The equation of a tangent to the hyperbola 16 | vedclass

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(A) The given equation of the hyperbola is $16x^2 - 25y^2 - 96x + 100y - 356 = 0$. Rearranging the terms,we get $16(x^2 - 6x) - 25(y^2 - 4y) = 356$. C

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$x - y + 2 = 0$

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(A) The given equation of the hyperbola is $16x^2 - 25y^2 - 96x + 100y - 356 = 0$. Rearranging the terms,we get $16(x^2 - 6x) - 25(y^2 - 4y) = 356$. Completing the square,$16(x - 3)^2 - 144 - 25(y - 2)^2 + 100 = 356$. $16(x - 3)^2 - 25(y - 2)^2 = 400$. Dividing by $400$,we get $\frac{(x - 3)^2}{25} - \frac{(y - 2)^2}{16} = 1$. Here,$a^2 = 25$ and $b^2 = 16$. The tangent makes an angle of $45^{\circ}$ with the transverse axis,so the slope $m = 	an(45^{\circ}) = 1$. The equation of the tangent to the hyperbola $\frac{(X)^2}{a^2} - \frac{(Y)^2}{b^2} = 1$ is $Y = mX \pm \sqrt{a^2m^2 - b^2}$,where $X = x - 3$ and $Y = y - 2$. Substituting the values,$y - 2 = 1(x - 3) \pm \sqrt{25(1)^2 - 16}$. $y - 2 = x - 3 \pm \sqrt{9}$. $y - 2 = x - 3 \pm 3$. Case $1$: $y - 2 = x - 3 + 3$ $\Rightarrow y = x + 2$ $\Rightarrow x - y + 2 = 0$. Case $2$: $y - 2 = x - 3 - 3$ $\Rightarrow y = x - 4$ $\Rightarrow x - y - 4 = 0$. Comparing with the options,$x - y + 2 = 0$ is the correct choice.

The equation of a tangent to the hyperbola $16x^2 - 25y^2 - 96x + 100y - 356 = 0$ which makes an angle $45^{\circ}$ with its transverse axis is

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