Complete Question
The complete question is shown on the first and second uploaded image
Answer:
The derived [tex]h = \frac{b^{2} - 2a^{2} }{2a^{2}}[/tex]
Step-by-step explanation:
Step One : Consider the ellipse in equilibrium.
Looking at the ellipse in equilibrium i.e when the ellipse has settled down on the concave support (represented by a parabola ) as shown on the third uploaded image.
Step Two : Consider the ellipse equation.
Generally the equation of the ellipse is given as
[tex]\frac{x^{2}}{a^{2}} + \frac{(y-h)^{2}}{b^{2}}\\[/tex]
Also the base on which it rest at equilibrium i.e the parabola is represented by [tex]y = x^{2} -1[/tex]
Substituting the value of y in the ellipse equation we have
[tex]\frac{x^{2}}{a^{2}}+ \frac{(x^{2}-1-h)^{2}}{b^{2}}=1[/tex]
Let [tex]x^{2} = t[/tex]
So the equation becomes
[tex]\frac{t}{a^{2}} + \frac{(t -1 -h)^{2}}{b^{2}} =1[/tex]
Rearranging, we get :
[tex]a^{2} t^{2} + (b^{2}- 2a^{2}(h +1))t + a^{2}((h +1 )^{2} -b^{2} =0[/tex]
This equation above is a quadratic equation or a bi-quadratic equation in x as t = [tex]x^{2}[/tex]
Step Three : Relate the equation an the graph on the third uploaded image
We can see that from the graph , if A and B are the two values of x for which the points is made , then A + B = 0 (because they are symmetric in nature)
From Vieta's Roots(Vieta's formula is a formula that shows the relationship between the coefficients of a polynomial and the sum of its roots )
[tex]ax^{2} + bx + c =0[/tex]
with A and B as roots
A+B = [tex]\frac{-b}{a}[/tex]
But A + B = 0
So [tex]\frac{-b}{a}[/tex] = 0
or we can say that
[tex]\frac{b^{2} - 2a^{2}(h +1 )}{a^{2}}[/tex]
Rearranging we get
[tex]h = \frac{b^{2}- 2a^{2}}{2a^{2}}[/tex]
