Mathematics · 2019
JEE Main · 8 April 2019, Shift 2 · Q77
Given that the slope of the tangent to a curve y=y(x) at any point (x, y) is (2 y)/x^2. If the curve passes through the centre of the circle…
Given that the slope of the tangent to a curve $\displaystyle y=y(x)$ at any point $\displaystyle (x, y)$ is $\displaystyle \frac{2 y}{x^2}$. If the curve passes through the centre of the circle $\displaystyle x^2+y^2-2 x-2 y=0$, then its equation is :
Official answer
(1)
$\displaystyle x \log _{\mathrm{e}}|y|=2(x-1)$
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JEE Main 2019 Mathematics question, with the answer from NTA’s final answer key. Where our answers come from.