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  • Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 17 Sub Question 1 Maths Textbook Solution.

Provide Solution For R.D.Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 17 Sub Question 1 Maths Textbook Solution.

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Answer: \text { The points at which the tangents are parallel to } \mathrm{x} \text { -axis are }(0,5) \&(0,-5)

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

Given:\frac{x^{2}}{4}+\frac{y^{2}}{25}=1 \quad \rightarrow(1)

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

\frac{d}{d x}\left(x^{n}\right)=n x^{n-1}, \frac{d}{d x}(\text { constant })=0

\frac{2 x^{2-1}}{4}+\frac{2 y}{25} \frac{d y}{d x}=0

\frac{2 x}{4}+\frac{2 y}{25} \frac{d y}{d x}=0

\frac{x}{2}+\frac{2 y}{25} \frac{d y}{d x}=0

\frac{2 y}{25} \frac{d y}{d x}=\frac{-x}{2}

\frac{d y}{d x}=\frac{-x}{2} \times \frac{25}{2 y}

\frac{d y}{d x}=\frac{-25 x}{4 y}

Now the tangent is parallel to the x-axis if the slope of the tangent is zero

\frac{-25 x}{4 y}=0

\text { This is possible if } x=0

\text { Then, } \frac{x^{2}}{4}+\frac{y^{2}}{25}=1 \text { for } x=0

\begin{aligned} &\frac{0}{4}+\frac{y^{2}}{25}=1 \\ &y^{2}=25 \\ &y=\pm 5 \end{aligned}

\text { Substitute } y=\pm 5 \text { in } \frac{-25 x}{4 y}

\begin{array}{ll} \text { Put } y=5 & , \quad \text { Put } y=-5 \\ \frac{-25 x}{4(5)}=0 & , \quad \frac{-25 x}{4(-5)}=0 \\ \frac{-5 x}{4}=0 & , \quad \frac{5 x}{4}=0 \end{array}

x=0 \quad, \quad x=0

Thus the points at which the tangents are parallel to x-axis are 0,5&0,-5


 

 

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