Question

If the line $x\cos \alpha +y\sin \alpha =P$ touches the curve ${\left(\frac{x}{a}\right)}^m+{\left(\frac{y}{b}\right)}^m=1,$ then $(a\cos \alpha {)}^{\frac{m}{m-1}}+(b\sin \alpha {)}^{\frac{m}{m-1}}=$

• A. $P$
• B. $P^{\frac{m}{m+1}}$
• C. $P^{\frac{m}{m-1}}$
• D. $-P$

Answer 1

The correct option is C $P^{\frac{m}{m-1}}$

Given, equation ${\left(\frac{x}{a}\right)}^m+{\left(\frac{y}{b}\right)}^m=1$
Differentiating the equation of curve w.r.t $x$, we get
$m{\left(\frac{x}{a}\right)}^{m-1}\cdot \frac{1}{a}+m{\left(\frac{y}{b}\right)}^{m-1}\cdot \frac{1}{b}\cdot \frac{dy}{dx}=0$
On simplifying, we get
$\frac{dy}{dx}=\frac{-b^m{x}^{m-1}}{a^m{y}^{m-1}}$
Hence, at any point $P(x_1,y_1)$ on the curve
Slope of tangent = ${\left(\frac{dy}{dx}\right)}_{(x_1,y_1)}=\frac{-b^m{x}_1^{m-1}}{a^m{y}_1^{m-1}}$
$\therefore$ equation of tangent at $P$ is
$y-y_1=\frac{-b^m{x}_1^{m-1}}{a^m{y}_1^{m-1}}(x-x_1)$
$\Rightarrow \frac{y{y}_1^{m-1}}{b^m}-\frac{y_1^m}{b^m}=-\frac{x{x}_1^{m-1}}{a^m}+\frac{x_1^m}{a^m}$
i.e $\frac{x}{a}\cdot {\left(\frac{x_1}{a}\right)}^{m-1}+\frac{y}{b}\cdot {\left(\frac{y_1}{b}\right)}^{m-1}={\left(\frac{x_1}{a}\right)}^m+{\left(\frac{y_1}{b}\right)}^m=1$
(Since $P$ lies on the curve at any point)

Hence, the equation of the tangent at $P(x_1,y_1)$ on the curve is,
$\frac{x}{a}\cdot {\left(\frac{x_1}{a}\right)}^{m-1}+\frac{y}{b}\cdot {\left(\frac{y_1}{b}\right)}^{m-1}=1\cdots \left(i\right)$
and $x\cos \alpha +y\sin \alpha =P\cdots \left(ii\right)$

If Eqn. $\left(ii\right)$ is the tangent, then coefficients of Eqs. $\left(i\right)$ and $\left(ii\right)$ must be proprtional for point $(x_1,y_1)$
i.e. $\frac{\cos \alpha }{\frac{1}{a}\cdot {\left(\frac{x_1}{a}\right)}^{m-1}}=\frac{\sin \alpha }{\frac{1}{b}\cdot {\left(\frac{y_1}{b}\right)}^{m-1}}=\frac{P}{1}$

This gives
$\frac{x_1}{a}={\left(\frac{a\cos \alpha }{P}\right)}^{\frac{1}{m-1}}$, $\frac{y_1}{b}={\left(\frac{b\sin \alpha }{P}\right)}^{\frac{1}{m-1}}$
Since point $P(x_1,y_1)$ lies on the curve,
${\left(\frac{x_1}{a}\right)}^m+{\left(\frac{y_1}{b}\right)}^m=1$
i.e.
${\left(\frac{a\cos \alpha }{P}\right)}^{\frac{m}{m-1}}+{\left(\frac{b\sin \alpha }{P}\right)}^{\frac{m}{m-1}}=1$
i.e. ${\left(a\cos \alpha \right)}^{\frac{m}{m-1}}+{\left(b\sin \alpha \right)}^{\frac{m}{m-1}}=P^{\frac{m}{m-1}}$