The slope of normal at any point (x,y), X > 0, y > 0 on the curve y = y(x) is given by <math><mrow><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mi>x</mi><mi>y</mi><mo>−</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>1</mn></mrow></mfrac><mo>.</mo></mrow></math> If the curve passes through the point (1, 1), then e.y (e) is equal to

? −dxdy=x2xy−x2y2−1∴dydx=xy−x2y2−1x2Let xy=v⇒xdydx+y=dvdxPut x = 1, y = 1 ⇒tan−1=c⇒c=π4∴tan−1 (xy)=lnx=π4e (y (e))=tan (1+π4)=tan1+11−tan1

The slope of normal at any point (x,y), X &gt; 0, y &gt; 0 on the curve y = y(x) is given by&nbsp;<math><mrow><mfrac><mrow><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup></mrow><mrow><mi>x</mi><mi>y</mi><mo>−</mo><msup><mrow><mi>x</mi></mrow><mrow><mn>2</mn></mrow></msup><msup><mrow><mi>y</mi></mrow><mrow><mn>2</mn></mrow></msup><mo>−</mo><mn>1</mn></mrow></mfrac><mo>.</mo></mrow></math>&nbsp;If the curve passes through the point (1, 1), then e.y (e) is equal to

Ask & Answer Question

Option 1 -

$\frac{1 - t a n (1)}{1 + t a n (1)}$

Option 2 -

tan(1)

Option 3 -

Option 4 -

$\frac{1 + t a n (1)}{1 - t a n (1)}$

2 Views | Posted 2 months ago

Answered by

Payal Gupta | Contributor-Level 10

2 months ago

Correct Option - 4

Detailed Solution:

$? - \frac{d x}{d y} = \frac{x^{2}}{x y - x^{2} y^{2} - 1}$

$∴ \frac{d y}{d x} = \frac{x y - x^{2} y^{2} - 1}{x^{2}}$

$L e t x y = v \Rightarrow x \frac{d y}{d x} + y = \frac{d v}{d x}$

Put x = 1, y = 1 $\Rightarrow t a n^{- 1} = c \Rightarrow c = \frac{π}{4}$

$∴ t a n^{- 1} (x y) = l n x = \frac{π}{4}$

$e (y (e)) = t a n (1 + \frac{π}{4}) = \frac{t a n 1 + 1}{1 - t a n 1}$