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

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

Hint: Differentiate on both sides

Given: x^{2}+3y=3which 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}

\thereforethe 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)

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