Latest JEE PYQs (NAT) (2025 onwards)

Hint

(d)

To find the value of \(9\left(x_1 y_1+x_2 y_2+x_3 y_3\right)\), we need to determine the coordinates of the points of tangency \(P, Q\), and \(R\) on the line \(L: x+y=3\).
Step 1: Find Point \(P\) (Tangency on Circle \(C\))
The circle \(C\) is given by \(x^2+(y-1)^2=2\). The line is \(y=3-x\). Substituting this into the circle equation:
\(
\begin{gathered}
x^2+(3-x-1)^2=2 \\
x^2+(2-x)^2=2 \Longrightarrow x^2+4-4 x+x^2=2 \\
2 x^2-4 x+2=0 \Longrightarrow(x-1)^2=0
\end{gathered}
\)
Thus, \(x_1=1\). Substituting back into the line equation, \(y_1=3-1=2\). Point \(P=(1,2)\).
Step 2: Find Points \(Q\) and \(R\)
Points \(Q\left(x_2, y_2\right)\) and \(R\left(x_3, y_3\right)\) lie on the line \(x+y=3\). Since \(P\) is the midpoint of \(Q R\), we have:
\(
\begin{aligned}
& \frac{x_2+x_3}{2}=1 \Longrightarrow x_2+x_3=2 \\
& \frac{y_2+y_3}{2}=2 \Longrightarrow y_2+y_3=4
\end{aligned}
\)
We are given the distance \(P Q=\frac{2 \sqrt{2}}{3}\). Since \(Q\) lies on the line \(x+y=3\) (which has a slope of -1, or an angle of \(135^{\circ}\)), we can use the parametric form of a line:
\(
x=x_1 \pm r \cos \theta, \quad y=y_1 \pm r \sin \theta
\)
For \(\theta=135^{\circ}, \cos \theta=-\frac{1}{\sqrt{2}}\) and \(\sin \theta=\frac{1}{\sqrt{2}}\). With \(r=\frac{2 \sqrt{2}}{3}\) :
\(
\begin{gathered}
x=1 \pm \frac{2 \sqrt{2}}{3}\left(-\frac{1}{\sqrt{2}}\right)=1 \mp \frac{2}{3} \\
y=2 \pm \frac{2 \sqrt{2}}{3}\left(\frac{1}{\sqrt{2}}\right)=2 \pm \frac{2}{3}
\end{gathered}
\)
This gives us two points:
\(\left(\frac{1}{3}, \frac{8}{3}\right)\)
\(\left(\frac{5}{3}, \frac{4}{3}\right)\)
Step 3: Assigning \(Q\) and \(R\)
\(E_1\) has its major axis on the \(x\)-axis (equation \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\) with \(a>b\) ). \(E_2\) has its major axis on the \(y\)-axis (equation \(\frac{x^2}{A^2}+\frac{y^2}{B^2}=1\) with \(B>A\) ). For a line \(y=m x+c\) to touch \(\frac{x^2}{a^2}+ \frac{y^2}{b^2}=1\), the point of tangency is \(\left(\frac{-a^2 m}{c}, \frac{b^2}{c}\right)\). Here, \(m=-1\) and \(c=3\).
For \(E_1: x_2=\frac{a^2}{3}, y_2=\frac{b^2}{3}\). Since \(a>b\), it follows \(x_2>y_2\). Looking at our points, \(\left(\frac{5}{3}, \frac{4}{3}\right)\) satisfies \(x>y\). So, \(Q=\left(\frac{5}{3}, \frac{4}{3}\right)\).
For \(E_2: x_3=\frac{A^2}{3}, y_3=\frac{B^2}{3}\). Since \(B>A\), it follows \(y_3>x_3\). Looking at our points, \(\left(\frac{1}{3}, \frac{8}{3}\right)\) satisfies \(y>x\). So, \(R=\left(\frac{1}{3}, \frac{8}{3}\right)\).
Step 4: Final Calculation
Now we calculate \(x_1 y_1+x_2 y_2+x_3 y_3\) :
\(x_1 y_1=1 \times 2=2\)
\(x_2 y_2=\frac{5}{3} \times \frac{4}{3}=\frac{20}{9}\)
\(x_3 y_3=\frac{1}{3} \times \frac{8}{3}=\frac{8}{9}\)
Summing them up:
\(
x_1 y_1+x_2 y_2+x_3 y_3=2+\frac{20}{9}+\frac{8}{9}=2+\frac{28}{9}
\)
Multiply by 9:
\(
9\left(2+\frac{28}{9}\right)=18+28=46
\)

