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प्रश्न
Find the equation of tangent and normal to the curve given by x – 7 cos t andy = 2 sin t, t ∈ R at any point on the curve
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उत्तर
x = 7 cos t and y = 2 sin t, t ∈ R
Differentiating w.r.t. ‘t’,
`("d"x)/"dt"` = – 7 sin t and `("d"y)/"dt"` = 2 cos t
Slope of the tangent ‘m’
`("d"y)/("d"x) = (("d"y)/("dt"))/(("d"x)/("d"t"))`
= `(2 cot"t")/(- 7 sin "t")`
Any point on the curve is (7 Cos t, 2 sin t)
Equation of tangent is y – y1 = m (x – x1)
y – 2 sint = `- (2cot"t")/(7sin"t")` (x – 7 cos t)
7y sin t – 14 sin2t = – 2x cos t + 14 cos2t
2x cos t + 7 y sin t – 14(sin2t + cos2t) = 0
2x cos t + 7y sin t – 14 = 0
Now slope of normal is `- 1/3 = (7sin"t")/(2cos"t")`
Equation of normal is y – y1 = `- 1/"m"` (x – x1)
y – 2 sin t = `(7sin"t")/(2cos"t")` (x – 7 cos t)
2y cos t – 4 sin t cos t = 7x sin t – 49 sin t cos t 7x sin t – 2y cos t – 45 sin t cos t = 0
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