If one of the lines given by ax^2+2hxy+by^2=0 bisects an angle between coordinate axes then show that (a+b)^2=4h^2
Answers
Given: One of the lines given by ax^2+2hxy+by^2=0 bisects an angle between coordinate axes.
To find: Show that (a+b)^2=4h^2.
Solution:
- Now we have given the equation of pair of lines as : ax^2+2hxy+by^2 =0
- So it is given thatthe lines bisects the coordinate axis.
- So then, let the points (x,x) and (−x,x) satisfy the equation.
- Lets consider some cases.
- Case 1: Let the point (x,x) satisfy the equation ax^2+2hxy+by^2 =0 , we get:
ax^2 +2hx(x)+bx^2 =0
Taking x^2 common, we get:
x^2 (a+2h+b)=0
a+b = −2h .....................(i)
- Case 2: Let the point (x,−x) satisfy the equation ax^2+2hxy+by^2 =0 , we get:
ax^2 +2hx.(−x)+b(−x)^2 =0
Taking x^2 common, we get:
x^2 (a−2h+b)=0
a+b = 2h ....................(ii)
- Now from both the equations, we get a + b = ±2
- So squaring on both sides, we get:
(a+b)^2 = 4h^2
Hence proved.
Answer:
So from above solution we have proved that (a+b)^2 = 4h^2
Answer:
Given: One of the lines given by ax^2+2hxy+by^2=0 bisects an angle between coordinate axes.
To find: Show that (a+b)^2=4h^2.
Solution:
Now we have given the equation of pair of lines as : ax^2+2hxy+by^2 =0
So it is given thatthe lines bisects the coordinate axis.
So then, let the points (x,x) and (−x,x) satisfy the equation.
Lets consider some cases.
Case 1: Let the point (x,x) satisfy the equation ax^2+2hxy+by^2 =0 , we get:
ax^2 +2hx(x)+bx^2 =0
Taking x^2 common, we get:
x^2 (a+2h+b)=0
a+b = −2h .....................(i)
Case 2: Let the point (x,−x) satisfy the equation ax^2+2hxy+by^2 =0 , we get:
ax^2 +2hx.(−x)+b(−x)^2 =0
Taking x^2 common, we get:
Step-by-step explanation: