Let a tangent drawn at any point on the ellipse \(\frac{x^2}{25}+\frac{y^2}{16}=1\) cut the \(X\)-axis at…
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
\((5,4)\)
\((4,5)\)
\((0,0)\)
\((0,5)\)
Solution
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)
So, point \(Q\left(\frac{5}{\cos \theta}, 0\right) \Rightarrow\) Point \(R(0,5 \sec \theta)\)
Now, equation of circle with \(Q R\) as its diameter is
\(x(x-5 \sec \theta)+y(y-5 \sec \theta)=0\)
\(\Rightarrow \quad x^2+y^2-(5 \sec \theta) x-(5 \sec \theta) y=0\)
The above circle passes through the origin. Hence, option (c) is correct.