JEE Practice Questions (NAT)

Hint

(a) The given lines \(7 x+4 y=168\) and \(5 x+3 y=121\) intersect at \(P(20,7)\)

\(
\begin{aligned}
&\therefore \text { Area of shaded region }\\
&\begin{aligned}
A & =\frac{1}{2}\left(42-40 \frac{1}{3}\right) 20 \\
& =\frac{1}{2}\left(\frac{5}{3}\right) 20=\frac{50}{3} \text { (square units) }
\end{aligned}
\end{aligned}
\)
\(
3 A / 10=5
\)

Hint

\(
\begin{aligned}
& \text { (b) } x^2 y^2-9 x^2-25 y^2+225=0 \\
& \Rightarrow x^2\left(y^2-9\right)-25\left(y^2-9\right)=0 \\
& \Rightarrow \quad\left(y^2-9\right)\left(x^2-25\right)=0
\end{aligned}
\)

\(
\therefore \quad \text { Area } A=10 \times 6=60 \text { sq. units. }
\)
\(
A / 10=6
\)

Hint

(c) Lines \((k+1) x+8 y=4 k\) and \(k x+(k+3) y=3 k-1\) are coincident then we can compare ratio of coefficients
\(
\begin{aligned}
& \Rightarrow \frac{k+1}{k}=\frac{8}{k+3}=\frac{4 k}{3 k-1} \\
& \Rightarrow k^2+4 k+3=8 k \text { and } 24 k-8=4 k^2+12 k \\
& \Rightarrow(k-3)(k-1)=0 \text { and }(k-2)(k-1)=0 \\
& \Rightarrow k=1
\end{aligned}
\)

Hint

(d) Equation of angle bisector of angle \(A\)
\(
\frac{3 x+4 y}{5}= \pm \frac{4 x+3 y}{5} \Rightarrow x= \pm y
\)
equation of internal bisector is \(x=-y\)
since \(h\) and \(k\) lie on the line \(x=-y\)
\(
\Rightarrow h+k=0
\)

Hint

(a) As \(H, G\) and \(S\) are collinear
\(
\begin{aligned}
& \therefore\left|\begin{array}{ccc}
4 & b & 1 \\
b & 2 b-8 & 1 \\
-4 & 8 & 1
\end{array}\right|=0 \\
& \Rightarrow\left|\begin{array}{ccc}
4 & b & 1 \\
b-4 & b-8 & 0 \\
-(b+4) & 16-2 b & 0
\end{array}\right|=0 \\
& \Rightarrow(b-4)(16-2 b)+(b+4)(b-8)=0 \\
& \Rightarrow 2(b-4)(8-b)+(b+4)(b-8)=0 \\
& \Rightarrow(8-b)[(2 b-8)-(b+4)]=0 \\
& \Rightarrow(8-b)(b-12)=0
\end{aligned}
\)

Also
\(
\therefore \quad \frac{-8+4}{3}=b \quad \Rightarrow b=\frac{-4}{3}
\)
And \(\frac{16+b}{3}=2 b-8 \Rightarrow b=8\)
But no common value of ‘ \(b\) ‘ is possible

Hint

(d) Using section formula \(A\left(\frac{3 k-5}{k+1}, \frac{5 k+1}{k+1}\right)\)
Area of triangle \(A B C\) is 2 sq. units
\(
\begin{aligned}
& \Rightarrow \frac{1}{2}\left|\begin{array}{ccc}
1 & 5 & 1 \\
7 & -2 & 1 \\
\frac{3 k-5}{k+1} & \frac{5 k+1}{k+1} & 1
\end{array}\right|= \pm 2 \\
& \text { Operating } R_2 \rightarrow R_2-R_1 ; R_3 \rightarrow R_3-R_1 \\
& \Rightarrow\left|\begin{array}{ccc}
1 & 5 & 1 \\
6 & -7 & 0 \\
\frac{3 k-5}{k+1}-1 & \frac{5 k+1}{k+1}-5 & 0
\end{array}\right|= \pm 4 \\
& \Rightarrow 6\left(\frac{5 k+1-5 k-5}{k+1}\right)+7\left(\frac{3 k-5-k-1}{k+1}\right)= \pm 4 \\
& \Rightarrow-24+7(2 k-6)= \pm 4(k+1) \\
& \Rightarrow k=7 \text { or } k=\frac{31}{9}
\end{aligned}
\)

