Let the normal at a point P on the curve y^(2) – 3x^(2) + y + 10 = 0 intersect the y-axis at ( 0,3 / 2 ) . If m is the slope of the tangent at P to the curve, then |m| is equal to

Given curve: y^2 - 3x^2 + y + 10 = 0 => 2y (dy / dx) - 6x + (dy / dx) = 0 => (dy / dx) = (6x / 2y + 1) Let P be (x_1, y_1) Slope of tangent at P = (6x_1 / 2y_1 + 1) Therefore Slope of normal at P = - (2y_1 + 1 / 6x_1) => Equation of normal: (y - y_1) = - ( (2y_1 + 1 / 6x_1) )(x - x_1) This normal passes through point ( 0, (3 / 2) ). Therefore ( (3 / 2) - y_1 ) = - ( (2y_1 + 1 / 6x_1) )(0 - x_1) => y_1 = 1 Put y_1 = 1 in the equation of the curve, then we get x_1 = ± 2 => |m| = slope of tangent = | (6x_1 / 2y_1 + 1) | = (12 / 3) = 4

Let the normal at a point P on the curve y2 3x2 y 10 0 ... | null (Mathematics)

Given curve:

y^2 - 3x^2 + y + 10 = 0

\Rightarrow 2y \frac{dy}{dx} - 6x + \frac{dy}{dx} = 0

\Rightarrow \frac{dy}{dx} = \frac{6x}{2y + 1}

Let P be

(x_1, y_1)

Slope of tangent at

P = \frac{6x_1}{2y_1 + 1}

\therefore \text{Slope of normal at } P = - \frac{2y_1 + 1}{6x_1}

\Rightarrow \text{Equation of normal: } (y - y_1) = - \left( \frac{2y_1 + 1}{6x_1} \right)(x - x_1)

This normal passes through point

\left( 0, \frac{3}{2} \right).

\therefore \left( \frac{3}{2} - y_1 \right) = - \left( \frac{2y_1 + 1}{6x_1} \right)(0 - x_1)

\Rightarrow y_1 = 1

Put

y_1 = 1

in the equation of the curve, then we get

x_1 = \pm 2

\Rightarrow |m| = \text{slope of tangent} = \left| \frac{6x_1}{2y_1 + 1} \right| = \frac{12}{3} = 4

Explanation

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