MCA NIMCET Ellipse Previous Year Question Paper and Solution with PDF

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MCA NIMCET Previous Year Questions (PYQs)

MCA NIMCET Ellipse PYQ

Qus : 1 MCA NIMCET PYQ

The eccentricity of an ellipse whose center is at the origin is $\frac{1}{2}$. If one of its directrices is $x=-4$, find the equation of the normal to the ellipse at the point $\left(1,\frac{3}{2}\right)$.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2026 PYQ

Solution

For ellipse,

$e=\frac{1}{2}$

Directrix is $x=-\frac{a}{e}$

Given, $-\frac{a}{e}=-4$

$\frac{a}{e}=4$

$a=4e=4\cdot \frac{1}{2}=2$

Now, $b^2=a^2(1-e^2)$

$b^2=4\left(1-\frac{1}{4}\right)$

$b^2=3$

So ellipse is $\frac{x^2}{4}+\frac{y^2}{3}=1$

Equation of normal to ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$ at point $(x_1,y_1)$ is $\frac{a^2x}{x_1}-\frac{b^2y}{y_1}=a^2-b^2$

Here, $a^2=4,\ b^2=3,\ x_1=1,\ y_1=\frac{3}{2}$

So, $\frac{4x}{1}-\frac{3y}{3/2}=4-3$

$4x-2y=1$

Answer: $4x-2y=1$

Qus : 2 MCA NIMCET PYQ

The equation of ellipse with major axis along the x–axis and passes through the point $(4,3)$ and $(-1,4)$.

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2012 PYQ

Solution

Ellipse form:

$\displaystyle \frac{x^2}{a^2} + \frac{y^2}{b^2} = 1,\quad a^2 > b^2$

Using points: $\displaystyle \frac{16}{a^2} + \frac{9}{b^2} = 1$

$\displaystyle \frac{1}{a^2} + \frac{16}{b^2} = 1$

Solve gives:

$\displaystyle 7x^2 + 15y^2 = 247$

Qus : 3 MCA NIMCET PYQ

If S and S' are foci of the ellipse $\frac{x^2}{a^2}+\frac{y^2}{b^2}=1$, B is the end of the minor axis and BSS' is an equilateral triangle, then the eccentricity of the ellipse is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

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, $$\sqrt{a^2e^2 + a^2(1-e^2)} = 2ae,$$ $$\sqrt{a^2} = 2ae,$$ $$a = 2ae \;\;\Rightarrow\;\; e = \tfrac{1}{2}.$$

Final Answer: The eccentricity of the ellipse is $$\boxed{\tfrac{1}{2}}.$$

Qus : 4 MCA NIMCET PYQ

Eccentricity of ellipse $ 9x^{2} + 5y^{2} - 30y = 0 $

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2011 PYQ

Solution

Complete square: $ 5(y^{2} - 6y) = 5[(y-3)^{2} - 9] $ Thus: $ 9x^{2} + 5(y-3)^{2} = 45 $ Divide by 45: $ \frac{x^{2}}{5} + \frac{(y-3)^{2}}{9} = 1 $ Major axis along $y$ $ a^{2} = 9,\ b^{2} = 5 $ Eccentricity: $ e = \sqrt{1 - \frac{b^{2}}{a^{2}}} = \sqrt{1 - \frac{5}{9}} = \frac{2}{3} $

Qus : 5 MCA NIMCET PYQ

Equation of the tangent from the point (3,−1) to the ellipse 2x² + 9y² = 3 is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2019 PYQ

Solution

Qus : 6 MCA NIMCET PYQ

If (4, 3) and (12, 5) are the two foci of an ellipse passing through the origin, then the eccentricity of the ellipse is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

Solution

For any point $ P $ on the ellipse: $ PF_1 + PF_2 = 2a $

$$OF_1 = \sqrt{(4-0)^2 + (3-0)^2} = \sqrt{25} = 5$$

$$\boxed{e = \dfrac{\sqrt{17}}{9}}$$

Qus : 7 MCA NIMCET PYQ

The line 3x + 5y = k touches the ellipse 16 x2 + 25 y2 = 400, if k is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

Solution

Qus : 8 MCA NIMCET PYQ

The sum of the focal distances of any point on the ellipse $\frac{x^{2}}{a^{2}} +\frac{y^{2}}{b^{2}} =1$ with eccentricity $e$ is given by

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2013 PYQ

Solution

Qus : 9 MCA NIMCET PYQ

The equation $3x^2 + 10xy + 11y^2 + 14x + 12y + 5 = 0$ represents

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2024 PYQ

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.

Qus : 10 MCA NIMCET PYQ

The eccentricity of an ellipse, with its center at the origin is $\frac{1}{3}$ . If one of the directrices is $x=9$, then the equation of ellipse is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2022 PYQ

Solution

Qus : 11 MCA NIMCET PYQ

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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 12 MCA NIMCET PYQ

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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2010 PYQ

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)

Qus : 13 MCA NIMCET PYQ

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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2021 PYQ

Solution

Qus : 14 MCA NIMCET PYQ

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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2015 PYQ

Solution

Qus : 15 MCA NIMCET PYQ

The tangent to an ellipse x² + 16y² = 16 and making angel 60° with X-axis is:

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2020 PYQ

Solution

Qus : 16 MCA NIMCET PYQ

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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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2014 PYQ

Solution

Qus : 17 MCA NIMCET PYQ

Area of the greatest rectangle that can be inscribed in the ellipse is

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MCA NIMCET Previous Year PYQ MCA NIMCET NIMCET 2016 PYQ

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:**
$A_{max} = 2ab \times 1$
$\therefore \boxed{A_{max} = 2ab}$

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Rule for Classifying Conics Using Discriminant

? Based on value of \( \Delta \):

Example:

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