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Question: Answered & Verified by Expert
If $y=m x+4(n>0)$ is a tangent to the hyperbola $\frac{x^2}{25}-\frac{y^2}{9}=1$, then the point of contact of this tangent is
MathematicsHyperbolaTS EAMCETTS EAMCET 2023 (13 May Shift 1)
Options:
  • A $\left(-\frac{25}{4},-\frac{9}{4}\right)$
  • B $\left(\frac{25}{4}, \frac{9}{4}\right)$
  • C $(1,5)$
  • D $\left(-\frac{1}{2}, \frac{7}{2}\right)$
Solution:
1549 Upvotes Verified Answer
The correct answer is: $\left(-\frac{25}{4},-\frac{9}{4}\right)$
$\begin{aligned} & \text y=m x+4 \\ & \Rightarrow \frac{x^2}{25}-\frac{y^2}{9}=1 \\ & \because c^2=a^2 m^2-b^2 \Rightarrow 16=25 m^2-9 \\ & \Rightarrow m^2=1 \Rightarrow m=1 \\ & \therefore \text { Point of contact }\left(\frac{-a^2 m}{\sqrt{a^2 m^2-b^2}}, \frac{-b^2}{\sqrt{a^2 m^2-b^2}}\right) \\ & =\left(\frac{-25}{\sqrt{25-9}}, \frac{-9}{\sqrt{25-9}}\right)=\left(\frac{-25}{4}, \frac{-9}{4}\right) .\end{aligned}$

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