Tangents to ellipse

Find the magnitude of the angle at which the ellipse x² + 5 y² = 5 is visible from the point P[5, 1].

Correct answer:

A = 26.5651 °

Step-by-step explanation:

x^{2} + 5 y^{2} = 5 P_{0} = 5; P_{1} = 1 x^{2} / 5 + (y / 1)^{2} = 1 a = 5 = 5 ≐ 2.2361 b = 1 x_{0} = 0 y_{0} = 0 p_{1}, p_{2} : y = k_{x} + q b^{2} - q^{2} + a^{2} \cdot k^{2} = 0 1 - q^{2} + 5 \cdot k^{2} = 0 P_{1} = k \cdot P_{0} + q 1 - (P_{1} - k * P_{0})^{2} + 5 * k^{2} = 0 1 - (1 - k \cdot 5)^{2} + 5 \cdot k^{2} = 0 - 20 k^{2} + 10 k = 0 20 k^{2} - 10 k = 0 a = 20; b = - 10; c = 0 D = b^{2} - 4 a c = 1 0^{2} - 4 \cdot 20 \cdot 0 = 100 D > 0 k_{1, 2} = \frac{- b \pm D}{2 a} = \frac{1 0 \pm 1 0 0}{4 0} k_{1, 2} = \frac{1 0 \pm 1 0}{4 0} k_{1, 2} = 0.25 \pm 0.25 k_{1} = 0.5 k_{2} = 0 Factored form of the equation: 20 (k - 0.5) k = 0 q_{1} = P_{1} - k_{1} \cdot P_{0} = 1 - 0.5 \cdot 5 = - \frac{3}{2} = - 1.5 q_{2} = P_{1} - k_{2} \cdot P_{0} = 1 - 0 \cdot 5 = 1 φ_{1} = \frac{1 8 0 °}{π} \cdot arctan (k_{1}) = \frac{1 8 0 °}{π} \cdot arctan 0.5 ≐ 26.5651^{\circ} φ_{2} = \frac{1 8 0 °}{π} \cdot arctan (k_{2}) = \frac{1 8 0 °}{π} \cdot arctan 0 = 0^{\circ} A = ∣ φ_{1} - φ_{2} ∣ = ∣ 26.5651 - 0 ∣ = 26.565 1^{\circ} = 26 ° 3 3^{'} 54 "

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