Answer to Question #247493 in Calculus for Lian

Question #247493
Find an equation of the line tangent to the curve y = 3x
2 − 1 and
parallel to the line 2x − y + 3 =
1
Expert's answer
2021-10-07T12:07:10-0400

Given equation of curve is "y=3x^2-1"

"\\therefore \\frac{dy}{dx}=6x"

So, slope of tangent at any point "(x_1,y_1)" is: "6x_1"

Now, according to question, this tangent is parallel to the given line "2x-y+3=0"

So, "6x_1=2\\\\"

"\\Rightarrow x_1=\\frac{1}{3}"

"\\therefore y_1=3x_1^2-1=3(\\frac{1}{3})^2-1=-\\frac{2}{3}"

So, equation of tangent is: "y+\\frac{2}{3}=2(x-\\frac{1}{3})\\\\"

"\\Rightarrow 3y+2=6x-2\\\\\n\\Rightarrow 6x-3y-4=0"


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