The equation of a tangent to the hyperbola \(16 x^2-25 y^2-96 x+100 y-356=0\) which makes an angle…
The equation of a tangent to the hyperbola \(16 x^2-25 y^2-96 x+100 y-356=0\) which makes an angle \(45^{\circ}\) with its transverse axis is
- \(x-y+2=0\)
- \(x-y+4=0\)
- \(x+y+2=0\)
- \(x+y+4=0\)
Solution
Given equation of hyperbola
\(\begin{aligned}
& 16 x^2-25 y^2-96 x+100 y-356=0 \\
& \Rightarrow \quad \frac{(x-3)^2}{25}-\frac{(y-2)^2}{16}=1 \quad \ldots (i)
\end{aligned}\)
Now equation of tangent to the hyperbola (i) having slope ' 1 ' is
\(\begin{aligned}
y-2 & =1(x-3)+\sqrt{25(1)-16} \\
\Rightarrow \quad y-2 & =x-3+3 \\
y-2 & =x-3+3 \text { or } y-2=x-3-3 \\
\text { or } \quad x-y-4 & =0 \\
\Rightarrow \quad x-y+2 & =0
\end{aligned}\)
Hence, option (1) is correct.
Asked in: AP EAMCET 2019 (20 Apr Shift 1)
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