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Question: The locus of the image of the focus of the ellipse \[\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1,...

The locus of the image of the focus of the ellipse x225+y29=1,  (a>b),\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1,\;(a > b), with respect to any of the tangents to the ellipse is:
A. (x+4)2+y2=100{\left( {x + 4} \right)^2} + {y^2} = 100
B. (x+2)2+y2=50{\left( {x + 2} \right)^2} + {y^2} = 50
C. (x−4)2+y2=100{\left( {x - 4} \right)^2} + {y^2} = 100
D. (x−2)2+y2=50{(x - 2)^2} + {y^2} = 50

Explanation

Solution

Hint : To find the locus of the image of the focus of the given ellipse with respect to any of its tangent, first of all find coordinate of the focus of the ellipse and then let the coordinate of the locus to be (h,  k)(h,\;k) and then find the equation accordingly.

Complete step-by-step answer :
In order to find the locus of the image of the focus of the given ellipse x225+y29=1,  (a>b),\dfrac{{{x^2}}}{{25}} + \dfrac{{{y^2}}}{9} = 1,\;(a > b), we will first find coordinate of its focus,
From the given equation, we get
a=5  and  b=9a = 5\;{\text{and}}\;b = 9
Therefore eccentricity will be given as
e=1−(ba)2=1−(35)2=1−925=1625=45e = \sqrt {1 - {{\left( {\dfrac{b}{a}} \right)}^2}} = \sqrt {1 - {{\left( {\dfrac{3}{5}} \right)}^2}} = \sqrt {1 - \dfrac{9}{{25}}} = \sqrt {\dfrac{{16}}{{25}}} = \dfrac{4}{5}
So coordinates of the focus will be given as
f≡(ae,  0)≡(5×45,  0)≡(4,  0)f \equiv (ae,\;0) \equiv \left( {5 \times \dfrac{4}{5},\;0} \right) \equiv \left( {4,\;0} \right)
Now, let us the coordinates of locus of image of its focus be (h,  k)(h,\;k) , then the locus will be given as
(h±42)2+(k2)2=52{\left( {\dfrac{{h \pm 4}}{2}} \right)^2} + {\left( {\dfrac{k}{2}} \right)^2} = {5^2}
Which is the equation of an auxiliary circle in which the locus of image of the focus will lie, solving the equation further and then replacing h  and  kh\;{\text{and}}\;k with x  and  yx\;{\text{and}}\;y we will get
(h±4)2+k2=25×4 (h±4)2+k2=100 (x±4)2+y2=100   {\left( {h \pm 4} \right)^2} + {k^2} = 25 \times 4 \\\ {\left( {h \pm 4} \right)^2} + {k^2} = 100 \\\ {\left( {x \pm 4} \right)^2} + {y^2} = 100 \;
Therefore (x±4)2+y2=100{\left( {x \pm 4} \right)^2} + {y^2} = 100 is the required equation of the locus of the image of the focus of the given ellipse. So option A. and C. both are the correct options.
So, the correct answer is “Option A and C”.

Note : A plane curve circling two focal points is called an ellipse; it is typically oval-shaped. An ellipse is drawn by drawing a point moving in a plane such that the amount of its distance from the two focal points is constant. When the two focal points become the same, the ellipse is turned into a circle.