Provide Solution For  R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15 .1 Question 7 Maths Textbook Solution.

Answer: $\text { The required points are }(2,-2) \&(-2,2)$

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

Given: $\text { The curve } x y+4=0$

Solution:

$\inline \text { If a tangent line to the curve } y=f(x) \text { makes an angle } \theta \text { with } x \text { -axis in the positive direction, then }$

$\inline \frac{d y}{d x}=\text { The slope of\: tangent }=\tan \theta$

$\inline x y+4=0$

$\inline \text { Differentiating the above with respect to } x$$\inline x \frac{d}{d x}(y)+y \frac{d}{d x}(x)+\frac{d}{d x}(4)=0$

$\inline x \frac{d y}{d x}+y=0$

$\inline \therefore \frac{d}{d x}(\text { constant })=0$

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

$\frac{d y}{d x}=\frac{-y}{x} \quad \rightarrow(1)$

$\text { Also, } \frac{d y}{d x}=\tan 45^{\circ}=1 \quad \rightarrow(2)$

$\text { From (1)\&(2) we get }$

$\begin{gathered} \frac{-y}{x}=1 \\ x=-y \end{gathered}$

Substituting x=-y in xy+4=0

\begin{aligned} &x(-x)+4=0 \\ &-x^{2}+4=0 \\ &x^{2}=4 \\ &x=\pm 2 \end{aligned}

$\text { So when } x=2, y=-2$

$\text { and when } x=-2, y=2$

$\text { Thus the required points are }(2,-2) \&(-2,2)$