First slide

Tangents and normals

Question

The coordinates of the point P on the curve $y^{2} = 2 x^{3}$ , the tangent at which is perpendicular to the line 4x – 3y+2 = 0, are given by

Moderate

A

$(2, 4)$
B

$(1, \sqrt{2})$
C

$(1 / 2, - 1 / 2)$
D

$(1 / 8, - 1 / 16)$

Solution

$\begin{aligned} Differentiating y^{2} = 2 x^{3}, we get \\ 2 y \frac{d y}{d x} = 6 x^{2} \Rightarrow \frac{d y}{d x} = \frac{3 x^{2}}{y} \end{aligned}$
The slope of the line 4x – 3 y+2 =0 is 4/3. Therefore , the
coordinates of P (x, y) must satisfy.
$\frac{3 x^{2}}{y} \cdot \frac{4}{3} = - 1 \Rightarrow 4 x^{2} = - y$
$\begin{array}{l} Also, y^{2} = 2 x^{3} . Solving these, we get x = 1 / 8 and \\ y = - 1 / 16 (clearly x \neq 0) \end{array}$

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