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Explain Solution R.D. Sharma Class 12 Chapter 15 Tangents and Normals Exercise15.1 Question 14 maths Textbook Solution.

Answers (1)

Answer: The requried point are $\left ( 2,3 \right )$ & $\left (- 2,-3 \right )$

Hint: $\text { The slope of the tangent is the limit of } \frac{\Delta y}{\Delta x} \text { as } \Delta x \text { approaches zero. }$

GIven:

$\text { Equation of the curve is } x^{2}+y^{2}=13 \rightarrow(1)$

$\text { Equation of the line is } 2 x+3 y=7 \quad \rightarrow(2)$

Solution:

$\text { Differentiating eqn(1) with respect to } x$

$\begin{aligned} &\therefore \frac{d}{d x}\left(\mathrm{x}^{n}\right)=n x^{n-1}, \frac{d}{d x}(\text { constant })=0 \\ &2 x^{2-1}+2 y^{2-1} \frac{d y}{d x}=0 \end{aligned}$

$\begin{aligned} &2 x+2 y \frac{d y}{d x}=0 \\ &2 y \frac{d y}{d x}=-2 x \\ &\frac{d y}{d x}=\frac{-x}{y} \end{aligned}$

$\text { Slope } \mathrm{m}_{1} \text { for }(1)=\frac{d y}{d x}=\frac{-x}{y} \quad \rightarrow(3)$

$\text { Differentiating eqn(1) with respect to } x$

$2+3 \frac{d y}{d x}=0$

$3 \frac{d y}{d x}=-2$

$\frac{d y}{d x}=\frac{-2}{3}$

$\text { Slope } \mathrm{m}_{2}=\frac{d y}{d x}=\frac{-2}{3}$

ATQ

$m_{1}=m_{2} \ldots .\{\text { as the tangents are parallel }\}$

$\frac{-x}{y}=\frac{-2}{3}$

$x=\frac{2}{3} y$

$\text { When put } x=\frac{2}{3} y \text { in eqn (1) } x^{2}+y^{2}=13$

$\left(\frac{2}{3} y\right)^{2}+y^{2}=13$

$\frac{4}{9} y^{2}+y^{2}=13$

L.C.M.of 1ans 9 is 9

$\frac{4 y^{2}+9 y^{2}}{9}=13$

$\frac{13 y^{2}}{9}=13$

$y^{2}=9$

$y=\pm 3$

Put $y=\pm 3 \text { in } x=\frac{2}{3} y$

$x=\frac{2}{3}(\pm 3)$

$x=\pm 2$

Thus the requried points are $\left ( 2,3 \right )$ & $\left (- 2,-3 \right )$

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