Question

2. Find the equation of the line tangent to the curve with parametric equations x = 4 cos θ y = 9 sin θ at θ = π 4 .

Answer 1

Answer:

$9x+4y=36\sqrt{2}$

Step-by-step explanation:

The given equations are

$x=4\cos \theta$

$y=9\sin \theta$

Differentiate with respect to θ .

$\dfrac{dx}{d\theta}=-4\sin \theta$

$\dfrac{dy}{d\theta}=9\cos \theta$

$\dfrac{dy}{dx}=\dfrac{dy}{d\theta}\times \dfrac{d\theta}{dx}=\dfrac{9\cos \theta}{-4\sin \theta}=-\dfrac{9}{4}\cot \theta$

At θ = π/4 ,

$\dfrac{dy}{dx}=-\dfrac{9}{4}\cot (\dfrac{\pi}{4})=-\dfrac{9}{4}$

$x=4\cos (\dfrac{\pi}{4})=4(\dfrac{1}{\sqrt{2}})=\dfrac{4}{\sqrt{2}}$

$y=9\sin (\dfrac{\pi}{4})=9(\dfrac{1}{\sqrt{2}})=\dfrac{9}{\sqrt{2}}$

Slope of the tangent line is -9/4 and point of tangency is $(\dfrac{4}{\sqrt{2}},\dfrac{9}{\sqrt{2}})$.

The equation of tangent line is

$y-y_1=m(x-x_1)$

where, m is slope.

$y-\dfrac{9}{\sqrt{2}}=-\dfrac{9}{4}(x-\dfrac{4}{\sqrt{2}})$

$y-\dfrac{9}{\sqrt{2}}=-\dfrac{9}{4}(x)+\dfrac{9}{\sqrt{2}}$

$\dfrac{9}{4}(x)+y=\dfrac{9}{\sqrt{2}}+\dfrac{9}{\sqrt{2}}$

$\dfrac{9}{4}(x)+y=\dfrac{18}{\sqrt{2}}$

Multiply both sides by 4.

$9x+4y=\dfrac{72}{\sqrt{2}}$

$9x+4y=36\sqrt{2}$

Therefore, the equation of the line tangent is $9x+4y=36\sqrt{2}$.