Question

find the equations of tangent and normal to the curve y= 3x² - 3x -5 where the tangent is parallel to the line 3x - y + 1 = 0​

Answer 1

$\large\underline{\sf{Solution-}}$

Given curve is

$\rm \: y = {3x}^{2} - 3x - 5 - - - - (1)$

Let assume that the tangent touches the curve at P(x, y).

Now,

On differentiating both sides w. r. t. x, we get

$\rm \: \dfrac{d}{dx} y = \dfrac{d}{dx}({3x}^{2} - 3x - 5)$

We know,

$\boxed{\tt{ \: \: \dfrac{d}{dx} {x}^{n} \: = \: {nx}^{n - 1} \: }}$

So, using this, we get

$\rm \: \dfrac{dy}{dx} = 6x - 3 - - - (2)$

$\rm\implies \:\rm \: Slope \: of \: tangent \: at \: (x,y) = 6x - 3 - - - (3)$

Now, It is given that tangent is parallel to the line 3x - y + 1 = 0.

We know,

Two lines having slope m and M are parallel iff m = M

So, it implies

Slope of tangent = Slope of line 3x - y + 1 = 0

$\rm\implies \:6x - 3 = 3$

$\rm \: 6x = 3 + 3$

$\rm \: 6x = 6$

$\rm\implies \:x = 1 - - - - (4)$

On substituting the value of x in equation (1), we get

$\rm \: y = 3 - 3 - 5$

$\rm\implies \:y = - 5$

So, it means tangent touches the curve at the point P (1, - 5).

And,

$\rm \: Slope \: of \: tangent (m)\: at \: (1, - 5) = 6 - 3 = 3$

We know,

Point Slope Form :- Equation of line which passes through the point (a, b) and having slope m is y - b = m(x - a).

Now, Equation of tangent which passes through the point (1, - 5) and having slope m = 3, is

$\rm \: y - ( - 5) = 3(x - 1)$

$\rm \: y + 5= 3x - 3$

$\rm \: y = 3x - 8$

$\rm\implies \:3x - y - 8 = 0$

Now,

We know, Normal and tangent are perpendicular to each other.

We know, Two lines having slope m and M are perpendicular iff Mm = - 1.

So,

$\rm\implies \:Slope \: of \: normal \: = \: - \: \dfrac{1}{3}$

So, Equation of normal which passes through the point (1, - 5) and having slope - 1/3 is

$\rm \: y - ( - 5) = - \dfrac{1}{3} (x - 1)$

$\rm \: y + 5 = - \dfrac{1}{3} (x - 1)$

$\rm \: 3y + 15 = - x + 1$

$\rm\implies \:x + 3y + 14 = 0$

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Additional Information

Let y = f(x) be any curve, then line which touches the curve y = f(x) exactly at one point say P is called tangent and that very point P, if we draw a perpendicular on tangent, that line is called normal to the curve at P.

2. If tangent is parallel to x - axis, its slope is 0.

3. If tangent is parallel to y - axis, its slope is not defined.

4. Two lines having slope M and m are parallel, iff M = m.

5. If two lines having slope M and m are perpendicular, iff Mm = - 1.

$\begin{gathered}\boxed{\begin{array}{c|c} \bf f(x) & \bf \dfrac{d}{dx}f(x) \\ \\ \frac{\qquad \qquad}{} & \frac{\qquad \qquad}{} \\ \sf k & \sf 0 \\ \\ \sf sinx & \sf cosx \\ \\ \sf cosx & \sf - \: sinx \\ \\ \sf tanx & \sf {sec}^{2}x \\ \\ \sf cotx & \sf - {cosec}^{2}x \\ \\ \sf secx & \sf secx \: tanx\\ \\ \sf cosecx & \sf - \: cosecx \: cotx\\ \\ \sf \sqrt{x} & \sf \dfrac{1}{2 \sqrt{x} } \\ \\ \sf logx & \sf \dfrac{1}{x}\\ \\ \sf {e}^{x} & \sf {e}^{x} \end{array}} \\ \end{gathered}$