Application of Derivatives — Class 12 Maths Solution

Given, line is $x\cos \alpha + y\sin \alpha = p$ …….(i)
and curve is $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}} = 1$
$\Rightarrow$ ${b^2}{x^2} + {a^2}{y^2} = {a^2}{b^2}$ …….(ii)

Now, differentiating Eq. (ii) w.r.t. $x$, we get
${b^2} \cdot 2x + {a^2} \cdot 2y \cdot \frac{{dy}}{{dx}} = 0$
$\Rightarrow$ $\frac{{dy}}{{dx}} = \frac{{ - 2{b^2}x}}{{2{a^2}y}} = \frac{{ - x{b^2}}}{{y{a^2}}}$

……..(iii)

From Eq. (i), $y\sin \alpha = p - x\cos \alpha$
$\Rightarrow$ $y = - x\cot \alpha + \frac{p}{{\sin \alpha }}$
Thus, slope of the line is $( - \cot \alpha )$.

So, the given equation of line will be tangent to the Eq. (ii), if $\left( { - \frac{x}{y} \cdot \frac{{{b^2}}}{{{a^2}}}} \right) = ( - \cot \alpha )$

$\Rightarrow$ $\quad \frac{x}{{{a^2}\cos \alpha }} = \frac{y}{{{b^2}\sin \alpha }} = k$ [say]

$\Rightarrow$ $\quad x = k{a^2}\cos \alpha$
and $y = {b^2}k\sin \alpha$

So, the line $x\cos \alpha + y\sin \alpha = p$ will touch the curve $\frac{{{x^2}}}{{{a^2}}} + \frac{{{y^2}}}{{{b^2}}}$ at the point
$\left( {k{a^2}\cos \alpha ,k{b^2}\sin \alpha } \right)$.

Therefore from Eq. (i),

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

$\Rightarrow$ ${a^2}{\cos ^2}\alpha + {b^2}{\sin ^2}\alpha = \frac{p}{k}$

$\Rightarrow$ ${\left( {{a^2}{{\cos }^2}\alpha + {b^2}{{\sin }^2}\alpha } \right)^2} = \frac{{{p^2}}}{{{k^2}}}$ ……..(iv)

From Eq. (ii), ${b^2}{k^2}{a^4}{\cos ^2}\alpha + {a^2}{k^2}{b^4}{\sin ^2}\alpha = {a^2}{b^2}$

$\Rightarrow$ ${k^2}\left( {{a^2}{{\cos }^2}\alpha + {b^2}{{\sin }^2}\alpha } \right) = 1$

$\Rightarrow$ $\left( {{a^2}{{\cos }^2}\alpha + {b^2}{{\sin }^2}\alpha } \right) = \frac{1}{{{k^2}}}$ …….(v)

On dividing Eq. (iv) by Eq. (v), we get
${a^2}{\cos ^2}\alpha + {b^2}{\sin ^2}\alpha = {p^2}$

Hence proved.