NIMCET 2027 Ellipse Previous Year Questions (PYQs) – Page 1 of 2
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Solution
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Solution
The foci are at: $$S(ae,0), \quad S'(-ae,0)$$ and the end of the minor axis is: $$B(0,b), \quad \text{where } b^2 = a^2(1-e^2).$$
In an equilateral triangle ∆BSS′: $$BS = SS'.$$
Now, $$BS = \sqrt{(ae)^2 + b^2}, \quad SS' = 2ae.$$ Hence, $$\sqrt{a^2e^2 + b^2} = 2ae.$$
But, $$b^2 = a^2(1-e^2).$$
So, $$\sqrt{a^2e^2 + a^2(1-e^2)} = 2ae,$$ $$\sqrt{a^2} = 2ae,$$ $$a = 2ae \;\;\Rightarrow\;\; e = \tfrac{1}{2}.$$
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Solution
For any point \( P \) on the ellipse: \( PF_1 + PF_2 = 2a \)
$$OF_1 = \sqrt{(4-0)^2 + (3-0)^2} = \sqrt{25} = 5$$
$$\boxed{e = \dfrac{\sqrt{17}}{9}}$$
Solution
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Solution
Rule for Classifying Conics Using Discriminant
Given the equation: \( Ax^2 + Bxy + Cy^2 + Dx + Ey + F = 0 \)
Compute: \( \Delta = B^2 - 4AC \)
? Based on value of \( \Delta \):
- Ellipse: \( \Delta < 0 \) and \( A \ne C \), \( B \ne 0 \) → tilted ellipse
- Circle: \( \Delta < 0 \) and \( A = C \), \( B = 0 \)
- Parabola: \( \Delta = 0 \)
- Hyperbola: \( \Delta > 0 \)
Example:
For the equation: \( 3x^2 + 10xy + 11y^2 + 14x + 12y + 5 = 0 \)
\( A = 3 \), \( B = 10 \), \( C = 11 \) →
\( \Delta = 10^2 - 4(3)(11) = 100 - 132 = -32 \)
Since \( \Delta < 0 \), it represents an ellipse.
Solution
NIMCET 2027
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NIMCET 2027
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