Hint

(a)

We have, \(x+y=3\) and \(x-y=3\) are tangents
∴ The centre of both circles will lie on \(x\)-axis
Equation of circle is
\(
\therefore \quad(x-\alpha)^2+y^2=r^2
\)
Hence, centre is \(\mathrm{C}(\alpha, 0)\).
\(
r=P C=\sqrt{(\alpha+9)^2+16} \dots(i)
\)
Also, \(\left|\frac{\alpha-3}{\sqrt{2}}\right|=r \dots(ii)\)
From (i) and (ii), we get
\(
\begin{aligned}
& \sqrt{(\alpha+9)^2+16}=\left|\frac{\alpha-3}{\sqrt{2}}\right| \\
& \Rightarrow \quad \alpha=-5 \text { or }-37 \\
& r=\left|\frac{-5-3}{\sqrt{2}}\right| \text { or }\left|\frac{-37-3}{\sqrt{2}}\right|=4 \sqrt{2} \text { or } 20 \sqrt{2} \\
& \therefore \text { Now, }\left|r_1^2-r_2^2\right|=\left|(4 \sqrt{2})^2-(20 \sqrt{2})^2\right|=768
\end{aligned}
\)

Explanation: Step 1: Identify the Bisector Line for the Centers
The two lines \(L_1: x+y-3=0\) and \(L_2: x-y-3=0\) intersect at \((3,0)\). A circle touching both lines must have its center (\(h, k\)) on an angle bisector of these lines. The bisectors are:
\(
\frac{x+y-3}{\sqrt{2}}= \pm \frac{x-y-3}{\sqrt{2}}
\)
This gives two possible lines for the center: \(y=0\) (the \(x\)-axis) or \(x=3\).
Since the point (\(-9,4\)) is far from the line \(x=3\), the circles passing through it must be centered on the x -axis. Thus, the center is \((h, 0)\).
Step 2: Define the Radius in terms of the Center
The radius \(r\) is the perpendicular distance from the center \((h, 0)\) to the line \(x+y-3=0\) :
\(
r=\frac{|h+0-3|}{\sqrt{2}} \Longrightarrow \sqrt{2} r=|h-3|
\)
Since the point \((-9,4)\) lies to the left of the intersection \((3,0)\), we know \(h<3\). Therefore:
\(
h-3=-\sqrt{2} r \Longrightarrow h=3-\sqrt{2} r
\)
Step 3: Set up the Circle Equation using the Point
The distance from the center \((h, 0)\) to the point \((-9,4)\) is the radius \(r\). Using the distance formula:
\(
\begin{gathered}
(h-(-9))^2+(0-4)^2=r^2 \\
(h+9)^2+16=r^2
\end{gathered}
\)
Substitute \(h=3-\sqrt{2} r\) :
\(
\begin{gathered}
(3-\sqrt{2} r+9)^2+16=r^2 \\
(12-\sqrt{2} r)^2+16=r^2
\end{gathered}
\)
Step 4: Solve the Quadratic for \(r\)
Expanding the equation:
\(
\begin{gathered}
144-24 \sqrt{2} r+2 r^2+16=r^2 \\
r^2-24 \sqrt{2} r+160=0
\end{gathered}
\)
Let the radii of the two circles be \(r_1\) and \(r_2\). From the properties of quadratic equations:
Sum of roots \(\left(r_1+r_2\right): 24 \sqrt{2}\)
Product of roots \(\left(r_1 r_2\right): 160\)
Step 5: Calculate the Absolute Difference of Squares
We need to find \(\left|r_1^2-r_2^2\right|\), which factors into \(\left(r_1+r_2\right)\left|r_1-r_2\right|\). First, find \(\left(r_1-r_2\right)^2\) :
\(
\begin{gathered}
\left(r_1-r_2\right)^2=\left(r_1+r_2\right)^2-4 r_1 r_2=(24 \sqrt{2})^2-4(160) \\
\left(r_1-r_2\right)^2=1152-640=512
\end{gathered}
\)
So, \(\left|r_1-r_2\right|=\sqrt{512}=16 \sqrt{2}\).
\(
\left|r_1^2-r_2^2\right|=(24 \sqrt{2})(16 \sqrt{2})=24 \times 16 \times 2=768
\)

