NIMCET Ellipse Previous Year Questions (PYQs) – Page 2 of 2
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The locus of the point of intersection of tangents to the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ which meet right angles is
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Solution
If any tangent to the ellipse $\frac{X^2}{a^2} + \frac{Y^2}{b^2} = 1$ intercepts equal length $l$ on both axes, then $l =$
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Solution
Tangent intercept form: x/A + y/B = 1
Equal intercepts → A = B = l
Condition for tangent to ellipse:
1 = a²/A² + b²/B² = a²/l² + b²/l²
⇒ l² = a² + b²
⇒ l = √(a² + b²)
The eccentric angle of the extremities of latus-rectum of the ellipse $\frac{{x}^2}{{a}^2}^{}+\frac{{y}^2}{{b}^2}^{}=1$ are given by
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Solution
The foci of the ellipse $\frac{x^{2}}{16}+\frac{y^{2}}{b^{2}}=1$ and the hyperbola $\frac{x^{2}}{144}-\frac{y^{2}}{{81}}=\frac{1}{25}$ coincide, then the value of $b^{2}$ is
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Solution
The tangent to an ellipse x2 + 16y2 = 16 and making angel 60° with X-axis is:
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Solution
The condition that the line lx + my + n = 0 becomes a tangent to the ellipse $\frac{x^{2}}{a^{2}}+\frac{y^{2}}{b^{2}}=1$ , is
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Solution
Area of the greatest rectangle that can be inscribed in the ellipse is
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Solution
Given ellipse: $\dfrac{x^2}{a^2} + \dfrac{y^2}{b^2} = 1$
**Setting up the Rectangle:**
Let $P(a\cos\theta,\ b\sin\theta)$ be a point on the ellipse
Then the rectangle has:
Length $= 2a\cos\theta$
Breadth $= 2b\sin\theta$
**Area of Rectangle:**
$A = 2a\cos\theta \times 2b\sin\theta$
$A = 4ab\sin\theta\cos\theta$
$A = 2ab\sin 2\theta$
**Maximizing the Area:**
$A$ is maximum when $\sin 2\theta$ is maximum
$\Rightarrow \sin 2\theta = 1$
$\Rightarrow 2\theta = 90^\circ$
$\Rightarrow \theta = 45^\circ$
**Maximum Area:**
$A_{max} = 2ab \times 1$
$\therefore \boxed{A_{max} = 2ab}$
**Setting up the Rectangle:**
Let $P(a\cos\theta,\ b\sin\theta)$ be a point on the ellipse
Then the rectangle has:
Length $= 2a\cos\theta$
Breadth $= 2b\sin\theta$
**Area of Rectangle:**
$A = 2a\cos\theta \times 2b\sin\theta$
$A = 4ab\sin\theta\cos\theta$
$A = 2ab\sin 2\theta$
**Maximizing the Area:**
$A$ is maximum when $\sin 2\theta$ is maximum
$\Rightarrow \sin 2\theta = 1$
$\Rightarrow 2\theta = 90^\circ$
$\Rightarrow \theta = 45^\circ$
**Maximum Area:**
$A_{max} = 2ab \times 1$
$\therefore \boxed{A_{max} = 2ab}$
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