Join the Most Relevant JEE Main 2025 Test Series & get 99+ percentile! Join Now
Search any question & find its solution
Question: Answered & Verified by Expert
The value of $m$ for which $y=m x+6$ is a tangent to the hyperbola $\frac{x^2}{100}-\frac{y^2}{49}=1$, is
MathematicsHyperbolaJEE Main
Options:
  • A $\sqrt{\frac{17}{20}}$
  • B $\sqrt{\frac{20}{17}}$
  • C $\sqrt{\frac{3}{20}}$
  • D $\sqrt{\frac{20}{3}}$
Solution:
2862 Upvotes Verified Answer
The correct answer is: $\sqrt{\frac{17}{20}}$
If $y=m x+c$ touches $\frac{x^2}{a^2}-\frac{y^2}{b^2}=1$, then $c^2=a^2 m^2-b^2$. Here $c=6, a^2=100, b^2=49$ $\therefore 36=100 m^2-49 \Rightarrow 100 m^2=85 \Rightarrow m=\sqrt{\frac{17}{20}}$.

Looking for more such questions to practice?

Download the MARKS App - The ultimate prep app for IIT JEE & NEET with chapter-wise PYQs, revision notes, formula sheets, custom tests & much more.