The normal at any point \) of curve meets the -axis at and is the foot of the ordinate at If }{1+x^{2}}, \) then equation of such curve, given that it passes through the point (3,1) is:

Maths · The solution of differential equation by the method of separation of variables

The normal at any point \) of curve meets the -axis at and is the foot of the or

The normal at any point \( \boldsymbol{P}(\boldsymbol{x}, \boldsymbol{y}) \) of \( \mathbf{a} \) curve meets the \( x \) -axis at \( Q \) and \( N \) is the foot of the ordinate at \( P \) If \( N Q=\frac{x\left(1+y^{2}\right)}{1+x^{2}}, \) then equation of such curve, given that it passes through the point (3,1) is:

A. \( x^{2}-y^{2}=8 \)
B. \( x^{2}+2 y^{2}=11 \)
C. \( x^{2}-5 y^{2}=4 \)
D. \( x^{2}+3 y^{2}=7 \)

Step-by-step solution

The normal at P meets x-axis at Q, and N is foot of ordinate. NQ = |y dy/dx|. Given NQ = x(1+y^2)/(1+x^2). Assuming positive branch, we get y dy/dx = x(1+y^2)/(1+x^2). Separating variables: y dy/(1+y^2) = x dx/(1+x^2). Integrate: (1/2)ln(1+y^2) = (1/2)ln(1+x^2) + c => 1+y^2 = K(1+x^2). Using (3,1): 2 = K*10 => K=1/5 => 1+y^2 = (1/5)(1+x^2) => x^2 - 5y^2 = 4.

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Step-by-step solution

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