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Tangent of a Curve y =f(x)

Find the leng...

Question

Find the length of tangent to $y^{2} = 4 a x$ at $(a t^{2}, 2 a t)$

2 a t \sqrt{(1 + t^{2})}

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t \sqrt{(1 + t^{2})}

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\sqrt{2 a t (1 + t^{2})}

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\sqrt{(1 + t^{2})}

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Solution

The correct option is A $2 a t \sqrt{(1 + t^{2})}$
The equation of parabola is $y^{2} = 4 a x$
The length of tangent is $y \sqrt{1 + {(\frac{d x}{d y})}^{2}}$
Differentiate the parabola equation with respect to $y$ , we get $\frac{d x}{d y} = \frac{y}{2 a}$
The length of tangent is $y \sqrt{1 + {(\frac{d x}{d y})}^{2}} = y \sqrt{1 + {(\frac{y}{2 a})}^{2}}$
Now substitute $y = 2 a t$ , the length of tangent is $2 a t \sqrt{1 + {(\frac{2 a t}{2 a})}^{2}} = 2 a t \sqrt{1 + t^{2}}$
Therefore, the correct option is $A$

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