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Need Solution for R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.2 Question 13 Sub Question 2 Maths Textbook Solution.

Answers (1)


Equation of tangent \Rightarrow 36y+12x-227=0

Hint: Differentiate both with respect to x


perpendicular to 5y-15x=13


the equation of the line is 5y-15x=13

5y-15x=13    \therefore y=3x+\frac{13}{5}

This is of the form y=mx+c

\therefore the\: \: slope\: \: of\: \: the\: \: line\: \: is\: \: 3

If a tangent is perpendicular to the line 5y-15x=13, then the slope of the tangent is

\begin{aligned} &-\frac{1}{\text { slope of the line }}=-\frac{1}{3} \\ &\Rightarrow 2 x-2=-\frac{1}{3} \\ &\Rightarrow 2 x=-\frac{1}{3}+2 \\ &\Rightarrow 2 x=\frac{5}{3} \\ &\Rightarrow x=\frac{5}{6} \end{aligned}


\Rightarrow y=\frac{25}{36}-\frac{10}{6}+7=\frac{217}{36}

Thus, the equation of the tangent passing through [5/6,217/36] is given by

\begin{aligned} &\Rightarrow y-\frac{217}{36}=-\frac{1}{3}\left[x-\frac{5}{6}\right] \Rightarrow \frac{36 y-217}{36}=-\frac{1}{18}(6 x-5)\\ &\Rightarrow 36 y-217=-2(6 x-5) \Rightarrow 36 y-217=-12 x+10\\ &\Rightarrow 36 y+12 x-227=0\\ &\text { the equation of the tangent is } 36 y+12 x-227=0 \end{aligned}(Ans)


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