Question

Simplify this....
${x}^{2} + 2xy \sec( \alpha ) + {y}^{2} = 0$
please do it... :)​

Answer 1

Answer:

Angles between the lines is α

Step-by-step explanation:

Let us take the two lines given below that represent the given equation.

y − m₁x = 0 and y − m₂x = 0

A combined equation of these two lines will be:

(y − m₁x) x (y − m₂x) = 0

y₂ − (m₁ + m₂)xy + m - 2x² = 0

Comparing this with the given equation:

m₁ + m₂ = −2secα

m₁m₂ = 1

∴ (m₁−m₂)² = (m₁+m₂)² − 4m₁m₂

⇒ (m₁ − m₂)² = 4sec²α − 4 = 4(sec²α − 1)

⇒ (m₁ − m₂)² = 4tan²α

⇒ |m₁ − m₂| = 2tanα

Now, angle between the lines is:

tanα=∣(m₁ − m₂) / 1 + m₁m₂∣

tanα=∣2tanα/(1 + 1)∣ = tanα

∴Angle = α

Hope this helps!

Please give it a brainliest if you'd like too :D