Question

Find the curve which passes through the point (2, 0) such that segment of the tangent between the point of tangency & the y - axis has a constant length equal to 2.

Answer 1

Let $PT$ be tangent

Length of $PT$$=2$

$PT=x\sec \theta =x\sqrt{1-{\tan }^2\theta }=x\sqrt{1-y^2}(y=\tan \theta )\sqrt[x]{x1+(\frac{dy}{dx}{)}^2}=2=\frac{dy}{dx}=\sqrt[x]{x\frac{4-x^2}{x}}y=\int \sqrt{\frac{4-x^2}{x}dx+c}x=2\sin \theta \Rightarrow y=2\times \int \frac{{\cos }^2\theta }{\sin \theta }d\theta +c\Rightarrow y=\pm 2\log (\left|\frac{2-\sqrt{4-x^2}}{x}\right|+\sqrt{4-x^2})+c$

Gives a point that lies on $(2,0)$