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

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}