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  • Stuck here, help me understand: - The equation of a tangent to the parabola,, which makes an anglewith the positive direction of - Limit , continuity and differentiability - JEE Main

 

The equation of a tangent to the parabola, x^{2}=8y , which makes an angle \theta with the positive direction of x-axis, is :

  • Option 1)

     

    x=y\cot \theta +2\tan \theta

     

     

     

  • Option 2)

     

    x=y\cot \theta -2\tan \theta

  • Option 3)

     

    x=x\tan \theta +2\cot \theta

  • Option 4)

     

    x=x\tan \theta -2\cot \theta

Answers (1)

best_answer

 

Slope of the tangent -

Let y = f(x) is a curve then  dy / dx = f'(x) and at a particular point (h, k) it gives slope of tangent. From fig

M_{T}=\lim_{\delta x\rightarrow 0}\:\frac{(y+\delta y)-y}{(x+\delta x)-x}=\lim_{\delta x\rightarrow 0}\:\frac{\delta y}{\delta x}

- wherein

 

 

Equation of the tangent -

To find the equation of the tangent we need either one slope + one point or two points.

\therefore \:\:(y-y_{\circ})=m(x_{\circ }-y_{\circ })
 

or\:\:(y-y_{2})=\frac{y_{2}-y_{1}}{x_{2}-x_{1}}(x-x_{2})

- wherein

Where  (x_{\circ},y_{\circ})   is the point on the curve and M = MT  slope of tangent.

 

Equation of tangent to parabola , x^{2}=8y\\\\is \: \: y=mx -2m^{2} \: \: \: \: m=tan\Theta \\\\y=x\: tan\Theta -2tan^{2}\Theta \\\\x=y\: cot\Theta +2tan\Theta


Option 1)

 

x=y\cot \theta +2\tan \theta

 

 

 

Option 2)

 

x=y\cot \theta -2\tan \theta

Option 3)

 

x=x\tan \theta +2\cot \theta

Option 4)

 

x=x\tan \theta -2\cot \theta

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