If straight line x cos(alpha) + y sin(alpha) = p touches the curve x2/a2 + y2/b2 = 1 , then - Maths - Application of Derivatives

Question

If straight line x cos(alpha) + y sin(alpha) =
p touches the curve x2/a2 +
y2/b2 = 1 , then prove that
a2 cos2(alpha) + b2
sin2(alpha) = p2

Ajanta Trivedi · Accepted Answer

the given equation of the straight line is :

$x \cos α + y \sin α = p ............ (1) \Rightarrow y \sin α = p - x \cos α \Rightarrow y = \frac{p - x \cos α}{\sin α} = p \cos e c α - x \cot α$

line (1) is tangent to the curve $\frac{x^{2}}{a^{2}} + \frac{y^{2}}{b^{2}} = 1........... (2)$

$b^{2} x^{2} + a^{2} y^{2} = a^{2} b^{2} b^{2} x^{2} + a^{2} * (p \cos e c α - x \cot α)^{2} = a^{2} b^{2} b^{2} x^{2} + a^{2} [p^{2} \cos e c^{2} α + x^{2} \cot^{2} α - 2 p \cos e c α \cot α \cdot x] = a^{2} b^{2} x^{2} (b^{2} + a^{2} \cot^{2} α) - 2 a^{2} p \cos e c α \cot α \cdot x + a^{2} (p^{2} \cos e c^{2} α - b^{2}) = 0$

for tangent, it must have equal roots, i.e. $D = 0$

$4 a^{4} p^{2} \cos e c^{2} α \cot^{2} α - 4 (b^{2} + a^{2} \cot^{2} α) \cdot a^{2} (p^{2} \cos e c^{2} α - b^{2}) = 0 a^{2} p^{2} \cos e c^{2} α \cot^{2} α - (b^{2} p^{2} \cos e c^{2} α + a^{2} p^{2} \cot^{2} α \cos e c^{2} α - b^{4} - a^{2} b^{2} \cot^{2} α) = 0 b^{2} p^{2} \cos e c^{2} α - b^{4} - a^{2} b^{2} \cot^{2} α = 0 p^{2} * \frac{1}{\sin^{2} α} - b^{2} - a^{2} * \frac{\cos^{2} α}{\sin^{2} α} = 0$

$p^{2} - b^{2} \sin^{2} α - a^{2} \cos^{2} α = 0 \Rightarrow a^{2} \cos^{2} α + b^{2} \sin^{2} α = p^{2}$

which is the required result.

hope this helps you.

cheers!!

If straight line x cos(alpha) + y sin(alpha) = p touches the curve x2/a2 + y2/b2 = 1 , then prove that a2 cos2(alpha) + b2 sin2(alpha) = p2 .

If straight line x cos(alpha) + y sin(alpha) = p touches the curve x²/a² + y²/b² = 1 , then prove that a² cos²(alpha) + b² sin²(alpha) = p² .