#### Explain Solution R.D.Sharma Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 17 maths Textbook Solution.

Answer : Equation of the tangent is $4x-y=-13$

Hint: Differentiate on both sides

Given: $x^{2}+3y=3$which is parallel to $y-4x+5=0$

Solution: Given equation of the curve is

$x^{2}+3y=3$                    ....($i$)

On differentiating on both sides

$2x+3\frac{dy}{dx}=0$ or $\frac{dy}{dx}=-\frac{2x}{3}$

$\therefore$the slope of tangent $\left ( m \right )=-\frac{2x}{3}$

Given the equation of the line

$y-4x+5=0$ or $y=yx-5$

$which\: is \: of \: the\: form\: y=mx+c$

$\therefore slope \: of\: tangent\: =slope\: of\: line$

or $-\frac{2x}{3}=4$ or $-2x=12$

$\Rightarrow x=-6$

on putting $\Rightarrow x=-6$ in the Eq$\left ( i \right )$ we get

\begin{aligned} &(-6)^{2}+3 y=3 \text { or } 3 y=3-36 \\ &\text { or } 3 y=-33 \text { or } y=-11 \end{aligned}

so,the tanent is passing through point $\left ( -6,11 \right )$ and it has slope 4

Hence the required equation of the tangent is

\begin{aligned} &y+11=4(x+6)\\ &y+11=4 x+24\\ &4x-y=-13 \end{aligned}(Ans)