NIMCET Mathematics Previous Year Questions (PYQs) – Page 95 of 95
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Solution
Solution
Finding Foci of a Conic
Given Equation: \( x^2 + 2x - 4y^2 + 8y - 7 = 0 \)
Step 1: Complete the square
⇒ \( (x + 1)^2 - 4(y - 1)^2 = 4 \)
Rewriting: \( \frac{(x + 1)^2}{4} - \frac{(y - 1)^2}{1} = 1 \)
This is a horizontal hyperbola with:
- Center: \( (-1, 1) \)
- \( a^2 = 4 \), \( b^2 = 1 \)
- \( c = \sqrt{a^2 + b^2} = \sqrt{5} \)
✅ Foci: \( (-1 \pm \sqrt{5},\ 1) \)
Solution
Let distances from the lighthouse be $x$ (for $30^\circ$) and $y$ (for $45^\circ$). With height $h=100$ m:
$\tan 30^\circ=\dfrac{100}{x}\ \Rightarrow\ x=\dfrac{100}{\tan30^\circ}=100\sqrt{3}$, $\tan 45^\circ=\dfrac{100}{y}\ \Rightarrow\ y=\dfrac{100}{\tan45^\circ}=100$.
Ships are on opposite sides ⇒ distance $=x+y=100\sqrt{3}+100=100(\sqrt{3}+1)\approx \boxed{273.2\ \text{m}}$.
Solution
Solution
Solution
Solution
Two Towers – Height of the Second Tower
Let the first tower be \(AB\) (top \(A\), base \(B\)) and the second tower be \(CD\) (top \(C\), base \(D\)). The bases \(B\) and \(D\) are 25 m apart.
From the top \(A\): angle of depression to base \(D\) is \(60^\circ\) and to top \(C\) is \(30^\circ\).
- From right \(\triangle ABD\): \[ \tan 60^\circ=\frac{AB}{BD}=\frac{AB}{25}\;\Rightarrow\; AB=25\sqrt{3}. \]
- From right \(\triangle ACD\): vertical difference \(=AB-CD\) and horizontal \(=25\). \[ \tan 30^\circ=\frac{AB-CD}{25}=\frac{1}{\sqrt{3}} \;\Rightarrow\; AB-CD=\frac{25}{\sqrt{3}}. \]
- Substitute \(AB=25\sqrt{3}\): \[ 25\sqrt{3}-CD=\frac{25}{\sqrt{3}} \;\Rightarrow\; CD=25\!\left(\sqrt{3}-\frac{1}{\sqrt{3}}\right) =\frac{50}{\sqrt{3}}\;\text{m}. \]
Solution
Locus of Midpoint of Chords
Given Parabola: \( y^2 = 4x \)
Condition: Chords pass through the vertex \( (0, 0) \)
Let the other end of the chord be \( (x_1, y_1) \), so the midpoint is:
\( M = \left( \frac{x_1}{2}, \frac{y_1}{2} \right) = (h, k) \)
Since the point lies on the parabola: \( y_1^2 = 4x_1 \)
⇒ \( (2k)^2 = 4(2h) \)
⇒ \( 4k^2 = 8h \)
⇒ \( \boxed{k^2 = 2h} \)
✅ Locus of midpoints: \( y^2 = 2x \)
Solution
Given hyperbola: $\dfrac{x^2}{144} - \dfrac{y^2}{81} = \dfrac{1}{25}$ ...(ii)
**For Hyperbola:**
Rewriting (ii) in standard form:
$\dfrac{x^2}{\dfrac{144}{25}} - \dfrac{y^2}{\dfrac{81}{25}} = 1$
Here, $a^2 = \dfrac{144}{25}$ and $b^2 = \dfrac{81}{25}$
For hyperbola, $c^2 = a^2 + b^2$
$\Rightarrow c^2 = \dfrac{144}{25} + \dfrac{81}{25} = \dfrac{225}{25} = 9$
$\therefore c = 3$
So, foci of hyperbola $= (\pm 3,\ 0)$
**For Ellipse:**
Here, $a^2 = 16$
For ellipse, $c^2 = a^2 - b^2$
$\Rightarrow c^2 = 16 - b^2$
Since foci of ellipse and hyperbola coincide:
$\Rightarrow 16 - b^2 = 9$
$\Rightarrow b^2 = 16 - 9$
$\therefore \boxed{b^2 = 7}$
Solution
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