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Question

If the radius of the auxiliary circle of hyperbola $$\dfrac{x^2}{16} - \dfrac{y^2}{4} = 1$$ is $$r$$ and if the abscissa of the points with eccentric angle 60 degrees on the hyperbola and auxiliary circle respectively are $$p$$ and $$q$$, then what is the value of $$p + q -  r$$ ?


Solution

Equation of auxiliary circle of hyperbola $$\dfrac{x^2}{a^2} - \dfrac{y^2}{b^2} = 1 $$ is $$x^2 + y^2 = a^2$$.

In this case, $$a^2 = 16$$.

Therefore, Required equation is $$x^2 + y^2 = 16$$.

Corresponding points $$N(4 \cos60,4 \sin60)$$,  $$N(2,2\sqrt3)$$ on circle and

$$P(4 \sec60,2 \tan60)$$ , $$P(8,2\sqrt3)$$ on hyperbola.

Hence $$r = 4, p = 8, q = 2$$. So, $$p + q - r = 6$$.


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