AP EAMCET · Maths · Ellipse
The least intercept made by a tangent to the ellipse \(\frac{x^2}{64}+\frac{y^2}{49}=1\) with coordinate axes is
- A \(40\)
- B \(10\)
- C \(15\)
- D \(100\)
Answer & Solution
Correct Answer
(C) \(15\)
Step-by-step Solution
Detailed explanation
Given, ellispe : \(\frac{x^2}{64}+\frac{y^2}{49}=1\) Here, \(a^2=64 \Rightarrow a=8\) \(b^2=49 \Rightarrow b=7\) When intercept made by tangent on coordinate axes is minimum, then least value of intercept \(=a+b=15\)
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