TS EAMCET · Maths · Ellipse
The minimum length of the intercept between the coordinate axes made by a tangent of the ellipse \(\frac{x^2}{64}+\frac{y^2}{4}=1\) is
- A \(10\)
- B \(\frac{17}{2}\)
- C \(8\)
- D \(\frac{15}{2}\)
Answer & Solution
Correct Answer
(A) \(10\)
Step-by-step Solution
Detailed explanation
Equation of given ellipse is \(\frac{x^2}{64}+\frac{y^2}{4}=1\) Now, let a parametric point on ellipse (i) \(P(8 \cos \theta, 2 \sin \theta)\) So, equation of tangent to ellipse (i) at point \(P\) is…
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