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  • Please help! A curve is such that, distance of any tangent to it measured from origin is equal to distance of point of contact from y-axis. If curve also pans through (1,2) then equation will be

A curve is such that, distance of any tangent to it measured from origin is equal to distance of point of contact from y-axis. If curve also pans through (1,2) then equation will be

  • Option 1)

    x^2 + y = 5x

  • Option 2)

    x^2 - y^2 = 5x

  • Option 3)

    x^2 + y ^2= 5

  • Option 4)

    x^2 + y^2 = 5x

 

Answers (1)

best_answer

As we have learnt,

 

Equation of tangent at a point (x y) -

Y-y= \frac{dy}{dx}\left ( X-x \right )

-

 

 

Equation of tangent will be 

(Y -y) = \frac{\mathrm{d} y}{\mathrm{d} x}(X-x) \\*\Rightarrow\left ( \frac{\mathrm{d} y}{\mathrm{d} x} \right )X - Y + (y-x\frac{\mathrm{d} y}{\mathrm{d} x}) = 0

According to question -

\left | \frac{y-x\frac{\mathrm{d} y}{\mathrm{d} x}}{\sqrt{\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )^2 + 1}} \right | = |x|

Squaring both sides, we get 

y^2 + x^2\left (\frac{\mathrm{d} y}{\mathrm{d} x} \right )^2 -2xy\frac{\mathrm{d} y}{\mathrm{d} x} = x^2\left (\frac{\mathrm{d} y}{\mathrm{d} x}\right) + x^2 \\*\Rightarrow \frac{\mathrm{d} y}{\mathrm{d} x} = \frac{y^2 - x^2}{2xy} \Rightarrow 2xydy = y^2dx -x^2dx \\*\Rightarrow y^2dx-x(2y)dy = x^2dx \Rightarrow \frac{y^2dx - x(2y)dy}{x^2} = dx \\* d\left(\frac{y^2}{x}\right) =- dx \Rightarrow \frac{y^2}{x} + x = c \Rightarrow x^2 + y^2 = cx

\because (1,2) will satisfy, so, c = 5 so x^2 + y^2 =5x


Option 1)

x^2 + y = 5x

Option 2)

x^2 - y^2 = 5x

Option 3)

x^2 + y ^2= 5

Option 4)

x^2 + y^2 = 5x

Posted by

Himanshu

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