Let Two fixed points are A a - 0 and B - a - 0- If - A - B - then the locus of point C of triangle ABC will beA- x2-y2-2 x y tan-a2B- x2-y2-2 x y tan-a2C- x2-y2-2 x y -a2D- x2-y2-2 x y -a2

The correct option is D Given ∠ A ∠ B = θ nbsp;⇒ tan(A B) = tan θ .............(i) In right angled triangle CDA, tan A = k/a h and similarly in trian ...

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Standard XII

Physics

Relative Velocity in 2D

Let Two fixed...

Question

Let Two fixed points are A(a, 0) and B(-a, 0). If $∠ A - ∠ B = θ$ , then the locus of point C of triangle ABC will be

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Solution

The correct option is D

Given $∠ A - ∠ B = θ \Rightarrow t a n (A - B) = t a n θ$ .............(i)

In right angled triangle CDA, tan A = $\frac{k}{a - h}$

and similarly in triangle CDB, tan B = $\frac{k}{a + h}$

Also from (i), $\frac{t a n A - t a n B}{1 + t a n A . t a n B} = t a n θ$

Substituting the values of tan A and tan B, we get

$h^{2} - k^{2} + 2 h k c o t θ = a^{2}$

Hence the locus is $x^{2} - y^{2} + 2 x y c o t θ = a^{2}$ .

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Let Two fixed points are A(a, 0) and B(-a, 0). If ∠A−∠B=θ, then the locus of point C of triangle ABC will be

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