Hint

(d)

Step 1: Find the Center and Radius of Circle \(C\)
Let the center of circle \(C\) be ( \(\alpha, 0\) ) where \(\alpha>0\) (since it lies on the positive x-axis). Let its radius be \(r\).
Condition 1: The circle touches \(x-y+1=0\). The perpendicular distance from (\(\alpha, 0\)) to this line equals the radius:
\(
r=\frac{|\alpha-0+1|}{\sqrt{1^2+(-1)^2}}=\frac{\alpha+1}{\sqrt{2}}
\)
Squaring gives: \(r^2=\frac{(\alpha+1)^2}{2}\)
Condition 2: The circle cuts a chord of length \(\frac{4}{\sqrt{13}}\) on \(-3 x+2 y-1=0\). Let \(d\) be the perpendicular distance from \((\alpha, 0)\) to the line \(3 x-2 y+1=0\) :
\(
d=\frac{|3 \alpha-0+1|}{\sqrt{3^2+2^2}}=\frac{3 \alpha+1}{\sqrt{13}}
\)
The relationship between radius \(r\), distance \(d\), and chord length \(L\) is:
\(
r^2=d^2+\left(\frac{L}{2}\right)^2 \Longrightarrow r^2=\left(\frac{3 \alpha+1}{\sqrt{13}}\right)^2+\left(\frac{2}{\sqrt{13}}\right)^2
\)
\(
r^2=\frac{9 \alpha^2+6 \alpha+1+4}{13}=\frac{9 \alpha^2+6 \alpha+5}{13}
\)
Step 2: Solve for \(\alpha\) and \(r\)
Equate the two expressions for \(r^2\) :
\(
\begin{gathered}
\frac{(\alpha+1)^2}{2}=\frac{9 \alpha^2+6 \alpha+5}{13} \\
13\left(\alpha^2+2 \alpha+1\right)=2\left(9 \alpha^2+6 \alpha+5\right) \\
13 \alpha^2+26 \alpha+13=18 \alpha^2+12 \alpha+10 \\
5 \alpha^2-14 \alpha-3=0
\end{gathered}
\)
Solving the quadratic: \((5 \alpha+1)(\alpha-3)=0\). Since \(\alpha>0\), we have \(\alpha=3\).
Now, find \(r^2\) and the diameter \(D\) :
\(
r^2=\frac{(3+1)^2}{2}=\frac{16}{2}=8 \Longrightarrow r=2 \sqrt{2}
\)
Diameter \(D=2 r=4 \sqrt{2}\).
Step 3: Determine Hyperbola \(H\) parameters
For the hyperbola \(\frac{x^2}{\alpha^2}-\frac{y^2}{\beta^2}=1\) :
Transverse Axis Length (2 \(\alpha\)): Equal to diameter of \(C\).
\(
2 \alpha=4 \sqrt{2} \Longrightarrow \alpha=2 \sqrt{2} \Longrightarrow \alpha^2=8
\)
Focus (\(a e, 0\)): One focus is the center of \(C\), which is \((3,0)\).
\(
\alpha e=3
\)
Find \(\beta^2\) : Using the identity \(b^2=a^2\left(e^2-1\right)=(a e)^2-a^2\) :
\(
\beta^2=3^2-\alpha^2=9-8=1
\)
Step 4: Final Calculation
Substitute the values of \(\alpha^2\) and \(\beta^2\) :
\(
2 \alpha^2+3 \beta^2=2(8)+3(1)=19
\)

Hint

(b)

