The tangents drawn to the hyperbola 5x^2 - 9y | vedclass

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(C) The given hyperbola is $5x^2 - 9y^2 = 90$,which can be written as $\frac{x^2}{18} - \frac{y^2}{10} = 1$. Here,$a^2 = 18$ and $b^2 = 10$. The equat

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$x^2 - y^2 = 28$

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(C) The given hyperbola is $5x^2 - 9y^2 = 90$,which can be written as $\frac{x^2}{18} - \frac{y^2}{10} = 1$. Here,$a^2 = 18$ and $b^2 = 10$. The equation of a tangent to the hyperbola with slope $m$ is $y = mx \pm \sqrt{a^2m^2 - b^2}$,which is $y = mx \pm \sqrt{18m^2 - 10}$. If this tangent passes through $P(h, k)$,then $k - mh = \pm \sqrt{18m^2 - 10}$. Squaring both sides,we get $(k - mh)^2 = 18m^2 - 10$,which simplifies to $m^2(h^2 - 18) - 2mhk + (k^2 + 10) = 0$. Let the slopes be $m_1 = 	an \alpha$ and $m_2 = 	an \beta$. Since $\alpha + \beta = 90^\circ$,we have $\beta = 90^\circ - \alpha$,so $m_2 = 	an(90^\circ - \alpha) = \cot \alpha = \frac{1}{m_1}$. Thus,$m_1 m_2 = 1$. From the quadratic equation in $m$,the product of roots is $m_1 m_2 = \frac{k^2 + 10}{h^2 - 18}$. Setting this equal to $1$,we get $k^2 + 10 = h^2 - 18$,which implies $h^2 - k^2 = 28$. Replacing $(h, k)$ with $(x, y)$,the locus is $x^2 - y^2 = 28$.

The tangents drawn to the hyperbola $5x^2 - 9y^2 = 90$ through a variable point $P$ make the angles $\alpha$ and $\beta$ with its transverse axis. If $\alpha$ and $\beta$ are complementary angles,then the locus of $P$ is

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