Question

Find the equation of the tangent and the normal to the parabola x^2 - 4x - 8y + 12 = 0 at (4,3/2).

Answer 1

Answer:

$\huge\fcolorbox{blue}{pink}{☃αղsաҽɾ☃}$

$\bold{Equation \: of \: parabola \: is} \\ \\ ⇒ \: {x}^{2} - 4x - 8y + 12 = 0 \\ \\ ⇒ {(x - 2)}^{2} - 4 = 8y - 12 \\ \\ ⇒ \: {(x - 2)}^{2} = 8y - 8 \\ \\ ⇒ \: {(x - 2)}^{2} = 8(y - 1) \\ \\ ⇒ \: 4a =8 \\ \\ ⇒ \: a = \frac{8}{4} \\ \\ ⇒ \: a = 2$

$\bold{Equation \: of \: tangents \: at (x1, y1) \: is} \\ \\ ⇒ \: (x - 2)(x1 - 2) = 2a(y - 1 + y1 - 1) \\ \\ ⇒ \: (x - 2)(4 - 2) = 2a (y - 1 + \frac{3}{2} - 1 ) \\ \\ ⇒ \: 2(x - 2) = 4( \frac{2y - 1}{2} ) \\ \\ ⇒ \: x - 2y - 1 = 0$

$\bold{Equation \: of \: normal \: will \: be  \: y – y1 = m(x – x1) } \\ \\ \bold{ \: m-slope \: of \: normal \: Slope \: of \: tangent \: is: 1/2  \: Slope \: of \: normal \: is: -2 }$

$⇒ \: y - \frac{3}{2} = - 2(x - 4) \\ \\ ⇒ \: 2y - 3 = - 4x + 16 \\ \\ ⇒ \: 4x + 2y - 19 = 0$

$\bold \red {hope \: it \: helps \: u \: :) }$

Answer 2

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

Given curve is

$\rm :\longmapsto\: {x}^{2} - 4x - 8y + 12 = 0 - - - (1)$

and

$\rm :\longmapsto\:point \: \bigg(4, \dfrac{3}{2} \bigg)$

Let assume that the coordinate is represented as P.

Now,

$\rm :\longmapsto\: {x}^{2} - 4x - 8y + 12 = 0$

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

$\rm :\longmapsto\: \dfrac{d}{dx}( {x}^{2} - 4x - 8y + 12) = 0$

$\rm :\longmapsto\: \dfrac{d}{dx} {x}^{2} - 4\dfrac{d}{dx}x - 8\dfrac{d}{dx}y + \dfrac{d}{dx}12= 0$

$\rm :\longmapsto\:2x - 4 - 8\dfrac{dy}{dx} = 0$

$\rm :\longmapsto\:2x - 4 = 8\dfrac{dy}{dx}$

$\rm :\longmapsto\:x - 2 = 4\dfrac{dy}{dx}$

$\rm :\implies\:\dfrac{dy}{dx} = \dfrac{x - 2}{4}$

Now, slope of tangent to curve (1) at P is given by

$\rm :\longmapsto\:Slope \: of \: tangent, \: m = \bigg(\dfrac{dy}{dx}\bigg) _P$

$\rm \: = \: \: \dfrac{4 - 2}{4}$

$\rm \: = \: \: \dfrac{2}{4}$

$\rm \: = \: \: \dfrac{1}{2}$

Hence,

$\rm :\longmapsto\:Slope \: of \: tangent, \: m \: = \: \dfrac{1}{2}$

We know, that

Equation of tangent line passes through the point ( a, b ) having slope m is given by

$\red{ \boxed{ \bf{ \: y - b \: = \: m(x - a)}}}$

So,

Equation of tangent passes through the point (4, 3/2) and having slope 1/2 is

$\rm :\longmapsto\:y - \dfrac{3}{2} = \dfrac{1}{2} (x - 4)$

$\rm :\longmapsto\:\dfrac{2y - 3}{2} = \dfrac{1}{2} (x - 4)$

$\rm :\longmapsto\:2y - 3 = x - 4$

$\bf :\longmapsto\:x - 2y - 1 = 0$

Now,

We know that

Equation of normal passes through the point ( a, b ) and having slope m is given by

$\red{ \boxed{ \bf{ \: y - b \: = - \: \frac{1}{m} \: (x - a)}}}$

So,

Equation of normal passes through the point (4, 3/2) and having m = 1/2 is

$\rm :\longmapsto\:y - \dfrac{3}{2} = - \: \dfrac{1}{\dfrac{1}{2} } (x - 4)$

$\rm :\longmapsto\:y - \dfrac{3}{2} = - \: 2 (x - 4)$

$\rm :\longmapsto\: \dfrac{2y - 3}{2} = - \: 2 (x - 4)$

$\rm :\longmapsto\:2y - 3 = - 4(x - 4)$

$\rm :\longmapsto\:2y - 3 = - 4x + 16$

$\bf :\longmapsto\:4x + 2y - 19 = 0$

Hence,

$\begin{gathered}\begin{gathered}\bf\: Equation of-\begin{cases} &\sf{Tangent : \: x - 2y - 1 = 0} \\ \\ &\sf{Normal : \: 4x + 2y - 19 = 0} \end{cases}\end{gathered}\end{gathered}$

Additional Information :-

1. 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.