The point of the curve $y^{2} = 2 (x - 3)$ at which the normal is parallel to the line y - 2x + 1 = 0 is
[MP PET 1998]

(5, 2)

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(- \frac{1}{2}, - 2)

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(5, -2)

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(\frac{3}{2}, 2)

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Solution

The correct option is D $(\frac{3}{2}, 2)$
Let $y = x^{6} a n d z = x^{3}; ∴ \frac{d y}{d z} = \frac{d y / d x}{d z / d x} = \frac{6 x^{5}}{3 x^{2}} = 2 x^{3} .$

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Similar questions

y^{2} = 2 (x - 3)

y^{2} = 2 (x - 3)

y^{2} = 2 (x - 3)

y - 2 x + 1 = 0

The point of intersection of the line $\frac{x}{1} = \frac{y - 1}{2} = \frac{z + 2}{3}$ and the plane 2x + 3y + z = 0 is
[MP PET 1989]

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