Step 1: Establish the Circle’s Properties
Since the circle touches the \(x\)-axis at (\(a, 0\)), its center must lie on the vertical line \(x=a\). The radius \(r\) is the vertical distance from the center to the \(x\)-axis. Given the point of contact on the parabola is \((4,6)\) (in the first quadrant), the center is \((a, r)\)
Step 2: Find the Normal to the Parabola
The circle touches the parabola \(y^2=9 x\) at \(P(4,6)\). The normal to the parabola at this point must pass through the center of the circle \((a, r)\).
Differentiating the parabola:
\(
2 y \frac{d y}{d x}=9 \Longrightarrow \frac{d y}{d x}=\frac{9}{2 y}
\)
At \((4,6)\), the slope of the tangent is \(m_t=\frac{9}{12}=\frac{3}{4}\). The slope of the normal is \(m_n=-\frac{4}{3}\).
The equation of the normal line is:
\(
y-6=-\frac{4}{3}(x-4) \Longrightarrow 4 x+3 y=34
\)
Step 3: Relate the Center Coordinates
Substitute the center (\(a, r\)) into the normal equation:
\(
4 a+3 r=34 \Longrightarrow a=\frac{34-3 r}{4}
\)
Step 4: Use the Radius Condition
The distance from the center \((a, r)\) to the point of contact \((4,6)\) is equal to \(r\) :
\(
(a-4)^2+(r-6)^2=r^2
\)
\(
\begin{gathered}
a^2-8 a+16+r^2-12 r+36=r^2 \\
a^2-8 a-12 r+52=0
\end{gathered}
\)
Substitute \(a=\frac{34-3 r}{4}\) :
\(
\begin{gathered}
\left(\frac{34-3 r}{4}\right)^2-8\left(\frac{34-3 r}{4}\right)-12 r+52=0 \\
\frac{1156-204 r+9 r^2}{16}-68+6 r-12 r+52=0 \\
1156-204 r+9 r^2+16(-6 r-16)=0 \\
9 r^2-300 r+900=0 \Longrightarrow 3 r^2-100 r+300=0
\end{gathered}
\)
Step 5: Solve for \(\boldsymbol{r}\)
Using the quadratic formula:
\(
r=\frac{100 \pm \sqrt{10000-3600}}{6}=\frac{100 \pm 80}{6}
\)
This gives \(r=30\) or \(r=\frac{10}{3}\).
If \(r=\frac{10}{3}\), then \(a=\frac{34-10}{4}=6\) (rejected as \(a<0\)).
If \(r=30\), then \(a=\frac{34-90}{4}=-14\) (accepted).
The radius \(r\) is 30.

Hint

(c) Step 1: Analyze the Parabola and Focal Chord
The parabola is \(y^2=12 x\). Here, \(4 a=12\), so \(a=3\). The focus \(S\) is \((3,0)\). Let the coordinates of \(P\) be (\(a t_1^2, 2 a t_1\)) and \(Q\) be (\(a t_2^2, 2 a t_2\)). Since \(P Q\) is a focal chord, we have the property:
\(
t_1 t_2=-1
\)
Step 2: Use the Focal Chord Length Property
The focal distances for a parabola are \(S P=a\left(1+t_1^2\right)\) and \(S Q=a\left(1+t_2^2\right)\). Given \((S P)(S Q)=\frac{147}{4}:\)
\(
\begin{aligned}
& a\left(1+t_1^2\right) \cdot a\left(1+t_2^2\right)=\frac{147}{4} \\
& 9\left(1+t_1^2+t_2^2+t_1^2 t_2^2\right)=\frac{147}{4}
\end{aligned}
\)
Since \(t_1^2 t_2^2=(-1)^2=1\) :
\(
\begin{gathered}
9\left(2+t_1^2+t_2^2\right)=\frac{147}{4} \\
2+t_1^2+t_2^2=\frac{147}{36}=\frac{49}{12} \\
t_1^2+t_2^2=\frac{49}{12}-2=\frac{25}{12}
\end{gathered}
\)
Step 3: Determine the Center and Diameter of Circle \(C\)
The circle is drawn with \(P Q\) as the diameter.
Center \((h, k): h=\frac{a\left(t_1^2+t_2^2\right)}{2}=\frac{3(25 / 12)}{2}=\frac{25}{8} k=\frac{2 a\left(t_1+t_2\right)}{2}=3\left(t_1+t_2\right)\)
Diameter \(L\) : The length of the focal chord \(P Q=S P+S Q=a\left(2+t_1^2+t_2^2\right)=\frac{49}{4}\). The radius \(R=\frac{49}{8}\).
Step 4: Find \(k\) and the Circle Equation
We need \(\left(t_1+t_2\right)\). We know \(\left(t_1+t_2\right)^2=t_1^2+t_2^2+2 t_1 t_2=\frac{25}{12}-2=\frac{1}{12}\). Thus, \(t_1+t_2= \pm \frac{1}{\sqrt{12}}= \pm \frac{1}{2 \sqrt{3}}\). From the given circle equation, the \(y\)-coefficient is \(-64 \sqrt{3}\). In standard form \((x-h)^2+(y-k)^2=R^2\), the \(y\)-term is \(-2 k y\). Matching coefficients: \(k=\frac{64 \sqrt{3}}{2 \times 64}=\frac{\sqrt{3}}{2}\). Our calculated \(k=3\left(t_1+t_2\right)\). Let’s check: \(3\left(\frac{1}{2 \sqrt{3}}\right)=\frac{\sqrt{3}}{2}\). This matches perfectly.
The equation of the circle is:
\(
\begin{gathered}
\left(x-\frac{25}{8}\right)^2+\left(y-\frac{\sqrt{3}}{2}\right)^2=\left(\frac{49}{8}\right)^2 \\
x^2-\frac{25}{4} x+\frac{625}{64}+y^2-\sqrt{3} y+\frac{3}{4}=\frac{2401}{64}
\end{gathered}
\)
Multiply by 64 to match the format \(64 x^2+64 y^2-\alpha x-64 \sqrt{3} y=\beta\) :
\(
\begin{gathered}
64 x^2-16(25) x+625+64 y^2-64 \sqrt{3} y+48=2401 \\
64 x^2+64 y^2-400 x-64 \sqrt{3} y=2401-625-48 \\
64 x^2+64 y^2-400 x-64 \sqrt{3} y=1728
\end{gathered}
\)
Step 5: Final Calculation
Comparing the equation to \(64 x^2+64 y^2-\alpha x-64 \sqrt{3} y=\beta\) :
\(\alpha=400\)
\(\beta=1728\)
Therefore:
\(
\beta-\alpha=1728-400=1328
\)

