AP EAMCET · Maths · Ellipse
Let a tangent drawn at any point on the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) cut the \(X\)-axis at \(Q\). Let \(R\) be the image of \(Q\) with respect to \(y=x\). If \(S\) is a circle with \(Q R\) as its diameter, then the fixed point through which the circle \(S\) passes is
- A \((5,4)\)
- B \((4,5)\)
- C \((0,0)\)
- D \((0,5)\)
Answer & Solution
Correct Answer
(C) \((0,0)\)
Step-by-step Solution
Detailed explanation
The equation of given ellipse is \(\frac{x^2}{25}+\frac{y^2}{16}=1\). Let a point \(P(5 \cos \theta, 4 \sin \theta)\) on the ellipse, then equation of tangent to the ellipse at point \(P\) is \(x\left(\frac{\cos \theta}{5}\right)+y\left(\frac{\sin \theta}{4}\right)=1\) ...(i)…
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