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Miscellaneous Exercise 6(II) · Q144

Q.The normal lines to a given curve at each point (x,y)(x,y) on the curve pass through (2,0)(2,0). The curve passes through (2,3)(2,3). Find the equation of the curve.

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Let P(x,y)P(x,y) be any point on the curve. The normal at PP has slope −dxdy-\dfrac{dx}{dy} (negative reciprocal of the tangent slope dydx\dfrac{dy}{dx}), and it passes through (2,0)(2,0), so its slope also equals y−0x−2=yx−2\dfrac{y-0}{x-2}=\dfrac{y}{x-2}. Equating: −dxdy=yx−2-\dfrac{dx}{dy}=\dfrac{y}{x-2}, i.e. (x−2)dx=−y dy(x-2)dx=-y\,dy, i.e. (x−2)dx+y dy=0(x-2)dx+y\,dy=0. Integrating: (x−2)22+y22=c1\dfrac{(x-2)^2}{2}+\dfrac{y^2}{2}=c_1, i.e. (x−2)2+y2=c(x-2)^2+y^2=c. Through $( …

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