Hint

(b)

To find the value of \(a+d\), we first need to find the equation of the mirrored parabola and determine its intersection with the line \(y=-5\).
Step 1: Find the Mirror Image of the Parabola
The original parabola is \(y^2=4 x\). We need its reflection about the line \(L: x+y+4=0\). The reflection of a point (\(x_1, y_1\)) about the line \(a x+b y+c=0\) is given by:
\(
\frac{x-x_1}{a}=\frac{y-y_1}{b}=-2 \frac{a x_1+b y_1+c}{a^2+b^2}
\)
For the line \(x+y+4=0(a=1, b=1, c=4)\) :
\(
\frac{x-x_1}{1}=\frac{y-y_1}{1}=-2 \frac{x_1+y_1+4}{1^2+1^2}=-\left(x_1+y_1+4\right)
\)
This gives us:
\(x=x_1-x_1-y_1-4 \Longrightarrow x=-y_1-4 \Longrightarrow y_1=-(x+4)\)
\(y=y_1-x_1-y_1-4 \Longrightarrow y=-x_1-4 \Longrightarrow x_1=-(y+4)\)
Substitute \(x_1\) and \(y_1\) into the original parabola equation \(y_1^2=4 x_1\) :
\(
\begin{gathered}
(-(x+4))^2=4(-(y+4)) \\
(x+4)^2=-4(y+4)
\end{gathered}
\)
Step 2: Intersection Points \(A\) and \(B\)
The line is \(y+5=0\), so \(y=-5\). Substitute this into the mirrored parabola:
\(
\begin{gathered}
(x+4)^2=-4(-5+4) \\
(x+4)^2=-4(-1)=4 \\
x+4= \pm 2
\end{gathered}
\)
If \(x+4=2\), then \(x=-2\). Point \(A=(-2,-5)\).
If \(x+4=-2\), then \(x=-6\). Point \(B=(-6,-5)\).
Distance \(d\) : The distance between \(A(-2,-5)\) and \(B(-6,-5)\) is:
\(
d=|-2-(-6)|=4
\)
Step 3: Area a of \(\triangle S A B\)
The focus \(S\) of the original parabola \(y^2=4 x\) is at \((1,0)\). The coordinates of the triangle vertices are \(S(1,0), A(-2,-5)\), and \(B(-6,-5)\).
The area \(a\) is given by:
\(
\begin{gathered}
a=\frac{1}{2}\left|x_S\left(y_A-y_B\right)+x_A\left(y_B-y_S\right)+x_B\left(y_S-y_A\right)\right| \\
a=\frac{1}{2}|1(-5-(-5))+(-2)(-5-0)+(-6)(0-(-5))| \\
a=\frac{1}{2}|0+10-30| \\
a=\frac{1}{2}|-20|=10
\end{gathered}
\)
Step 4: Final Value
\(
a+d=10+4=14
\)

