NIMCET 2024 Mathematics PYQ — If and are the two focii of an ellipse passing through the origin… | Mathem Solvex | Mathem Solvex
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NIMCET 2024 — Mathematics PYQ
NIMCET | Mathematics | 2024
If (4,3) and (12,5) are the two focii of an ellipse passing through the origin, then the eccentricity of the ellipse is
Choose the correct answer:
A.
913
B.
1813
C.
1817
D.
917
(Correct Answer)
Correct Answer:
917
Explanation
To find the eccentricity (e) of the ellipse, we can use fundamental geometric properties relating the foci, any point on the ellipse, and the major axis.
Step 1: Find the distance between the two foci (2ae)
Let the two foci be S1=(4,3) and S2=(12,5).
In any ellipse, the distance between the two foci is equal to 2ae, where a is the semi-major axis and e is the eccentricity.
Using the distance formula d=(x2−x1)2+(y2−y1)2:
2ae=(12−4)2+(5−3)2
2ae=(8)2+(2)2
2ae=64+4=68
We can simplify 68 as 4×17=217:
2ae=217
ae=17— (Equation 1)
Step 2: Use the focal property of an ellipse to find the length of the major axis (2a)
A fundamental property of an ellipse states that the sum of the distances from any point P on the ellipse to its two foci is constant and equal to the length of the major axis (2a).
PS1+PS2=2a
The problem states that the ellipse passes through the origin. Therefore, we can take our point P as the origin, O(0,0):
OS1+OS2=2a
Now, calculate the distances from the origin (0,0) to each focus:
Distance to S1(4,3):
OS1=(4−0)2+(3−0)2=16+9=25=5
Distance to S2(12,5):
OS2=(12−0)2+(5−0)2=144+25=169=13
Substitute these values back to find 2a:
5+13=2a
18=2a
a=9— (Equation 2)
Step 3: Calculate the eccentricity (e)
Now substitute the value of a=9 from Equation 2 into Equation 1:
9e=17
e=917
Correct Answer
The correct option is (D)917.
Explanation
To find the eccentricity (e) of the ellipse, we can use fundamental geometric properties relating the foci, any point on the ellipse, and the major axis.
Step 1: Find the distance between the two foci (2ae)
Let the two foci be S1=(4,3) and S2=(12,5).
In any ellipse, the distance between the two foci is equal to 2ae, where a is the semi-major axis and e is the eccentricity.
Using the distance formula d=(x2−x1)2+(y2−y1)2:
2ae=(12−4)2+(5−3)2
2ae=(8)2+(2)2
2ae=64+4=68
We can simplify 68 as 4×17=217:
2ae=217
ae=17— (Equation 1)
Step 2: Use the focal property of an ellipse to find the length of the major axis (2a)
A fundamental property of an ellipse states that the sum of the distances from any point P on the ellipse to its two foci is constant and equal to the length of the major axis (2a).
PS1+PS2=2a
The problem states that the ellipse passes through the origin. Therefore, we can take our point P as the origin, O(0,0):
OS1+OS2=2a
Now, calculate the distances from the origin (0,0) to each focus:
Distance to S1(4,3):
OS1=(4−0)2+(3−0)2=16+9=25=5
Distance to S2(12,5):
OS2=(12−0)2+(5−0)2=144+25=169=13
Substitute these values back to find 2a:
5+13=2a
18=2a
a=9— (Equation 2)
Step 3: Calculate the eccentricity (e)
Now substitute the value of a=9 from Equation 2 into Equation 1:
