Question

Find the equation of the tangents drawn from the point (1,2) to the curve
y^2-2x^3 - 4y +8=0.​

Answer 1

Step-by-step explanation:

Given equation of the curve is:

$y^2 - 2x^3-4y+8=0\\</p><p>\therefore y^2-4y=2x^3-8$

Differentiating w.r.t. x on both sides

$2y \frac{dy}{dx} -4\frac{dy}{dx}=6x^2\\</p><p>(2y-1) \frac{dy}{dx}=6x^2 \\</p><p> \frac{dy}{dx}= \frac{6x^2}{2y-1}\\</p><p>\frac{dy}{dx}_{(1, 2)}= \frac{6\times (1)^2}{2\times 2 -1}\\</p><p>\frac{dy}{dx}_{(1, 2)}= \frac{6\times 1}{4 -1}\\</p><p>\frac{dy}{dx}_{(1, 2)}= \frac{6}{3}\\</p><p>\frac{dy}{dx}_{(1, 2)}= 2$

Equation of tangent is given as:

$y-y_1=\frac{dy}{dx}(x-x_1)\\</p><p>y-2=2(x-1)\\</p><p>y-2=2x-2\\</p><p>y = 2x\\ </p><p>2x - y =0$

Which is the required equation of the tangent.