TS EAMCET · Maths · Hyperbola
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
- A \(\left(-\frac{25}{4},-\frac{9}{4}\right)\)
- B \(\left(\frac{25}{4}, \frac{9}{4}\right)\)
- C \((1,5)\)
- D \(\left(-\frac{1}{2}, \frac{7}{2}\right)\)
Answer & Solution
Correct Answer
(A) \(\left(-\frac{25}{4},-\frac{9}{4}\right)\)
Step-by-step Solution
Detailed explanation
\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…
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