) 2 ∣3 ⋅ x − 4 ⋅ y + 9∣ ) 2 ; ( x − 2 ) 2 + ( y − 3 ) 2 = ( 3 ⋅ x − 4 ⋅ y + 9 ) 2 25 . (x - 2)^2 + (y - 3)^2 = \frac{(3 \cdot x - 4 \cdot y + 9)^2}{25}. ( x − 2 ) 2 + ( y − 3 ) 2 = 25 ( 3 ⋅ x − 4 ⋅ y + 9 ) 2 .
Simplifying
25 ⋅ ( x 2 − 4 ⋅ x + 4 + y 2 − 6 ⋅ y + 9 ) = ( ( 3 ⋅ x + 9 ) − 4 ⋅ y ) 2 ; 25 \cdot (x^2 - 4 \cdot x + 4 + y^2 - 6 \cdot y + 9) = ((3 \cdot x + 9) - 4 \cdot y)^2; 25 ⋅ ( x 2 − 4 ⋅ x + 4 + y 2 − 6 ⋅ y + 9 ) = (( 3 ⋅ x + 9 ) − 4 ⋅ y ) 2 ; 25 ⋅ ( x 2 + y 2 − 4 ⋅ x − 6 ⋅ y + 13 ) = ( 3 ⋅ x + 9 ) 2 + 2 ⋅ ( − 4 ⋅ y ) ⋅ ( 3 ⋅ x + 9 ) + ( − 4 ⋅ y ) 2 ; 25 \cdot (x^2 + y^2 - 4 \cdot x - 6 \cdot y + 13) = (3 \cdot x + 9)^2 + 2 \cdot (-4 \cdot y) \cdot (3 \cdot x + 9) + (-4 \cdot y)^2; 25 ⋅ ( x 2 + y 2 − 4 ⋅ x − 6 ⋅ y + 13 ) = ( 3 ⋅ x + 9 ) 2 + 2 ⋅ ( − 4 ⋅ y ) ⋅ ( 3 ⋅ x + 9 ) + ( − 4 ⋅ y ) 2 ; 25 ⋅ ( x 2 + y 2 − 4 ⋅ x − 6 ⋅ y + 13 ) = 9 ⋅ x 2 + 54 ⋅ x + 81 − 24 ⋅ x ⋅ y − 72 ⋅ y + 16 ⋅ y 2 ; 25 \cdot (x^2 + y^2 - 4 \cdot x - 6 \cdot y + 13) = 9 \cdot x^2 + 54 \cdot x + 81 - 24 \cdot x \cdot y - 72 \cdot y + 16 \cdot y^2; 25 ⋅ ( x 2 + y 2 − 4 ⋅ x − 6 ⋅ y + 13 ) = 9 ⋅ x 2 + 54 ⋅ x + 81 − 24 ⋅ x ⋅ y − 72 ⋅ y + 16 ⋅ y 2 ; 25 ⋅ x 2 + 25 ⋅ y 2 − 100 ⋅ x − 150 ⋅ y + 325 = 9 ⋅ x 2 + 16 ⋅ y 2 − 24 ⋅ x ⋅ y + 54 ⋅ x − 72 ⋅ y + 81 ; 25 \cdot x^2 + 25 \cdot y^2 - 100 \cdot x - 150 \cdot y + 325 = 9 \cdot x^2 + 16 \cdot y^2 - 24 \cdot x \cdot y + 54 \cdot x - 72 \cdot y + 81; 25 ⋅ x 2 + 25 ⋅ y 2 − 100 ⋅ x − 150 ⋅ y + 325 = 9 ⋅ x 2 + 16 ⋅ y 2 − 24 ⋅ x ⋅ y + 54 ⋅ x − 72 ⋅ y + 81 ; 16 ⋅ x 2 + 9 ⋅ y 2 + 24 ⋅ x ⋅ y − 154 ⋅ x − 78 ⋅ y + 244 = 0. 16 \cdot x^2 + 9 \cdot y^2 + 24 \cdot x \cdot y - 154 \cdot x - 78 \cdot y + 244 = 0. 16 ⋅ x 2 + 9 ⋅ y 2 + 24 ⋅ x ⋅ y − 154 ⋅ x − 78 ⋅ y + 244 = 0.
Now we get the following equation
16 ⋅ x 2 + 9 ⋅ y 2 + 24 ⋅ x ⋅ y − 154 ⋅ x − 78 ⋅ y + 244 = 0. 16 \cdot x^2 + 9 \cdot y^2 + 24 \cdot x \cdot y - 154 \cdot x - 78 \cdot y + 244 = 0. 16 ⋅ x 2 + 9 ⋅ y 2 + 24 ⋅ x ⋅ y − 154 ⋅ x − 78 ⋅ y + 244 = 0.
Answer: 16 ⋅ x 2 + 9 ⋅ y 2 + 24 ⋅ x ⋅ y − 154 ⋅ x − 78 ⋅ y + 244 = 0. 16 \cdot x^2 + 9 \cdot y^2 + 24 \cdot x \cdot y - 154 \cdot x - 78 \cdot y + 244 = 0. 16 ⋅ x 2 + 9 ⋅ y 2 + 24 ⋅ x ⋅ y − 154 ⋅ x − 78 ⋅ y + 244 = 0.
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#339914 on Dec 2023