AP EAMCET · Maths · Ellipse
If a tangent having slope \(\frac{1}{3}\) to the ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1(a>b)\) is a normal to the circle \((x+1)^2+(y+1)^2=1\), then \(a^2\) lies in the interval
- A \(\left(\frac{\sqrt{2}}{\sqrt{5}}, 2\right)\)
- B \(\left(\frac{2}{5}, 4\right)\)
- C \(\left(1, \frac{10}{9}\right)\)
- D \((3,5)\)
Answer & Solution
Correct Answer
(B) \(\left(\frac{2}{5}, 4\right)\)
Step-by-step Solution
Detailed explanation
Center of circle: \((x+1)^2+(y+1)^2=1 \implies C = (-1, -1)\) Tangent to ellipse \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with slope \(m=\frac{1}{3}\) is \(y = \frac{1}{3}x \pm \sqrt{\frac{a^2}{9}+b^2}\) Since the tangent is normal to the circle, it passes through the center…
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