NCERT Solutions
Class 11 Maths
Straight Lines

Ex.Misc. Q.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 +
= 1 is b2.
The equation of the given line is
+
= 1
=> bx cos θ + ay sin θ – ab = 0 ……………. (1)
Length of the perpendicular from the point {√ (a2 - b2), 0} to the line 1 is
p1 = |b cos θ × √ (a2 - b2) + a sin θ × 0 - ab| ÷ √ (b2 cos2 θ + a2 sin2 θ)
= |b cos θ × √ (a2 - b2) - ab| ÷ √ (b2 cos2 θ + a2 sin2 θ) ……. (2)
Length of the perpendicular from the point {-√ (a2 - b2), 0} to the line 1 is
p2 = |b cos θ × {-√ (a2 + b2)} + a sin θ × 0 - ab| ÷ √ (b2 cos2 θ + a2 sin2 θ)
= |b cos θ × √ (a2 + b2) + ab| ÷ √ (b2 cos2 θ + a2 sin2 θ) ………. (3)
Multiply equation (2) and (3), we get
p1p2 = {|b cos θ × √ (a2 - b2) - ab|× |b cos θ × √ (a2 + b2) + ab|} ÷ {√ (b2 cos2 θ + a2 sin2 θ)}2
= {|b cos θ × √ (a2 - b2) - ab|× |b cos θ × √ (a2 + b2) + ab|} ÷ (b2 cos2 θ + a2 sin2 θ)
= {|b cos θ × √ (a2 - b2)} 2 – (ab)2| ÷ (b2 cos2 θ + a2 sin2 θ)
= |b2 cos2 θ × (a2 - b2) – a2 b2| ÷ (b2 cos2 θ + a2 sin2 θ)
= |a2 b2 cos2 θ – b4 cos2 θ – a2 b2| ÷ (b2 cos2 θ + a2 sin2 θ)
= b2|a2 cos2 θ – b2 cos2 θ – a2| ÷ (b2 cos2 θ + a2 sin2 θ)
= b2|a2 cos2 θ – b2 cos2 θ – a2 sin2 θ - a2 cos2 θ| ÷ (b2 cos2 θ + a2 sin2 θ)
= b2|-(b2 cos2 θ + a2 sin2 θ) | ÷ (b2 cos2 θ + a2 sin2 θ)
= b2(b2 cos2 θ + a2 sin2 θ) ÷ (b2 cos2 θ + a2 sin2 θ)
= b2
Hence, p1p2 = b2