Hint

(b) \(x^2-3 y^2-2 x y+8 y-4 \equiv(x-3 y+2)(x+y-2)\)

Now \((-5,-1)\) lies on \(x-3 y+2=0\)
In limiting case line passing through \((-5,-1)\) can be parallel to \(x+y-2=0\)
i.e. \(m>-1\)
and maximum slope can occur if it passes through \((0,0)\)
i.e. \(m<\frac{1}{5} \Rightarrow m \in\left(-1, \frac{1}{5}\right)\)
\(\Rightarrow a=-1\) and \(b=\frac{1}{5}\)
\(\Rightarrow\left(a+\frac{1}{b}\right)=-1+5=4\)

Hint

\(
\begin{aligned}
&\text { (b) Given pair of lines } x^2-\left(y^2-4 y+4\right)=0\\
&\begin{aligned}
& \Rightarrow x^2-(y-2)=0 \\
& \Rightarrow(x+y-2)(x-y+2)=0
\end{aligned}
\end{aligned}
\)

\(
\text { Required area is } A=\frac{1 \cdot 1}{2}=\frac{1}{2}
\)
\(
16 A=8
\)

Hint

\(
\begin{aligned}
& \text { (d) Let } x=r \cos \theta ; y=r \sin \theta \\
& \Rightarrow 2 r \cos \theta+3 r \sin \theta=6 \\
& \Rightarrow r=\frac{6}{2 \cos \theta+3 \sin \theta} ; \text { and } r=\sqrt{x^2+y^2}
\end{aligned}
\)
for \(r\) to be minimum \(2 \cos \theta+3 \sin \theta\) must be maximum i.e. \(\sqrt{13}\)
\(
\therefore \quad r_{\min }=\frac{6}{\sqrt{13}}
\)
\(
\sqrt{13} \mathrm{~m}=6
\)

Hint

(c) Given vertices of triangle are \(O(0,0), B(6,8)\) and \(C(-4,3)\)
Slope of \(O B=\frac{8}{6}\)
Slope of \(O C=-\frac{3}{4}\)
\(
\therefore \quad \angle B O C=\frac{\pi}{2}
\)
\(\triangle \mathrm{OBC}\) is right angled at \(O\)
Circumcentre \(=\) midpoint of hypotenuse \(B C=\left(1, \frac{11}{2}\right)\)
Orthocentre = vertex \(O(0,0)\)
Required distance \(=\sqrt{\left(1+\frac{121}{4}\right)}=\frac{5 \sqrt{5}}{2}\) unit
\(
\frac{2}{\sqrt{5}} L=5
\)

Hint

(b) Any point on the line \(x+y=4\) is \((t, 4-4)\) where \(t \in R\)
Now distance of this point from the line \(4 x+3 y-10=0\) is 1
\(
\begin{array}{cc}
\Rightarrow & \frac{|4 t+3(4-t)-10|}{5}=1 \\
\Rightarrow & |t+2|=5 \\
\Rightarrow & t=3 \text { or } t=-7 \\
\Rightarrow & \text { sum of values is }-4
\end{array}
\)

Hint

(c) \(\text { Area of } \triangle O A B=\frac{1}{2}(1)(8)=4 \text { sq. units }\)

Equation of \(O B\) is \(y=\frac{1}{9} x\)
Hence point \(E\) is \(\left(C, \frac{C}{9}\right)\)
Now area of \(\triangle B D E\) is 2 square units.
\(
\begin{aligned}
& \Rightarrow \frac{1}{2}\left(1-\frac{C}{9}\right)(9-C)=2 \\
& \Rightarrow(9-C)^2=36 \\
& \Rightarrow 9-C= \pm 6 \\
& \Rightarrow C=3
\end{aligned}
\)

