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  • Explain solution rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 question 10

Explain solution rd sharma class 12 chapter 15 Tangents and Normals exercise 15.3 question 10

Answers (1)

Hence Prove,\begin{aligned} &a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha=P^{2} \end{aligned}

Hint –

The equation of tangent at p(x_{1},y_{1})

=\frac{x x_{1}}{a^{2}}-\frac{y y_{1}}{b^{2}}=1

Given –

x \cos \alpha+y \sin \alpha=\mathrm{p}---(\mathrm{i})

And the curve \frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1

  Or b^{2} x^{2}+a^{2} y^{2}=a^{2} b^{2}---(i i)

Let line and curve touches each other at point p(x_{1},y_{1})

=x_{1} \cos \alpha+y_{1} \sin \alpha=\mathrm{p}---(\mathrm{iii})

And b^{2} x_{1}^{2}+a^{2} y_{1}^{2}=a^{2} b^{2}---(i v)

Differentiating (ii), we get

As we know,\frac{d}{d x}\left(x^{n}\right)=^{n} x^{n-1}, \frac{d}{d x}(\text { constants })=0

\begin{aligned} &=2 b^{2} x+2 a^{2} y \frac{d y}{d x}=0 \\ &=\left(\frac{d y}{d x}\right)_{x_{1}, y_{1}}=\frac{-b^{2} x_{1}}{a^{2} y_{1}} \end{aligned}

Also, slope of line (i) is = \frac{-\cos \alpha}{\sin \alpha}

According to question,
\begin{aligned} &=\frac{-b^{2} x_{1}}{a^{2} y_{1}}=\frac{-\cos \alpha}{\sin \alpha} \\ &=\frac{x_{1}}{a^{2} \cos \alpha}=\frac{y_{1}}{b^{2} \sin \alpha}=\partial \\ \end{aligned}

\begin{aligned} &=x_{1}=\partial a^{2} \cos ^{2} \alpha, y c g f y \partial b^{2} \sin ^{2} \alpha=p \\ &=\partial=\frac{p}{a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha} \\ \end{aligned}

\begin{aligned} &=x_{1}=\frac{p a^{2} \cos \alpha}{a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha} \end{aligned} and \begin{aligned} y_{1}=\frac{p b^{2} \sin \alpha}{a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha} \end{aligned}

Putting these values in (iv), we get

\begin{aligned} &=\frac{b^{2} p^{2} a^{4} \cos ^{2} \alpha}{\left(\left(a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\right)\right)^{2}}+\frac{a^{2} p^{2} b^{4} \sin ^{2} \alpha}{\left(\left(a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\right)\right)^{2}}=a^{2} b^{2} \\ \end{aligned}

\begin{aligned}&=\frac{a^{2} b^{2} p^{2}\left(a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\right)}{\left(a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha\right)^{2}}=a^{2} b^{2} \\ &=a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha=P^{2} \end{aligned}

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