Courses
Courses for Kids
Free study material
Offline Centres
More
Store Icon
Store

Write the parametric equation of ellipse \[\dfrac{{{{(x - 3)}^2}}}{{25}} + \dfrac{{{{(y + 2)}^2}}}{{16}} = 1\].

seo-qna
Last updated date: 22nd Mar 2024
Total views: 393.9k
Views today: 6.93k
MVSAT 2024
Answer
VerifiedVerified
393.9k+ views
Hint: The standard equation for an ellipse \[\dfrac{{{{(x - h)}^2}}}{{{a^2}}} + \dfrac{{{{(y - k)}^2}}}{{{b^2}}} = 1\], where ( h, k ) is the centre of the ellipse. The parametric form for an ellipse is F(t) = ( x(t), y(t)) where \[x(t) = r\cos (t) + h\] and \[y(t) = r\sin (t) + k\].

Complete step-by-step answer:
Given equation of ellipse is \[\dfrac{{{{(x - 3)}^2}}}{{25}} + \dfrac{{{{(y + 2)}^2}}}{{16}} = 1\] ----( 1 )
So, when we compare this equation of ellipse with the standard equation of ellipse we will get 3 as h and -2 as k ( as the general formula of ellipse has ( y - k ) but here 2 is positive so we had to take this 2 as -2 because ( y – (-2)) = ( y + 2 ) )
Now let us assume \[(x - 3)\]as P and \[(y + 2)\]as Q
Now putting the value of \[(x - 3)\]and \[(y + 2)\] in equation 1.
\[ \Rightarrow \dfrac{{{{(P)}^2}}}{{25}} + \dfrac{{{{(Q)}^2}}}{{16}} = 1\] ( \[\sqrt {25} = 5\] and \[\sqrt {16} = 4\] )
\[ \Rightarrow \dfrac{{{{(P)}^2}}}{{{5^2}}} + \dfrac{{{{(Q)}^2}}}{{{4^2}}} = 1\] and this ellipse is in the form of \[\dfrac{{{x^2}}}{{{a^2}}} + \dfrac{{{y^2}}}{{{b^2}}} = 1\]
So, from above a = 5 and b = 4
And we know that the parametric ellipse for any function is \[x = a\cos \theta + h\] and \[y = b\sin \theta + k\] here r = a and b.
Now let us understand this concept very clearly that here we had already included h and k in the parametric ellipse equation so we had to directly put the values of ( a, b ) and ( h, k ) from equation ( 1 ) to this parametric ellipse equation.
Now as we had solved above we get ( a = 5 , b = 4 ) and ( h = 3, k = -2 ) so putting these values in a parametric ellipse equation.
So, \[x = 5\cos \theta + 3\] and \[y = 4\sin \theta + ( - 2)\]
So, the parametric equation for ellipse \[\dfrac{{{{(x - 3)}^2}}}{{25}} + \dfrac{{{{(y + 2)}^2}}}{{16}} = 1\] is \[x = 5\cos \theta + 3\] and \[y = 4\sin \theta - 2\].

Note :- whenever we come up with this type of situation in which we are given an ellipse equation and we had to find the equation of parametric for the same ellipse then we always had to compare the given equation with the general equation and if we had to find a general equation from a parametric equation then we must divide the parametric equation by a first and then divide the equation by b and in the end we had to square the equation and add them up to obtain a standard equation of the ellipse.