Hint

(c) To solve this, we need to find the center of the circle from the parabola’s focus and then use the intersection point of the lines to determine \(\lambda\).
Step 1: Find the Focus of the Parabola
The given parabola is \(y^2=4 x+16\). Let’s rewrite it in standard form:
\(
y^2=4(x+4)
\)
Comparing this to \(Y^2=4 a X\) :
\(a=1\)
\(Y=y, X=x+4\)
The focus of \(Y^2=4 a X\) is \((a, 0)\). Thus, for our parabola:
\(X=1 \Longrightarrow x+4=1 \Longrightarrow x=-3\)
\(Y=0 \Longrightarrow y=0\)
The center of the circle \(C\) is \(S(-3,0)\).
Step 2: Determine the Points of Intersection
The radius of circle \(C\) is 5. The equation of the circle is:
\(
(x+3)^2+(y-0)^2=5^2 \Longrightarrow(x+3)^2+y^2=25
\)
The circle passes through the intersection of:
\(3 x-y=0 \Longrightarrow y=3 x\)
\(x+\lambda y=4\)
Substitute \(y=3 x\) into the circle equation to find the \(x\)-coordinates of the intersection points:
\(
\begin{gathered}
(x+3)^2+(3 x)^2=25 \\
x^2+6 x+9+9 x^2=25 \\
10 x^2+6 x-16=0 \Longrightarrow 5 x^2+3 x-8=0
\end{gathered}
\)
Factoring the quadratic:
\(
(5 x+8)(x-1)=0
\)
The intersection points on the circle are:
Case 1: \(x=1\). Then \(y=3(1)=3\). Point is \((1,3)\).
Case 2: \(x=-8 / 5\). Then \(y=3(-8 / 5)=-24 / 5\). Point is \((-8 / 5,-24 / 5)\).
Step 3: Solve for \(\lambda\)
Now, substitute these points into the second line equation \(x+\lambda y=4\) :
For (1, 3):
\(
1+\lambda(3)=4 \Longrightarrow 3 \lambda=3 \Longrightarrow \lambda=1
\)
For (-8/5, -24/5):
\(
-\frac{8}{5}+\lambda\left(-\frac{24}{5}\right)=4
\)
Multiply by 5:
\(
-8-24 \lambda=20 \Longrightarrow-24 \lambda=28 \Longrightarrow \lambda=-\frac{28}{24}=-\frac{7}{6}
\)
Step 4: Calculate \(12 \lambda_1+29 \lambda_2\)
Given \(\lambda_1<\lambda_2\) :
\(\lambda_1=-\frac{7}{6}\)
\(\lambda_2=1\)
Now, compute the final value:
\(
12\left(-\frac{7}{6}\right)+29(1)=15
\)

Hint

(c) Step 1: Analyze the initial ellipse \(\boldsymbol{E}_1\)
The equation of \(E_1\) is \(\frac{x^2}{9}+\frac{y^2}{4}=1\). Comparing this with the standard form \(\frac{x^2}{a^2}+\frac{y^2}{b^2}=1\), we have:
Semi-major axis \(a_1=3\)
Semi-minor axis \(b_1=2\)
Area \(A_1=\pi a_1 b_1=\pi(3)(2)=6 \pi\)
Eccentricity e: \(b_1^2=a_1^2\left(1-e^2\right) \Rightarrow 4=9\left(1-e^2\right) \Rightarrow 1-e^2=\frac{4}{9}\)
Step 2: Establish the relationship between subsequent ellipses
For any ellipse \(E_i\), let the semi-major axis be \(a_i\) and the semi-minor axis be \(b_i\). Since all ellipses have the same eccentricity, the ratio of the axes remains constant:
\(
\frac{b_i}{a_i}=\sqrt{1-e^2}=\sqrt{\frac{4}{9}}=\frac{2}{3} \Longrightarrow b_i=\frac{2}{3} a_i
\)
According to the problem, the length of the minor axis of \(E_i\) is the length of the major axis of \(E_{i+1}\) :
\(
2 b_i=2 a_{i+1} \Longrightarrow a_{i+1}=b_i
\)
Substituting \(b_i=\frac{2}{3} a_i\), we get \(a_{i+1}=\frac{2}{3} a_i\). Consequently, \(b_{i+1}=\frac{2}{3} a_{i+1}=\frac{2}{3} b_i\).
Step 3: Determine the common ratio of the area series
The area of the \(i\)-th ellipse is \(A_i=\pi a_i b_i\). The ratio of successive areas is:
\(
r=\frac{A_{i+1}}{A_i}=\frac{\pi a_{i+1} b_{i+1}}{\pi a_i b_i}=\frac{a_{i+1}}{a_i} \cdot \frac{b_{i+1}}{b_i}=\frac{2}{3} \cdot \frac{2}{3}=\frac{4}{9}
\)
Since \(r=\frac{4}{9}<1\), the sum of areas \(\sum_{i=1}^{\infty} A_i\) forms an infinite geometric series.
Step 4: Calculate the sum and final expression
The sum of the infinite geometric series is:
\(
\sum_{i=1}^{\infty} A_i=\frac{A_1}{1-r}=\frac{6 \pi}{1-\frac{4}{9}}=\frac{6 \pi}{\frac{5}{9}}=\frac{54 \pi}{5}
\)
Now, we find the required value:
\(
\frac{5}{\pi}\left(\sum_{i=1}^{\infty} A_i\right)=\frac{5}{\pi}\left(\frac{54 \pi}{5}\right)=54
\)

