# If the straight-line x cos α + y sin α = p touches the curve

Question:

If the straight-line cos α + sin α = touches the curve x2/a2 + y2/b2 = 1, then prove that a2 cos2 α + b2 sin2 α = p2.

Solution:

The given curve is x2/a2 + y2/b2 = 1 and the straight-line cos α + sin α = p

Differentiating equation (i) w.r.t. x, we get

$\frac{1}{a^{2}} \cdot 2 x+\frac{1}{b^{2}} \cdot 2 y \cdot \frac{d y}{d x}=0$

$\Rightarrow \frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{b^{2}}{a^{2}} \cdot \frac{x}{y}$

So the slope of the curve $=\frac{-b^{2}}{a^{2}} \cdot \frac{x}{y}$

Now differentiating eq. (ii) w.r.t. $x$, we have

$\cos \alpha+\sin \alpha \cdot \frac{d y}{d x}=0$

$\therefore \quad \frac{d y}{d x}=\frac{-\cos \alpha}{\sin \alpha}=-\cot \alpha$

So, the slope of the straight line $=-\cot \alpha$ If the line is the tangent to the curve, then

$\frac{-b^{2}}{a^{2}} \cdot \frac{x}{y}=-\cot \alpha \Rightarrow \frac{x}{y}=\frac{a^{2}}{b^{2}} \cdot \cot \alpha \Rightarrow x=\frac{a^{2}}{b^{2}} \cot \alpha \cdot y$

Now from eq. (ii) we have $x \cos \alpha+y \sin \alpha=p$

$\frac{a^{2}}{b^{2}} \cdot \cot \alpha \cdot y \cdot \cos \alpha+y \sin \alpha=p$

$a^{2} \cot \alpha \cdot \cos \alpha y+b^{2} \sin \alpha y=b^{2} p$

$a^{2} \frac{\cos \alpha}{\sin \alpha} \cdot \cos \alpha y+b^{2} \sin \alpha y=b^{2} p$

$a^{2} \cos ^{2} \alpha y+b^{2} \sin ^{2} \alpha y=b^{2} \sin \alpha p$

\begin{aligned} a^{2} \cos ^{2} \alpha y+b^{2} \sin ^{2} \alpha y &=b^{2} \sin \alpha p \\ a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha &=\frac{b^{2}}{y} \cdot \sin \alpha \cdot p \end{aligned}

$\Rightarrow a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha=p \cdot p \quad\left[\because \frac{b^{2}}{y} \sin \alpha=p\right]$

$\frac{1}{a^{2}} \cdot 2 x+\frac{1}{b^{2}} \cdot 2 y \cdot \frac{d y}{d x}=0$

$\Rightarrow \frac{x}{a^{2}}+\frac{y}{b^{2}} \frac{d y}{d x}=0 \Rightarrow \frac{d y}{d x}=-\frac{b^{2}}{a^{2}} \cdot \frac{x}{y}$

So the slope of the curve $=\frac{-b^{2}}{a^{2}} \cdot \frac{x}{y}$

Now differentiating eq. (ii) w.r.t. $x$, we have

$\cos \alpha+\sin \alpha \cdot \frac{d y}{d x}=0$

$\therefore \quad \frac{d y}{d x}=\frac{-\cos \alpha}{\sin \alpha}=-\cot \alpha$

So, the slope of the straight line $=-\cot \alpha$ If the line is the tangent to the curve, then

$\frac{-b^{2}}{a^{2}} \cdot \frac{x}{y}=-\cot \alpha \Rightarrow \frac{x}{y}=\frac{a^{2}}{b^{2}} \cdot \cot \alpha \Rightarrow x=\frac{a^{2}}{b^{2}} \cot \alpha \cdot y$

Now from eq. (ii) we have $x \cos \alpha+y \sin \alpha=p$

$\frac{a^{2}}{b^{2}} \cdot \cot \alpha \cdot y \cdot \cos \alpha+y \sin \alpha=p$

$a^{2} \cot \alpha \cdot \cos \alpha y+b^{2} \sin \alpha y=b^{2} p$

$a^{2} \frac{\cos \alpha}{\sin \alpha} \cdot \cos \alpha y+b^{2} \sin \alpha y=b^{2} p$

$a^{2} \cos ^{2} \alpha y+b^{2} \sin ^{2} \alpha y=b^{2} \sin \alpha p$

$a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha=\frac{b^{2}}{y} \cdot \sin \alpha \cdot p$

$\Rightarrow a^{2} \cos ^{2} \alpha+b^{2} \sin ^{2} \alpha=p \cdot p \quad\left[\because \frac{b^{2}}{y} \sin \alpha=p\right]$

Therefore, a2 cos2 α + b2 sin2 α = p2.