Q

# Find the equation of the tangent to the curve y = 3x − 2 which is parallel to the line 4x − 2y + 5 = 0 .

25) Find the equation of the tangent to the curve $y = \sqrt{3x-2}$ which is parallel to the line $4x - 2y + 5 = 0 .$

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Parellel to line $4x - 2y + 5 = 0 \Rightarrow y = 2x + \frac{5}{2}$ means the slope of tangent and slope of line is equal
We know that the equation of line is
y = mx + c
on comparing with the given equation we get the slope of line m = 2 and c = 5/2
Now, we know that the  slope of the tangent at a given point to given curve is given by $\frac{dy}{dx}$
Given the equation of curve is
$y = \sqrt{3x-2}$
$\frac{dy}{dx} = \frac{1}{2}.\frac{3}{\sqrt{3x-2}}=\frac{3}{2\sqrt{3x-2}}$
$\frac{3}{2\sqrt{3x-2}} = 2\\ 3^2 = (4\sqrt{3x-2})^2\\ 9 = 16(3x-2)\\ 3x-2=\frac{9}{16}\\ 3x = \frac{9}{16} +2\\ 3x= \frac{41}{16}\\ x = \frac{41}{48}$
Now, when

$x = \frac{41}{48}$ ,  $y = \sqrt{3x-2} \Rightarrow y = \sqrt{3\times\frac{41}{48}-2 } = \sqrt{\frac{41}{16}-2}=\sqrt\frac{9}{16 } = \pm \frac{3}{4}$

but y cannot be -ve so we take only positive value
Hence, the coordinates are

$\left ( \frac{41}{48},\frac{3}{4} \right )$
Now, equation of tangent paasing through

$\left ( \frac{41}{48},\frac{3}{4} \right )$ and with slope m = 2 is
$y - y_1=m(x-x_1)\\ y-\frac{3}{4}=2(x-\frac{41}{48})\\ 48y-36=2(48x-41)\\ 48x-24y=41-18\\ 48x-24y=23$
Hence, equation of tangent paasing through $\left ( \frac{41}{48},\frac{3}{4} \right )$ and with slope m = 2 is   48x - 24y = 23

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