Question

The equation of the normal to the curve x = a cos³ θ, y = a sin³ θ at the point θ = π/4 is __________ .

विकल्प

• x = 0

• y = 0

• x = y

• x + y = a

Answer 1

${\displaystyle x=y}$

 

$$\text{ Here,}$$

$$x=a{\cos }^3\theta \text{ and }y=a{\sin }^3\theta$$

$$\frac{dx}{d\theta }=-3a{\cos }^2\theta \sin \theta \text{ and }\frac{dy}{d\theta }=3a{\sin }^2\theta \cos \theta$$

$$\therefore \frac{dy}{dx}=\frac{\frac{dy}{d\theta }}{\frac{dx}{d\theta }}=\frac{3a{\sin }^2\theta \cos \theta }{-3a{\cos }^2\theta \sin \theta }=-\tan \theta$$

$$\text{ Now,}$$

$$\text{ Slope of the tangent }={\left(\frac{dy}{dx}\right)}_{\theta }=\frac{\pi }{4}=-\text{ tan}\frac{\pi }{4}=-1$$

$$(x_1,y_1)=(a{\cos }^3\frac{\pi }{4},a{\sin }^3\frac{\pi }{4})=(\frac{a}{2\sqrt{2}},\frac{a}{2\sqrt{2}})$$

$$\therefore \text{ Equation of the normal }$$

$$=y-y_1=\frac{-1}{m}(x-x_1)$$

$$\Rightarrow y-\frac{a}{2\sqrt{2}}=1(x-\frac{a}{2\sqrt{2}})$$

$$\Rightarrow x=y$$