Hint

(a) Step 1: Determine the parameters \(\boldsymbol{a}\) and \(\boldsymbol{e}\)
A hyperbola in standard form \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) has a focus at \((-a e, 0)\) and a corresponding directrix \(x=-\frac{a}{e}\). From the given data:
Focus: \(-a e=-5 \Rightarrow a e=5\)
Directrix: \(x=-\frac{9}{5} \Rightarrow \frac{a}{e}=\frac{9}{5}\)
Multiplying the two equations:
\(
(a e) \cdot\left(\frac{a}{e}\right)=5 \cdot \frac{9}{5} \Rightarrow a^2=9 \Rightarrow a=3
\)
Dividing the two equations:
\(
\frac{a e}{a / e}=\frac{5}{9 / 5} \Rightarrow e^2=\frac{25}{9} \Rightarrow e=\frac{5}{3}
\)
Step 2: Determine \(b^2\) and the hyperbola equation
Using the relation \(b^2=a^2\left(e^2-1\right)\) :
\(
b^2=9\left(\frac{25}{9}-1\right)=9\left(\frac{16}{9}\right)=16
\)
The equation of the hyperbola is:
\(
\frac{x^2}{9}-\frac{y^2}{16}=1
\)
Step 3: Find the coordinates of point \(\boldsymbol{P}\)
The point \(P(\alpha, 2 \sqrt{5})\) lies on the hyperbola. Substituting \(y=2 \sqrt{5}\) :
\(
\begin{gathered}
\frac{\alpha^2}{9}-\frac{(2 \sqrt{5})^2}{16}=1 \Rightarrow \frac{\alpha^2}{9}-\frac{20}{16}=1 \\
\frac{\alpha^2}{9}=1+\frac{5}{4}=\frac{9}{4} \Rightarrow \alpha^2=\frac{81}{4}
\end{gathered}
\)
Step 4: Calculate the product of focal distances \(p\)
For any point (\(x, y\)) on a hyperbola, the focal distances are \(r_1=|e x+a|\) and \(r_2=|e x-a|\). The product \(p\) is:
\(
p=\left|e^2 x^2-a^2\right|
\)
Substituting \(e^2=\frac{25}{9}, x^2=\alpha^2=\frac{81}{4}\), and \(a^2=9\) :
\(
\begin{gathered}
p=\left|\left(\frac{25}{9}\right)\left(\frac{81}{4}\right)-9\right|=\left|\frac{25 \cdot 9}{4}-9\right| \\
p=\left|\frac{225}{4}-\frac{36}{4}\right|=\frac{189}{4}
\end{gathered}
\)
The value of \(4 p\) is:
\(
4 p=4 \times \frac{189}{4}=189
\)

Hint

(c)

