Find-the-angle-between-y-2-8x-and-x-2-y-2-16-at-their-point-of-intersection-for-which-y-is-positive-

Question Number 130294 by bramlexs22 last updated on 24/Jan/21

Find the angle between y^{2} = 8 x

and x^{2} + y^{2} = 16 at their point of

intersection for which y is positive

Answered by benjo_mathlover last updated on 24/Jan/21

Find the point of intersection

i . e solve y^{2} = 8 x \land y^{2} = 16 - x^{2}

we have x^{2} + 8 x - 16 = 0 \to {\begin{cases} x = 1.657 \\ x = - 9.657 \leftarrow not a real point of intersection \end{cases}

when x = 1.657, y = 3.641

coordinates of intersection are

P (1.657, 3.641)

Now we have to find \frac{dy}{dx} for each

of the two curves . Do that

(a) y^{2} = 8 x \to \frac{dy}{dx} = \frac{4}{y} = \frac{4}{3.641} = 1.099

\tan θ_{1} = 1.099 \to θ_{1} = 47 ° 42^{'}

(2) x^{2} + y^{2} = 16 \to \frac{dy}{dx} = - \frac{x}{y} = - 0.4551

\tan θ_{2} = - 0.4551 \to θ_{2} = - 24 ° 28^{'}

Finally θ = θ_{1} - θ_{2} = 72 ° 10^{'}