Question

The value of λ² is _______ ?

Answer 1

Correct Answer: 9

Step 1: Let point P = (x, y).
Distance from P to line L₁: √2·x + y − 1 = 0 is:
D₁ = |√2·x + y − 1| / √(2 + 1) = |√2·x + y − 1| / √3
Distance from P to line L₂: √2·x − y + 1 = 0 is:
D₂ = |√2·x − y + 1| / √(2 + 1) = |√2·x − y + 1| / √3

Step 2: Given D₁·D₂ = λ², so:
(|√2·x + y − 1||√2·x − y + 1|)/3 = λ²
⇒ |√2·x + y − 1||√2·x − y + 1| = 3λ² ...........(1)

Step 3: The line y = 2x + 1 meets the locus.
Substitute y = 2x + 1 into equation (1):
Left term: √2·x + (2x + 1) − 1 = (√2 + 2)x
Right term: √2·x − (2x + 1) + 1 = (√2 − 2)x
So the equation becomes:
|(√2 + 2)x|·|(√2 − 2)x| = 3λ²
⇒ x²·|(√2 + 2)(√2 − 2)| = 3λ²

Step 4: Simplify product using identity:
(√2 + 2)(√2 − 2) = (2 − 4) = −2
So: |−2x²| = 3λ² ⇒ 2x² = 3λ² ⇒ x² = (3λ²)/2 ...........(2)

Step 5: The points of intersection are at x = ±√[(3λ²)/2]
Using y = 2x + 1:
So coordinates of R and S: (±√[(3λ²)/2], 2·(±√[(3λ²)/2]) + 1)
Now compute RS² using distance formula:
RS² = (Δx)² + (Δy)²
= (2√[(3λ²)/2])² + (4√[(3λ²)/2])²
= 4·(3λ²)/2 + 16·(3λ²)/2 = 6λ² + 24λ² = 30λ²

Step 6: Given RS = √270, so RS² = 270
⇒ 30λ² = 270 ⇒ λ² = 9

Final Answer: λ² = 9

Answer 2

Step 1: From previous part, we found that λ² = 9.
So the equation of the locus C is:
|√2·x + y − 1||√2·x − y + 1| = 27 ...........(1)

Step 2: Let’s again consider the line y = 2x + 1, which intersects C at points R and S such that RS = √270.
We already found x-coordinates of R and S are ±√[(3λ²)/2] = ±√[(3×9)/2] = ±√13.5
So coordinates of R and S are:
R = (−√13.5, −2√13.5 + 1), S = (√13.5, 2√13.5 + 1)

Step 3: Find midpoint M of RS (perpendicular bisector passes through this point).
x-coordinate of M = (−√13.5 + √13.5)/2 = 0
y-coordinate of M = (−2√13.5 + 1 + 2√13.5 + 1)/2 = 2/2 = 1
So M = (0, 1)

Step 4: Slope of RS is:
m_RS = (2√13.5 + 1 − (−2√13.5 + 1)) / (√13.5 − (−√13.5))
= (4√13.5)/(2√13.5) = 2

So slope of perpendicular bisector = −1/2
Equation of perpendicular bisector using point-slope form through M = (0, 1):
y − 1 = (−1/2)x ⇒ y = −x/2 + 1 ...........(2)

Step 5: Now find intersection points of line (2) with locus (1):
Substitute y = −x/2 + 1 into equation (1):
Left term: √2·x + y − 1 = √2·x − x/2 ⇒ (2√2 − 1)x/2
Right term: √2·x − y + 1 = √2·x + x/2 ⇒ (2√2 + 1)x/2
Now:
|(2√2 − 1)x/2| · |(2√2 + 1)x/2| = 27

Step 6: Product of two expressions:
[(2√2 − 1)(2√2 + 1) = (2√2)² − 1² = 8 − 1 = 7]
So equation becomes:
(7x²)/4 = 27 ⇒ x² = (108/7)

Step 7: Now find R′ and S′ using x = ±√(108/7)
Then y = −x/2 + 1 ⇒ y = ∓√(108/7)/2 + 1
So coordinates of R′ and S′ are:
(±√(108/7), ∓√(108/7)/2 + 1)

Step 8: Use distance formula:
D = [(2√(108/7))]² + [(√(108/7))]² = 4·(108/7) + (108/7) = (432 + 108)/7 = 540/7 = 77.14

Final Answer: D = 77.14