TS EAMCET · Maths · Ellipse
If any tangent drawn to the ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\) touches one of the circles \(x^2+y^2=\alpha^2\), then the range of \(\alpha\) is
- A \(9 \leq \alpha \leq 16\)
- B \(16 \leq \alpha \leq 25\)
- C \(3 \leq \alpha \leq 4\)
- D \(4 \leq \alpha \leq 6\)
Answer & Solution
Correct Answer
(C) \(3 \leq \alpha \leq 4\)
Step-by-step Solution
Detailed explanation
Tangent to ellipse \(\frac{x^2}{16}+\frac{y^2}{9}=1\) is \(y=mx \pm \sqrt{16m^2+9}\). Distance from origin to tangent to circle \(x^2+y^2=\alpha^2\) is \(\alpha\). \(\alpha = \frac{|\pm \sqrt{16m^2+9}|}{\sqrt{m^2+1}}\). \(\alpha^2 = \frac{16m^2+9}{m^2+1}\).…
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