Given: Points ( a 2 − b 2 , 0 ) (\sqrt{a^2 - b^2}, 0) ( a 2 − b 2 , 0 ) ( − a 2 − b 2 , 0 ) (-\sqrt{a^2 - b^2}, 0) ( − a 2 − b 2 , 0 ) x a cos θ + y b sin θ = 1 \frac{x}{a}\cos\theta + \frac{y}{b}\sin\theta = 1 a x cos θ + b y sin θ = 1
Step 1: Rewrite the line equation in standard form A x + B y + C = 0 Ax + By + C = 0 A x + B y + C = 0
Multiplying by a b ab ab
b x cos θ + a y sin θ = a b bx\cos\theta + ay\sin\theta = ab b x cos θ + a y sin θ = ab Or:
b x cos θ + a y sin θ − a b = 0 bx\cos\theta + ay\sin\theta - ab = 0 b x cos θ + a y sin θ − ab = 0 Here A = b cos θ A = b\cos\theta A = b cos θ B = a sin θ B = a\sin\theta B = a sin θ C = − a b C = -ab C = − ab
Step 2: Find perpendicular distance from first point ( a 2 − b 2 , 0 ) (\sqrt{a^2 - b^2}, 0) ( a 2 − b 2 , 0 )
Using the formula:
d = ∣ A x 1 + B y 1 + C ∣ A 2 + B 2 d = \frac{|Ax_1 + By_1 + C|}{\sqrt{A^2 + B^2}} d = A 2 + B 2 ∣ A x 1 + B y 1 + C ∣ d 1 = ∣ b cos θ ⋅ a 2 − b 2 + 0 − a b ∣ b 2 cos 2 θ + a 2 sin 2 θ = ∣ b cos θ a 2 − b 2 − a b ∣ b 2 cos 2 θ + a 2 sin 2 θ d_1 = \frac{|b\cos\theta \cdot \sqrt{a^2 - b^2} + 0 - ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}} = \frac{|b\cos\theta\sqrt{a^2 - b^2} - ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}} d 1 = b 2 cos 2 θ + a 2 sin 2 θ ∣ b cos θ ⋅ a 2 − b 2 + 0 − ab ∣ = b 2 cos 2 θ + a 2 sin 2 θ ∣ b cos θ a 2 − b 2 − ab ∣ Step 3: Find perpendicular distance from second point ( − a 2 − b 2 , 0 ) (-\sqrt{a^2 - b^2}, 0) ( − a 2 − b 2 , 0 )
d 2 = ∣ b cos θ ⋅ ( − a 2 − b 2 ) + 0 − a b ∣ b 2 cos 2 θ + a 2 sin 2 θ = ∣ − b cos θ a 2 − b 2 − a b ∣ b 2 cos 2 θ + a 2 sin 2 θ d_2 = \frac{|b\cos\theta \cdot (-\sqrt{a^2 - b^2}) + 0 - ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}} = \frac{|-b\cos\theta\sqrt{a^2 - b^2} - ab|}{\sqrt{b^2\cos^2\theta + a^2\sin^2\theta}} d 2 = b 2 cos 2 θ + a 2 sin 2 θ ∣ b cos θ ⋅ ( − a 2 − b 2 ) + 0 − ab ∣ = b 2 cos 2 θ + a 2 sin 2 θ ∣ − b cos θ a 2 − b 2 − ab ∣ Step 4: Calculate the product d 1 × d 2 d_1 \times d_2 d 1 × d 2
d 1 × d 2 = ∣ b cos θ a 2 − b 2 − a b ∣ × ∣ − b cos θ a 2 − b 2 − a b ∣ b 2 cos 2 θ + a 2 sin 2 θ d_1 \times d_2 = \frac{|b\cos\theta\sqrt{a^2 - b^2} - ab| \times |-b\cos\theta\sqrt{a^2 - b^2} - ab|}{b^2\cos^2\theta + a^2\sin^2\theta} d 1 × d 2 = b 2 cos 2 θ + a 2 sin 2 θ ∣ b cos θ a 2 − b 2 − ab ∣ × ∣ − b cos θ a 2 − b 2 − ab ∣ Using ∣ u − v ∣ × ∣ − ( u + v ) ∣ = ∣ u 2 − v 2 ∣ |u-v| \times |-(u+v)| = |u^2 - v^2| ∣ u − v ∣ × ∣ − ( u + v ) ∣ = ∣ u 2 − v 2 ∣ u = b cos θ a 2 − b 2 u = b\cos\theta\sqrt{a^2 - b^2} u = b cos θ a 2 − b 2 v = a b v = ab v = ab
Numerator:
∣ b 2 cos 2 θ ( a 2 − b 2 ) − a 2 b 2 ∣ = ∣ a 2 b 2 cos 2 θ − b 4 cos 2 θ − a 2 b 2 ∣ |b^2\cos^2\theta(a^2 - b^2) - a^2b^2| = |a^2b^2\cos^2\theta - b^4\cos^2\theta - a^2b^2| ∣ b 2 cos 2 θ ( a 2 − b 2 ) − a 2 b 2 ∣ = ∣ a 2 b 2 cos 2 θ − b 4 cos 2 θ − a 2 b 2 ∣ = ∣ − a 2 b 2 sin 2 θ − b 4 cos 2 θ ∣ = b 2 ( a 2 sin 2 θ + b 2 cos 2 θ ) = |-a^2b^2\sin^2\theta - b^4\cos^2\theta| = b^2(a^2\sin^2\theta + b^2\cos^2\theta) = ∣ − a 2 b 2 sin 2 θ − b 4 cos 2 θ ∣ = b 2 ( a 2 sin 2 θ + b 2 cos 2 θ ) Step 5: Simplify the product.
d 1 × d 2 = b 2 ( a 2 sin 2 θ + b 2 cos 2 θ ) b 2 cos 2 θ + a 2 sin 2 θ = b 2 d_1 \times d_2 = \frac{b^2(a^2\sin^2\theta + b^2\cos^2\theta)}{b^2\cos^2\theta + a^2\sin^2\theta} = b^2 d 1 × d 2 = b 2 cos 2 θ + a 2 sin 2 θ b 2 ( a 2 sin 2 θ + b 2 cos 2 θ ) = b 2 ∴ Hence proved that the product of perpendicular distances equals b 2 b^2 b 2