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Find the equation of the tangent and the normal to the following curves at the indicated points: xy = c2 at (ct, c/t) |
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Answer» finding slope of the tangent by differentiating the curve \(y+x\frac{dy}{dx}=0\) \(\frac{dy}{dx}=-\frac{x}{y}\) m(tangent) at \((ct,\frac{c}{t})\) = \(\frac{1}{t^2}\) normal is perpendicular to tangent so, m1m2 = – 1 m(normal) at \((ct,\frac{c}{t})\) = t2 equation of tangent is given by y – y1 = m(tangent)(x – x1) \(y-\frac{c}{t}=-\frac{1}{t_2}(x-ct)\) equation of normal is given by y – y1 = m(normal)(x – x1) \(y-\frac{c}{t}={t^2}(x-ct)\) |
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