Question

Find the equation of the tangent to the curve y= root 3x-2 which is parralell

Answer 1

Question :-

Find the equation of the tangent to the curve y= root 3x-2 which is parralell to the line 4x - 2y +5= 0.

Answer:-

equation of tangent is→ 48x -24 y = 24

Step - by step explanation :-

Solution :-

$\bf{let \: (h ,\: k) \: be \: the \: point \: on \: curve \: } \\ \bf{from \: tangent}$

Given equation of tangent is↓

$\implies \: \bf{y \: = \sqrt{3x - 2} } \\ \\ \bf{differentiating \: with \: respect \: to \: x} \\ \\ \implies \: \bf{\frac{dy}{dx} = \frac{d {(3x - 2)}^{ \frac{1}{2} } }{dx} } \\ \\ \bf{ \implies \: \frac{dy}{dx} = \frac{3}{2 \sqrt{3x - 2} } }$

Now ,

$\bf{slope \: of \: tangent \: at \: (h, \: k) \: is} \\ \\ \bf{ \implies \: \frac{dy}{dx} | _{(h, \: k)} \: = \frac{3}{2 \sqrt{3h - 2} } }$

Now ,

Given tangent parallel to the line

4x -2 y + 5 = 0

condition :-

• Slope of tangent = slope of line

Given line ,

$\star \: \: \bf{4x - 2y + 5 = 0} \\ \\ \implies \: \bf{ y = 2x + \frac{5}{2} }$

Comparing this from y = mx + c

After comparing ,we get

Slope m = 2

Now , according to the condition

$\implies \: \bf{ \frac{3}{2 \sqrt{3h - 2} } = 2} \\ \\ \bf{ \implies \: 3 = 4 \sqrt{3h - 2} } \\ \\ \bf{ squairing \: on \: both \: sides \: } \\ \\ \implies \: \bf{ 9 = 16(3h - 2)} \\ \\ \implies \: \bf{3h - 2 = \frac{9}{16} } \\ \\ \implies \: \bf{ h \: = \frac{41}{16 \times 3} } \\ \\ \implies \: \boxed{\bf{h = \frac{41}{48} }}$

Now points (h ,k ) satisfied to the equation of curve ,

put x = h , y = k

$\implies \: \bf{ k = \sqrt{3h - 2} } \\ \\ \implies \: \bf{k = \sqrt{ \frac{41 - 32}{16} } = \sqrt{ \frac{ 9}{16} } } \\ \\ \implies \: \boxed{\bf{k = \frac{3}{4} }}$

Now equation of tangent having slope is 2 ,

$\bf{passing \: points \: ar e \: \bigg( \frac{41}{48} \:, \frac{3}{4} \bigg)}$

Now ,

$\implies \: \bf{y - \frac{3}{4} = 2 \big(x - \frac{41}{48 } \big)}$

On solving this we get,

$\implies \: \boxed{ \bf{48x - 24y = 23}}$

Hence ,

equation of tangent is→ 48x -24 y = 24