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Q.
The equation of the tangent to $4x^2 - 9y^2 = 36$ which is perpendicular to the straight line $5x + 2y - 10 = 0$ is
Application of Derivatives
Solution:
$4x^2-9y^2 = 36\quad\cdots\left(1\right)$
$\Rightarrow 8x - 18\, y \frac{dy}{dx} = 0$
$\Rightarrow \frac{dy}{dx} = \frac{4x}{9\,y}$
$\therefore $ Slope of the tangent $= \frac{4x}{9\,y}$.
Also, slope of line $5x + 2y - 10 = 0$ is $\frac{-5}{2}$
$\therefore $ Line is perpendicular to the tangent. So product of slopes $= - 1$
$\therefore \frac{4x}{9\,y} \times\left(-\frac{5}{2}\right) = -1$
$\Rightarrow y = \frac{10x}{9}\quad \cdots \left(2\right)$
Using $\left(2\right)$ in $\left(1\right)$, we get
$4x^{2} - \frac{100x^{2}}{9} = 36$
$\Rightarrow -64x^{2} = 324$
which gives imaginary $x$.
Hence, there is no point on the curve at which tangent is perpendicular to the given line.