The angle at which the curve $y={ x }^{ 2 }$ and the curve $x=\cfrac { 5 }{ 3 } \cos { t } ,y=\cfrac { 5 }{ 4 } \sin { t }$ intersect is

Given

$y={ x }^{ 2 }...(i)$

$\quad x=\cfrac { 5 }{ 3 } \cos { t } ;y=\cfrac { 5 }{ 4 } \sin { t } ...(ii)$

Which is parametric equation, we change this equation is caresian equation as follows

$\cos { t } =\cfrac { 3 }{ 5 } x;\sin { t } =\cfrac { 4 }{ 5 } y$

On Squaring and adding both i.e., $\cos{t}$ and $\sin {t}$ we get

$\cfrac { 9 }{ 25 } { x }^{ 2 }+\cfrac { 16 }{ 25 } { y }^{ 2 }=\cos ^{ 2 }{ t } +\sin ^{ 2 }{ t }$

$\Rightarrow 9{ x }^{ 2 }+16{ y }^{ 2 }=25....(iii)\quad \quad \left[ \because \cos ^{ 2 }{ \theta } +\sin ^{ 2 }{ \theta } =1 \right]$

$\therefore$ The intersection points at Eq.(i) and (iii) are $(1,1)$ and $(-1,1)$

Now, slope of tangent of Eq.(i) at point $(1,1)$ is

${ m }_{ 1 }=\cfrac { dy }{ dx } =2x\quad \therefore { m }_{ 1 }={ \left| \cfrac { dy }{ dx } \right| }_{ (1,1) }=2$

And slope of tangent of Eq. (iii) at point $(1,1)$ is

${ m }_{ 2 }=\cfrac { dy }{ dx } =-\cfrac { 9 }{ 16 }$

$\therefore$ Angle at point of intersection of Eqs. (i) and (iii) we get

${ \theta }_{ 1 }=\tan ^{ -1 }{ \left| \cfrac { { m }_{ 1 }-{ m }_{ 2 } }{ 1+{ m }_{ 1 }{ m }_{ 2 } } \right| } =\tan ^{ -1 }{ \cfrac { 41 }{ 2 } }$

similarly, slope of tangent of Eq. (i) at point $(-1,1)$

${ m }_{ 1 }={ \left| \cfrac { dy }{ dx } \right| }_{ (-1,1) }=-2$

And slope of tangent of Eq. (iii) at point $(-1,1)$

${ m }_{ 2 }=\cfrac { dy }{ dx } =\cfrac { 9 }{ 16 }$

$\therefore$ Angle at point of intersection of Eqs. (i) and (iii) we get

${ \theta }_{ 2 }=\tan ^{ -1 }{ \left| \cfrac {-2- \cfrac { 9 }{ 16 } }{ 1-\cfrac { 18 }{ 16 } } \right| } =\tan ^{ -1 }{ \cfrac { 41 }{ 2 } } \quad \quad$