Solveeit Logo
Login

Question

Mathematics Question on Slope of a line

The slope of normal at any point (x,y),x>0,y>0(x, y), x > 0, y > 0 on the curve y=y(x)y = y(x) is given by x2xy−x2y2−1\frac{x^2}{xy-x^2y2^-1} If the curve passes through the point (1, 1), then e⋅y(e)e · y(e) is equal to

A

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

B

tan(1)tan(1)

C

11

D

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

Answer

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

Explanation

Solution

∵−dxdy=x2xy−x2y2−1∵ -\frac{ dx}{dy} = \frac{x^2}{xy-x^2y^2-1}

dydx=x2y2−xy+1x2\frac{dy}{dx} = \frac{x^2y^2-xy+1}{x^2}

Assuming xy=v⇒y+xdydx=dvdxxy = v ⇒ y+x \frac{dy}{dx} = \frac{dv}{dx}

dvdx−y=(v2+v+1)yv\frac{dv}{dx}-y = \frac{(v^2+v+1)y}{v}

dvdx=v2+1x\frac{dv}{dx} =\frac{v^2+1}{x}

∵y(1)=1⇒tan–1(xy)=lnx\+tan–1(1)∵ y(1) = 1 ⇒ tan^{–1} (xy) = lnx \+ tan^{–1}(1)

Put xx = ee and yy = y(e)y(e) we get

tan–1(e⋅y(e))=1+tan–11tan^{–1} (e · y(e)) = 1 + tan^{–1} 1.

tan–1(e⋅y(e))–tan–11=1tan^{–1} (e · y(e)) – tan^{–1} 1 = 1

∴e(y(e))=1+tan(1)1−tan(1)∴ e(y(e)) = \frac{1+tan(1)}{1-tan(1)}

Hence, the correct option is (D): 1+tan(1)1−tan(1)\frac{1+tan(1)}{1-tan(1)}