# ISI2014-DCG-40

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Let the following two equations represent two curves $A$ and $B$. $$A: 16x^2+9y^2=144\:\: \text{and}\:\: B:x^2+y^2-10x=-21$$ Further, let $L$ and $M$ be the tangents to these curves $A$ and $B$, respectively, at the point $(3,0)$. Then the angle between these two tangents, $L$ and $M$, is

1. $0^{\circ}$
2. $30^{\circ}$
3. $45^{\circ}$
4. $90^{\circ}$
in Others
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For the curve A

\begin{align} 16x^{2}+9y^{2}&=144\\ \Rightarrow 16(2x)+9(2y)\frac{\mathrm{d}y}{\mathrm{d}x}&=0\\ \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} &=-\frac{16x}{9y}\\ \therefore \left.\begin{matrix} \frac{\mathrm{d}y}{\mathrm{d}x} \end{matrix}\right|_{(3,0)} &=-\frac{48}{0}\to -\infty\end{align}

$\therefore \mathrm{\theta}_L=\tan^{-1}(-\frac{48}{0})=90^{\circ}$

Now for the curve B

\begin{align} x^{2}+y^{2}-10x &=-21\\ \Rightarrow 2x+2y\frac{\mathrm{d}y}{\mathrm{d}x}-10&=0\\ \Rightarrow \frac{\mathrm{d}y}{\mathrm{d}x} &=-\frac{5-x}{y}\\ \therefore \left.\begin{matrix} \frac{\mathrm{d}y}{\mathrm{d}x} \end{matrix}\right|_{(3,0)} &=-\frac{2}{0}\to -\infty\end{align}

$\therefore \mathrm{\theta}_M=\tan^{-1}(-\frac{2}{0})=90^{\circ}$

$\therefore$ The angle between the tangents $L$ and $M$ is $\mathrm{\theta}_L-\mathrm{\theta}_M=90^{\circ}-90^{\circ}=0^{\circ}$

So the correct answer is A.

edited

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