Question

please slove the question
Find the equation of tangent to the curve y = x2 – 2x + 7 which is parallel to the line 2x – y +9= 0.​

Answer 1

please mark me brilliant

Answer 2

We Know that Slope of Tangent is

$\frac{dy}{dx}$

y = x² - 2x + 7

Differentiation w.r.t.x

$\frac{dy}{dx} = 2x - 2 \: \: \: \: - - - (1)$

Finding slope of line 2x - y + 9 = 0

=> 2x - y + 9 = 0

=> y = 2 x + 9

The Above Equation is of form y = mx + c where m is Slope of line Hence, Slope of line 2x - y + 9 is 2

Now,

Given tangent is parallel to 2x - y +9 = 0

Slope of tangent = Slope of line 2x – y + 9 = 0

$\frac{dy}{dx} = 2 \\ 2x - 2 = 2 \\ 2(x - 1) = 2 \\ x = 2$

Finding Y when x = 2

$y = x {}^{2} - 2x + 7 \\y = (2) {}^{2} - 2(2) + 7 \\y = 4 - 4 + 7 \\ y= 7$

We need to find ,

Equation of tangent passes through (2,7) & slope is 2

Using ⤵️

Y - y₁ = m (x − x₁ )

$(y - 7) = 2(x - 2) \\ y - 7 = 2x - 4 \\ y - 2x - 7 + 4 = 0 \\ y - 2x - 3 = 0$

Hence ,

Required Equation of tangent is y - 2x - 3 = 0

$\: \: \: \: \: \: \: \: Good \: \: morning \:$