Two curves are orthogonal if their tangent lines are perpendicular at each point of intersection. are the given families of curves orthogonal trajectories of each other? that is, is every curve in one family orthogonal to every curve in the other family? x2 + y2 = ax x2 + y2 = by

Hello,

[tex]P(x,y)=x^2+y^2-ax=0\\\\ \dfrac{\partial{P}}{\partial{x}}=2x-a\\\\ \dfrac{\partial{P}}{\partial{y}}=2y\\\\ Q(x,y)=x^2+y^2-by=0\\\\ \dfrac{\partial{Q}}{\partial{x}}=2x\\\\ \dfrac{\partial{Q}}{\partial{y}}=2y-b\\\\ \dfrac{\partial{P}}{\partial{x}}*\dfrac{\partial{Q}}{\partial{x}}+\dfrac{\partial{P}}{\partial{y}}*\dfrac{\partial{Q}}{\partial{y}}=(2x-a)*2x+2y(2y-b)\\ =4x^2+4y^2-2ax-2by\\ =2(2x^2+2y^2-ax-by)\\ =2*P(x,y)+Q(x,y))\\ =2*(0+0)=0\\\\ Answer YES [/tex]

Cricetus Cricetus · Answer 2 · 2021-10-07T09:50:11+02:00

Yes, the given curves are orthogonal. A further explanation is below.

Given equation is:

[tex]x^2+y^2=ax[/tex]

By differentiating both sides, we get

→ [tex]2x+2yy'=a[/tex]

→ [tex]y'=\frac{a-2x}{2y} = m_1[/tex]

again,

[tex]x^2+y^2=by[/tex]

By differentiating both sides, we get

→ [tex]2x+2yy' =by'[/tex]

→ [tex]y' = \frac{-2x}{2y-b}[/tex]

For both curves are orthogonal, we get

→ [tex]m_1 \ m_2 = -1[/tex]

By substituting the values, we get

→ [tex]\frac{(a-2x)}{2y} \ \frac{(-2x)}{2y-b} = -1[/tex]

→ [tex]-2ax +4x^2=-4y^2+2yb[/tex]

→ [tex]4(x^2+y^2)=2ax+2yb[/tex]

Since,

[tex]ax=x^2+y^2[/tex]
[tex]by=x^2+y^2[/tex]

then,

→ [tex]4(x^2+y^2) =2(x^2+y^2)+2(x^2+y^2)[/tex]

→ [tex]4x^2+4y^2=4x^2+4y^2[/tex] (true)

Thus the above response is appropriate.

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https://brainly.com/question/20308962