Question

The ratio of the length of tangent to the length of normal to the curve y = 3e5x at the intersection point of curve and y-axis, is

Answer 1

$\displaystyle\huge\red{\underline{\underline{Solution}}}$

FORMULA TO BE IMPLEMENTED

For any curve y = f(x)

$1. \: \sf{The \: length \: of \: the \: tangent \: at \: any \: point \: (x,y) \: is}$

$\displaystyle \: = \sf{ \: |y| \sqrt{1 + { \bigg( \frac{dx}{dy} \bigg)}^{2} } \: }$

$2. \: \sf{The \: length \: of \: the \: normal \: at \: any \: point \: (x,y) \: is}$

$\displaystyle \: = \sf{ \: |y| \sqrt{1 + { \bigg( \frac{dy}{dx} \bigg)}^{2} } \: }$

CALCULATION

Here the equation of the curve is

$\sf{ \: y = 3 {e}^{5x} \: } \: \: ........(1)$

Now the curve meet y axis

So x = 0

From Equation (1)

$\sf{y = 3 \times 1 = 3}$

So the point of intersection is ( 0, 3 )

Now Differentiating both sides of Equation (1) with respect to x we get

$\displaystyle \: \sf{ \: \frac{dy}{dx} = 15 {e}^{5x} \: } \: \:$

Now

$\displaystyle \: \sf{ \: \bigg[ \frac{dy}{dx} \bigg] _{(0,3)}= 15 \: } \: \:$

Hence

$\sf{The \: length \: of \: the \: tangent \: at \: any \: point \: (0,3) \: is}$

$\displaystyle \: = \sf{ \: |3| \sqrt{1 + { \bigg( \frac{1}{15} \bigg)}^{2} } \: }$

$\displaystyle \: = \sf{ \frac{ \sqrt{226} }{5} } \: \: \: unit$

Also

$\sf{The \: length \: of \: the \: normal \: at \: any \: point \: (0,3) \: is}$

$\displaystyle \: = \sf{ \: |3| \sqrt{1 + { \bigg( 15\bigg)}^{2} } \: }$

$\displaystyle \: = \sf{3 \sqrt{226} } \: \: unit$

Hence The ratio of the length of tangent to the length of normal to the curve is

$\displaystyle \: = \sf{ \frac{ \sqrt{226} }{5} } : 3 \sqrt{226}$

$\displaystyle \: = \sf{ \frac{ 1}{5} } : 3$

$\displaystyle \: = \sf{1 : 15\: }$

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