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Tangent to any point P of curve meets x-axi in T, Such that OP = PT (O is origin),  then curve is

  • Option 1)

    xy =c

  • Option 2)

    xy ^2=c

  • Option 3)

    x^2y =c

  • Option 4)

    x +y =c

 

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{dy}{dx}\left ( X-x \right ) at P (x, y)

For T, Y = 0

\\*\Rightarrow -y = \frac{dy}{dx}(X- x) \Rightarrow X = x -y\frac{\mathrm{d} x}{\mathrm{d} y} \\*\therefore T(x - y\frac{\mathrm{d} x}{\mathrm{d} y}, 0)

\because OP = PT \Rightarrow OP^2 = PT^2 \Rightarrow x^2 + y^2 = y^2\left(\frac{dx}{dy}\right)^2 + y^2 \\*\Rightarrow \frac{dx}{dy} = \frac{x}{y}\;or\;\frac{dx}{dy} = -\frac{x}{y}

On Integrating, we get

either y=cx\;or\;xy = c


Option 1)

xy =c

Option 2)

xy ^2=c

Option 3)

x^2y =c

Option 4)

x +y =c

Posted by

Himanshu

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