The slope of normal at any point \((x, y), x > 0, y > 0\) on the curve \(y = y(x)\) is given by \(\frac{x^2}{xy-x^2y2^-1}\) If the curve passes through the point (1, 1), then \(e · y(e)\) is equal to

Question

A line passing through the point \( A(9, 0) \) makes an angle of \( 30^\circ \) with the positive direction of the x-axis. If this line is rotated about \( A \) through an angle of \( 15^\circ \) in the clockwise direction, then its equation in the new position is:

Answer 1

To solve this problem, we need to analyze the given curve and the slope of the normal at any point \((x, y)\) on it. The normal to a curve at a point is perpendicular to the tangent at that point.

Given:

The slope of the normal to the curve at any point \((x, y)\) is \(\frac{x^2}{xy-x^2y^{-1}}\).

Using the relationship between the slopes of tangent and normal lines, if \(m\) is the slope of the tangent, then the slope of the normal is \(-\frac{1}{m}\). Hence, we can equate:

-\frac{1}{m} = \frac{x^2}{xy - \frac{x^2}{y}}

Simplifying the denominator:

xy - \frac{x^2}{y} = \frac{xy^2 - x^2}{y}

Therefore:

-\frac{1}{m} = \frac{x^2 y}{xy^2 - x^2}

Finding \(y' = m = \text{slope of tangent}\):

y' = -\frac{xy^2 - x^2}{x^2 y} = -\frac{x - \frac{x^2}{y^2}}{x^2 / y}

Substituting \(y = \frac{u}{x}\), where \(u = x y\):

m = \frac{u^2}{u^2 - x^2}\)

With this substitution, simplifying further yields:

y = u = \frac{x}{x - \frac{x}{u}}

Given that the curve passes through \((1, 1)\), substitute these values into our transformed equation:

y(1) = 1 = \tan^{-1}(1)

For \(e \cdot y(e)\), using the parametric conditions and transformations discussed above, we find:

Since y(e) = \tan(-1)\), we find that:

e \cdot y(e) = e \cdot \tan^{-1}\left(\frac{x}{1+x}\right)\right)

Hence, it simplifies to the option:

\(\frac{1+tan(1)}{1-tan(1)}\)

Answer 2