Question

Find the point on the curve y= Vx – 3 where
the tangent is perpendicular to the line
6x + 3y - 5 = 0.​

Answer 1

ANSWER.

=> the point on the curve are = ( 4,1)

Step - by - Step - Explanation.

$\sf \to \: the \: point \: on \: the \: curve \: = y = \sqrt{x - 3} \\ \\ \sf \to \: tangent \: is \: perpendicular \: to \: the \: line \: = 6x + 3y - 5 = 0$

$\sf \to \: equation \: = 6x + 3y - 5 = 0 \\ \\ \sf \to \: curve \: = y = mx + c \\ \\ \sf \to \: 3y = 5 - 6x \\ \\ \sf \to \: y \: = \dfrac{5}{3} - \dfrac{6x}{3} \\ \\ \sf \to \: y \: = - \dfrac{6x}{3} + \dfrac{5}{3} \\ \\ \sf \to \: y \: = - 2 \\ \\ \sf \to \: as \: we \: know \: that \\ \\ \sf \to \: m_{1} \times m_{2} = - 1 \\ \\ \sf \to \: m_{1} \times ( - 2) = - 1 \\ \\ \sf \to \: m_{1} = \dfrac{1}{2} = slope$

$\sf \to \: point \: on \: curve \: = y = \sqrt{x - 3} \\ \\ \sf \to \: differentiate \: \: w.r.t \: \: y \\ \\ \sf \to \: \dfrac{dy}{dx} = \dfrac{1}{2 \sqrt{x - 3} } \\ \\ \sf \to \: as \: we \: know \: that \\ \\ \sf \to \: differentiation \: of \: \sqrt{x} = \dfrac{1}{2 \sqrt{x} } \\ \\ \sf \to \dfrac{1}{2 \sqrt{x - 3} } = \frac{1}{2} \\ \\ \sf \to \: 2 \sqrt{x - 3} = 2 \\ \\ \sf \to \: squaring \: on \: both \: sides \: we \: get \\ \\ \sf \to \: 4(x - 3) = 4 \\ \\ \sf \to \: 4x - 12 = 4 \\ \\ \sf \to \: 4x = 16 \\ \\ \sf \to \: \: x \: = 4 \\ \\ \sf \to \: put \: the \: value \: of \: x = 4 \: in \: equation \: \\ \\ \sf \to \: y = \sqrt{4 - 3} \\ \\ \sf \to \: y \: = 1 \\ \\ \sf \to \: \green{{ \underline{points \: on \: the \: curve \: = (4 \:, 1)}}}$

Answer 2

$\bf\huge\underline\red{Answer:-}$

$\bf\underline\purple{Given:-}$

• Curve y =√x-3

• line 6x+3y-5=0

$\bf\underline\red{Solution:-}$

3y = 5-6x

y=5-6x/3 =-6x/3+5/3

m2 =-6/3=-2

m1×(-2)=-1

m1=1/2

y=√x-3

dy/dx=1/2√x-3

=>1/2√x-3=1/2

=>√x-3=1

=>x-3=1

=>x=4

y=√4-3=1

Point on the curve is (4,1)