If the equation of the tangent to the hyperbo | vedclass

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(C) The given equation of the hyperbola is $5x^2 - 9y^2 - 20x - 18y - 34 = 0$. Rearranging the terms: $5(x^2 - 4x) - 9(y^2 + 2y) = 34$. Completing the

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$2$ or $26$

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(C) The given equation of the hyperbola is $5x^2 - 9y^2 - 20x - 18y - 34 = 0$. Rearranging the terms: $5(x^2 - 4x) - 9(y^2 + 2y) = 34$. Completing the square: $5(x - 2)^2 - 20 - 9(y + 1)^2 + 9 = 34$,which simplifies to $5(x - 2)^2 - 9(y + 1)^2 = 45$. Dividing by $45$,we get $\frac{(x - 2)^2}{9} - \frac{(y + 1)^2}{5} = 1$. Here,$a^2 = 9$ and $b^2 = 5$. The slope of the tangent is $m = 	an(45^{\circ}) = 1$. The equation of a tangent with slope $m$ to the hyperbola $\frac{(x - h)^2}{a^2} - \frac{(y - k)^2}{b^2} = 1$ is $y - k = m(x - h) \pm \sqrt{a^2m^2 - b^2}$. Substituting the values: $y + 1 = 1(x - 2) \pm \sqrt{9(1)^2 - 5}$. $y + 1 = x - 2 \pm \sqrt{4} \implies y + 1 = x - 2 \pm 2$. Case $1$: $y + 1 = x - 2 + 2 \implies x - y - 1 = 0$. Here $b = -1, c = -1$. Then $b^2 + c^2 = (-1)^2 + (-1)^2 = 2$. Case $2$: $y + 1 = x - 2 - 2 \implies x - y - 5 = 0$. Here $b = -1, c = -5$. Then $b^2 + c^2 = (-1)^2 + (-5)^2 = 1 + 25 = 26$. Thus,$b^2 + c^2 = 2$ or $26$.

If the equation of the tangent to the hyperbola $5x^2 - 9y^2 - 20x - 18y - 34 = 0$ which makes an angle of $45^{\circ}$ with the positive $X$-axis is $x + by + c = 0$,then $b^2 + c^2 =$

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