Question

2.Plzzzz sove this question as soon as possible
see the attachement

Answer 1

$\bold{Given}\longrightarrow \\ curve \: c \: is \: given \: by \: parametric \: eqution \\ is \\ x = 2 {t}^{2} \: \: \: y = {3}^{t} \\ tangent \: drawn \: to \: C \: at \: point \: P \: (8 \: 9) \\ meets \: y \: axis \: at \: Q$

$\bold{concept \: used}\longrightarrow \\ 1) \dfrac{d}{dx} ( {a}^{x} ) = {a}^{x} log_{e}(a) \\ 2) \dfrac{d}{dx} ( \sqrt{x} ) = \dfrac{1}{2 \sqrt{x} } \\ 3)eqution \: of \: tangent \: at \: ( \alpha , \: \beta ) \\ (y - \beta ) = \dfrac{dy}{dx} at( \alpha , \: \beta ) \: \: (x - \alpha )$

$\bold{To \: prove}\longrightarrow \: y \: coordinate \: of \: q \: is \: \\ 9 - 9 log_{e}(3)$

Solution---> x= 2t² , y = 3ᵗ

$now \\ x = 2 {t}^{2} \\ = > {t}^{2} = \dfrac{x}{2} \\ = > t = \sqrt{ \dfrac{x}{2} } \\ now \\ y = {3}^{t} \\ = > y = {3}^{ \sqrt{ \frac{x}{2} } } \\ differentiating \: with \: respect \: to \: x \\ = > \dfrac{dy}{dx} = \dfrac{d}{dx} ( {3}^{ \sqrt{ \frac{x}{2} } } ) \\ = > \dfrac{dy}{dx} = {3}^{ \sqrt{ \frac{x}{2} } } log_{e}(3) \dfrac{d}{dx} ( \sqrt{ \dfrac{x}{2} \: } \: ) \\ = > \dfrac{dy}{dx} = {3}^{ \sqrt{ \frac{x}{2} } } log_{e}(3) \dfrac{1}{2 \sqrt{ \frac{x}{2} } } \dfrac{d}{dx} ( \dfrac{x}{2} \\ = > \dfrac{dy}{dx} = {3}^{ \sqrt{ \frac{x}{2} } } log_{e}(3) \dfrac{1}{ \sqrt{2} x} ( \dfrac{1}{2} ) \\ \dfrac{dy}{dx} = \dfrac{ log_{e}(3) {3}^{ \sqrt{ \frac{x}{2} } } }{2 \sqrt{2x} } \\ slope \: of \: tangent \: at \: point \: P \\ = log_{e}(3) \: \: {3}^{ \sqrt{ \frac{8}{2} } } \dfrac{1}{2 \sqrt{2(8)} } \\ = \dfrac{ log_{e}(3) \: {3}^{ \sqrt{4} } }{2 \sqrt{16} } \\ = \dfrac{ {3}^{2} log_{e}(3) }{2 \times 4} \\ = \dfrac{9 log_{e}(3) }{8}$

$equation \: of \: tangent \: at \: point \: (8 \: , \: 9)$

$(y - 9) = \dfrac{9}{8} log_{e}(3) (x - 8)$

$now \: tangent \: meet \: y \: axis \: at \: Q \: \\ at \: Q \: x = 0 \: so \: putting \: x = 0 \: in \: above \: equation$

$y - 9 = \dfrac{9}{8} log_{e}(3) (0 - 8)$

$= > y - 9 = - \dfrac{9}{8} \: \: 8 \: log_{e}(3)$

$= > y - 9 = - 9 log_{e}(3)$

$= > y = 9 - 9 log_{e}(3)$

$so \: y \: coordinate \: of \: Q \\ = 9 - 9log3$