Prove that the product of the lengths of the perpendiculars drawn from the points (√a² - b², 0) and (- √a² - b², 0) to the line x/a cosθ + y/b sinθ = 1 is b²

Solution:

The equation of the given line is

x/a cosθ + y/b sinθ = 1

bx cosθ + ay sinθ - ab = 0 ....(1)

Length of the perpendicular from point (√a² - b², 0) to the line (1) is

p₁ = |b cosθ (√a² - b²) + a sinθ (0) - ab|/√b²cos²θ + a²sin²θ

= |b cosθ (√a² - b²) - ab|/√b²cos²θ + a²sin²θ ....(2)

Length of the perpendicular from point (- √a² - b², 0) to the line (2) is

p₂ = |b cosθ (- √a² - b²) + a sinθ (0) - ab|/√b²cos²θ + a²sin²θ

= |b cosθ (√a² - b²) + ab|/√b²cos²θ + a²sin²θ ....(3)

On multiplying equations (2) and (3), we obtain

[|b cosθ (√a² - b²) - ab||b cosθ (√a² - b²) + ab|]/(√b²cos²θ + a²sin²θ)²

= [|b cosθ (√a² - b²) - ab||b cosθ (√a² - b²) + ab|]/(b²cos²θ + a²sin²θ)

= |(b cosθ √a² - b²)² - (ab)²|/(b²cos²θ + a²sin²θ)

= |b² cos²θ (a² - b²) - a²b²|/(b²cos²θ + a²sin²θ)

= |a²b² cos²θ - b⁴cos²θ - a²b²|/(b²cos²θ + a²sin²θ)

= b² |a² cos²θ - b² cos²θ - a²|/(b²cos²θ + a²sin²θ)

= b² |a² cos²θ - b² cos²θ - a²sin²θ - a²cos²θ|/(b²cos²θ + a²sin²θ) [∵ sin²θ + cos²θ = 1]

= b² |- (b² cos²θ + a²sin²θ)|/(b²cos²θ + a²sin²θ)

= b² (b² cos²θ + a²sin²θ)/(b²cos²θ + a²sin²θ)

= b²

Hence, proved

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NCERT Solutions Class 11 Maths Chapter 10 Exercise ME Question 23

Prove that the product of the lengths of the perpendiculars drawn from the points (√a² - b², 0) and (- √a² - b², 0) to the line x/a cosθ + y/b sinθ = 1 is b²

Summary:

We proved that the product of the lengths of the perpendiculars drawn from the points (√a² - b², 0) and (- √a² - b², 0) to the line x/a cosθ + y/b sinθ = 1 is b²