Find the equations of the tangent and the normal, to the curve at the point <img alt="\left ( x_{1},y

Find the equations of the tangent and the normal, to the curve $16x^{2}+9y^{2}= 145$ at the point $\left ( x_{1},y_{1} \right )$ , where $x_{1}= 2\, and\: y_{1}> 0\cdot$

Answers (1)

given curve is $16x^{2}+9y^{2}= 145---(i)$
$\because \left ( x_{1},y_{1} \right )$ lies on equation ---(i)
$\therefore 16x_{1}^{2}+9y_{1}^{2}= 145$
$\Rightarrow 16\left ( 2 \right )^{2}+9y_{1}^{2}= 145\; \; \left [ \because x_{1} = 2\left ( given \right )\right ]$
$\Rightarrow 9y_{1}^{2}= 145-64= 81$
$\Rightarrow y_{1}= 3\; \; \left [ \because y_{1} \geq 0\left ( given \right )\right ]$
$\therefore$ point of containt is (2,3)
Differentiating equation (1) w.r.t x, which will give us the slope of the tangent
$32x+18y\: \: \frac{dy}{dx}= 0$
$\Rightarrow \frac{dy}{dx}= \frac{-32x}{18y}$
Slope of the tangent at (2,3) $= \left [ \frac{dy}{dx} \right ]_{\left ( 2,3 \right )}$
$=\frac{-32\left ( 2 \right )}{18\left ( 3 \right )}= \frac{-64}{54}$
$= \frac{-32}{27}$
$\therefore$ Equation of tangent is
$y-3= m\left ( x-2 \right )$
$\Rightarrow y-3= \frac{-32}{27}\left ( x-2 \right )$
$\Rightarrow27y-81= -32x+64$
$\Rightarrow32x+27y= 145$
The slope of the normal $= \frac{-1}{slope\; of\; tangent}$
$\therefore$ Equation of normal is
$\Rightarrow y-3= \frac{27}{32}\left ( x-2 \right )$
$\Rightarrow 32y-96= 27x-54$
$\Rightarrow 27x-32y= -42$