Answer to Question #157497 in Differential Geometry | Topology for Bawe

Question #157497
Prove that the equation of the tangent and normal to the ellipse (x^2/a^2) + (y^2/b^2) = 1 at the point P (acosu , bsinu) are respectively
bxcosu + aysinu = ab and
axsinu + bycosu = (a^2 - b^2)sinucosu.
1
Expert's answer
2021-02-02T05:22:23-0500

The equation of tangent and normal to the ellipse "\\frac{x^2}{a^2}+\\frac{y^2}{b^2}=1 \\text{ at the point } (x_1,y_1) \\text{ are } \\frac{x_1x}{a}+\\frac{y_1y}{b}=1 \\text{ and } a^2y_1x-b^2x_1y-(a^2-b^2)x_1y_1=0 respectively.\\\\\n\\text{At } P(a\\cos u, b\\sin u), \\text{ the equation of tangent will be;}\\\\\n\\frac{ax\\cos u}{a}+\\frac{by\\sin u}{b}=1\\\\\nx\\cos u+y\\sin u=1\\\\\n\\text{Also, the equation of normal will be; }\\\\\na^2(b\\sin u) x-b^2(a\\cos u) y=(a^2-b^2)(ab\\sin u \\cos u) \\\\\na^2bx\\sin u-ab^2y\\cos u=(a^2-b^2)(ab\\sin u \\cos u) \\\\\n\\text{Divide through by ab}\\\\\nax\\sin u-by\\cos u=(a^2-b^2)\\sin u\\cos u"


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