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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 2 Maths Textbook Solution.

Provide Solution For R.D. Sharma Maths Class 12 Chapter 15 Tangents and Normals Exercise 15.1 Question 17 Sub Question 2 Maths Textbook Solution.

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Answer:\text { The points at which the tangents are parallel to } \mathrm{y} \text { -axis are }(2,0) \&(-2,0)

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

\begin{aligned} &\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} \& \frac{d x}{d y}=\frac{-4 y}{25 x} \end{aligned}

Now the tangent is parallel to the y-axis if the slope of the normal is zero
Slope of normal=\frac{-1}{\text { Slope of tangent }} \Rightarrow \frac{-1}{\frac{-25 x}{4 y}}=0

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

\text { This is possible if } y=0

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

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

x^{2}=4

x=\pm 2

Thus \text { The points at which the tangents are parallel to } \mathrm{y} \text { -axis are }(2,0) \&(-2,0)

 

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