Question

Please Answer this Question of Applications Of Derivatives Maths (M2) With Photo ​

Answer 1

Question :

Find the equation of tangent and normal to the curve $13x {}^{3} + 2x {}^{2}y + y {}^{3} = 1$ at ( 1,-2)

Solution :

Given curve :

$13x {}^{3} + 2x {}^{2}y + y {}^{3} = 1$

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First part of the question :

We to have to the find equation of tangent to the given curve :

First find slope of tangent :

$13x {}^{3} + 2x {}^{2}y + y {}^{3} = 1$ Differentiate the curve equation with respect to x

$\implies \: 39x {}^{2} + 2(2xy + x {}^{2} \times \frac{dy}{dx} ) + 3y {}^{2} \times \frac{dy}{dx} = 0$

$\implies \: 39x {}^{2} + 4xy + 2x {}^{2} \times \frac{dy}{dx} + 3y {}^{2} \times \frac{dy}{dx} = 0$

$\implies \: \frac{dy}{dx} (2x {}^{2} + 3y {}^{2} ) = - (39x {}^{2} + 4xy)$

$\frac{dy}{dx} = \frac{ - (39x {}^{2} +4xy) }{2x {}^{2} + 3y {}^{2} }$

at (1,-2)

$\frac{dy}{dx} = \frac{ - (39 \times 1 + 4 \times 1 \times - 2)}{2 \times 1 + 3( - 2) {}^{2} }$

$\implies \: \frac{dy}{dx} = m = \frac{ - 31}{14}$

Therefore slope of tangent = $\frac{-31}{14}$

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Now use two point form

Thus , equation of tangent:

$y + 2 = \frac{ - 31}{14} (x - 1)$

$\implies \: 31x + 14y - 3 = 0$

It' is the required equation of tangent .

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Second part of Question:

Find the equation of Normal :

${\purple{\boxed{\large{\bold{Slope\:of\:Normal=\frac{-1}{Slope\:of\:tangent }}}}}}$

Slope of Normal = $\frac{14}{31}$

Hence the equation of Normal at point (1,-2) and with slope 14/31 is

$y + 2 = \frac{14}{31} (x - 1)$

$\implies \: 14x - 31y - 76 = 0$

It is the required equation of Normal.