The point of contact of the tangent y = x + 2 | vedclass

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(C) The given hyperbola is $5x^2 - 9y^2 = 45$. Dividing by $45$,we get $\frac{x^2}{9} - \frac{y^2}{5} = 1$. Comparing with $\frac{x^2}{a^2} - \frac{y^

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$(-9/2, -5/2)$

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(C) The given hyperbola is $5x^2 - 9y^2 = 45$. Dividing by $45$,we get $\frac{x^2}{9} - \frac{y^2}{5} = 1$. Comparing with $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$,we have $a^2 = 9$ and $b^2 = 5$. The tangent line is $y = mx + c$,where $m = 1$ and $c = 2$. The point of contact $(x_1, y_1)$ for a tangent $y = mx + c$ to the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ is given by $\left( \frac{-a^2m}{c}, \frac{-b^2}{c} \right)$. Substituting the values: $x_1 = \frac{-9(1)}{2} = -\frac{9}{2}$ and $y_1 = \frac{-5}{2}$. Thus,the point of contact is $(-9/2, -5/2)$.

The point of contact of the tangent $y = x + 2$ to the hyperbola $5x^2 - 9y^2 = 45$ is

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