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  • The equation of a tangent to the hyperbola <img alt="\mathrm{16 x^2-25 y^2-96 x+ 100 y-356=0}" src="https://entrancecorner.oncodecogs.com/gif.latex?%5Cmathrm%7B16%20x%5E2-25%20y%5E2-96%20x&plus

The equation of a tangent to the hyperbola \mathrm{16 x^2-25 y^2-96 x+ 100 y-356=0}, which makes an angle \mathrm{\pi / 4} with the conjugate axis, is

Option: 1

\mathrm{y=x+2}


Option: 2

\mathrm{y=x-5}


Option: 3

\mathrm{y=x+3}


Option: 4

\mathrm{x=y+2}


Answers (1)

best_answer

The equation of the hyperbola is
\mathrm{ 16\left(x^2-6 x\right)-25\left(y^2-4 y\right)=356 \Rightarrow \frac{(x-3)^2}{25}-\frac{(y-2)^2}{16}=1 }
The equations of tangents of slope \mathrm{m=\tan \pi / 4=1} to this hyperbola are

\mathrm{ y-2=1 \times(x-3) \pm \sqrt{25 \times 1^2-16} }

                                      \mathrm{[Using \, y-k=m(x-h) \pm \sqrt{a^2 m^2-b^2} ]}

\mathrm{ \Rightarrow y-2=(x-3) \pm 3 \Rightarrow y=x+2 \text { and } y=x-4 }

Posted by

Suraj Bhandari

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