Find the slope of the normal to the curve $4 x^{3} + 6 x^{2} - 5 x y - 8 y^{2} + 9 x + 14 = 0$ T the point $- 2, 3$ .

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11

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\frac{9}{19}

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- \frac{19}{9}

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Solution

The correct option is D $- \frac{19}{9}$
The given curve is $4 x^{3} + 6 x^{2} - 5 x y - 8 y^{2} + 9 x + 14 = 0$

Differentiating above equation w.r.t $x$ , we get

$12 x^{2} + 12 x - 5 x \frac{d y}{d x} - 5 y - 16 y \frac{d y}{d x} + 9 = 0$

$\Rightarrow \frac{d y}{d x} = \frac{12 x^{2} + 12 x - 5 y + 9}{(5 x + 16 y)}$

At $(x, y) \equiv (- 2, 3)$

$m = \frac{d y}{d x} = \frac{12 (4) + 12 (- 2) - 5 (3) + 9}{5 (- 2) + 16 (3)} = \frac{18}{38} = \frac{9}{19}$ is the slope of tangent to the curve
Slope of normal is $- \frac{1}{m} = - \frac{19}{9}$

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Question 10.
On subtracting $\frac{5}{9}$ from $\frac{19}{9}$ , the result is
(a) $\frac{24}{9}$
(b) $\frac{14}{9}$
(c) $\frac{14}{18}$
(d) $\frac{14}{0}$

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Find the slope of the normal to the curve 4x3+6x2−5xy−8y2+9x+14=0T the point −2,3.

Find the slope of the normal to the curve $4 x^{3} + 6 x^{2} - 5 x y - 8 y^{2} + 9 x + 14 = 0$ T the point $- 2, 3$ .