Hint

(b) For \(P R=R Q\) to be minimum it should be the path of light

\(
\begin{aligned}
&\therefore \quad \angle P R A=\angle Q R M\\
&\text { From similar } \triangle P A R \text { and } \triangle Q M R\\
&\begin{aligned}
& \frac{A R}{R M}=\frac{P A}{Q M} \\
& \Rightarrow \frac{\alpha+3}{0-\alpha}=\frac{4}{1} \Rightarrow \alpha=-\frac{3}{5}
\end{aligned}
\end{aligned}
\)
\(
5|\alpha|=3
\)

Hint

(d) We know that the area of the triangle formed by joining the mid points of any triangle is one fourth of that triangle. Therefore required area is 8 .

Hint

(a) Lines \((2 a+b) \mathrm{x}+(a+3 b) y+(b-3 a)=0\) or \(a(2 x+ y-3)+b(x+3 y+1)=0\) are concurrent at point of intersection of lines \(2 x+y-3=0\) and \(x+3 y+1=0\) which is \((2,-1)\).
Now line \(m x+2 y+6=0\) must pass through this point \(\Rightarrow 2 m-2+6=0\) or \(m=-2\)
\(
|m|=2
\)

Hint

(c) Line \(3 x+2 y=24\) meets the axis at \(B(8,0)\) and \(A(0\), 12). Midpoint of \(A B\) is \(D(4,6)\)
Equation of perpendicular bisector of \(A B\) is
\(
2 x-3 y+10=0 \dots(1)
\)
Now line through \((0,-1)\) and parallel to \(x\)-axis is \(y=-1\)
Co-ordinates of \(C\) where line (1) meets \(y=-1\) is \(C\left(-\frac{13}{2},-1\right)\)
Now the area of triangle \(A B C\)
\(
\begin{aligned}
& \Delta=\frac{1}{2}\left|\begin{array}{ccc}
0 & 12 & 1 \\
8 & 0 & 1 \\
-\frac{13}{2} & -1 & 1
\end{array}\right| \\
& =\frac{1}{2}\left[0-12\left(8+\frac{13}{2}\right)+1(-8)\right] \\
& =\frac{1}{2}[-6(29)-8]=91
\end{aligned}
\)
\(
A / 13=7
\)

Hint

(b)

\(
\begin{aligned}
&P \text { is orthocenter }\\
&\begin{aligned}
& \Rightarrow A P \perp B C \\
& \Rightarrow\left(-\frac{1}{p}\right)\left(\frac{3+2}{2-1}\right)=-1 \\
& \Rightarrow \frac{5}{p} \Rightarrow p=5 \\
& \because B P \perp A C \\
& \Rightarrow \frac{27-2 q}{18+q}=-1 \Rightarrow q=27+18 \\
& \Rightarrow q=45 \\
& \therefore p+q=5+45=50
\end{aligned}
\end{aligned}
\)
\(
(p+q) / 10=5
\)

Note: The orthocenter of a triangle is the point where its three altitudes intersect. An altitude is a line segment drawn from a vertex perpendicular to the opposite side. The location of the orthocenter depends on the triangle’s type: it lies inside an acute triangle, coincides with the right-angle vertex in a right triangle, and is located outside an obtuse triangle. 

Hint

(d)

Since \(\angle B C A=90^{\circ}\)
Points \(A, O, B, C\) are concyclic
Let \(\angle A O C=\theta\)
\(\angle B O C=\angle B A C\)
\(\tan \left(\frac{\pi}{2}-\theta\right)=\frac{5}{12}\)
\(\frac{x}{y}=\frac{5}{12} \Rightarrow 12 x-5 y=0\)
Comparing with \(12 x-k y=0\), we get \(k=5\)