To solve this, we will use the properties of the hyperbola’s focus, latus rectum, and the geometry of right-angled triangles.
Step 1: Identify Hyperbola Parameters
For the standard hyperbola \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\) :
Foci: \(S(-a e, 0)\) and \(S^{\prime}(a e, 0)\).
We are given one focus at \(P(-3,0)\), so \(a e=3\).
Step 2: Geometry of the Latus Rectum
The other focus is \(S^{\prime}(3,0)\). The latus rectum through \(S^{\prime}\) consists of two points, \(L\) and \(L^{\prime}\), with coordinates:
\(L=\left(a e, \frac{b^2}{a}\right)=\left(3, \frac{b^2}{a}\right)\)
\(L^{\prime}=\left(a e,-\frac{b^2}{a}\right)=\left(3,-\frac{b^2}{a}\right)\)
The problem states that the latus rectum \(L L^{\prime}\) subtends a right angle at \(P(-3,0)\). Due to symmetry about the x -axis, the line segment \(P L\) must make an angle of \(45^{\circ}\) with the x -axis (the transverse axis).
This means the slope of \(P L\) is 1 :
\(
\begin{gathered}
m_{P L}=\frac{\frac{b^2}{a}-0}{3-(-3)}=\frac{b^2 / a}{6}=1 \\
\frac{b^2}{a}=6 \Longrightarrow b^2=6 a
\end{gathered}
\)
Step 3: Solve for \(a^2\) and \(b^2\)
We have two primary equations:
\(a e=3 \Longrightarrow a^2 e^2=9\)
\(b^2=a^2\left(e^2-1\right) \Longrightarrow a^2 e^2-a^2=b^2\)
Substitute \(a^2 e^2=9\) into the second equation:
\(
9-a^2=b^2
\)
Now substitute \(b^2=6 a\) :
\(
9-a^2=6 a \Longrightarrow a^2+6 a-9=0
\)
Using the quadratic formula for \(a(a>0)\) :
\(
a=\frac{-6+\sqrt{36-4(1)(-9)}}{2}=\frac{-6+\sqrt{72}}{2}=\frac{-6+6 \sqrt{2}}{2}=3 \sqrt{2}-3
\)
Now, calculate \(a^2\) and \(b^2\) :
\(a^2=(3 \sqrt{2}-3)^2=18+9-18 \sqrt{2}=27-18 \sqrt{2}\)
\(b^2=6 a=6(3 \sqrt{2}-3)=18 \sqrt{2}-18\)
Step 4: Find \(a^2 b^2\)
\(
a^2 b^2=(27-18 \sqrt{2})(18 \sqrt{2}-18)
\)
Factoring out 9 from the first term and 18 from the second:
\(
\begin{gathered}
a^2 b^2=9(3-2 \sqrt{2}) \cdot 18(\sqrt{2}-1)=162(3-2 \sqrt{2})(\sqrt{2}-1) \\
a^2 b^2=162(3 \sqrt{2}-3-4+2 \sqrt{2}) \\
a^2 b^2=162(5 \sqrt{2}-7)=810 \sqrt{2}-1134
\end{gathered}
\)
Step 5: Compare and Calculate \(\alpha+\beta\)
Given \(a^2 b^2=\alpha \sqrt{2}-\beta\) :
\(\alpha=810\)
\(\beta=1134\)
Summing them up:
\(
\alpha+\beta=810+1134=1944
\)

Hint

(c)

To find the value of \(\alpha+\beta+\gamma\), we first need to determine the standard equation of the hyperbola and then expand it into the general form.
Identify the Center and Direction
The foci are \(S^{\prime}(4,2)\) and \(S(8,2)\).
Since the \(y\)-coordinates are the same, the transverse axis is horizontal (parallel to the \(x\) -axis).
Center \((h, k)\) : The midpoint of the foci.
\(
(h, k)=\left(\frac{4+8}{2}, \frac{2+2}{2}\right)=(6,2)
\)
Equation of hyperbola is \(\frac{(x-6)^2}{a^2}-\frac{(y-2)^2}{4-a^2}=1\)
\(
\begin{aligned}
& \Rightarrow \quad\left(4-a^2\right)(x-6)^2-a^2(y-2)^2=a^2\left(4-a^2\right) \\
& \Rightarrow \quad\left(4-a^2\right) x^2-a^2 y^2+\left(12 a^2-48\right) x+4 a^2 y+144-44 a^2+a^4=0
\end{aligned}
\)
On comparing with \(3 x^2-y^2-\alpha x+\beta y+\gamma=0\), we get \(a^2=1\) and \(\alpha=36, \beta=4\) and \(\gamma=101 \therefore \quad \alpha+\beta+\gamma=141\)

Hint

(a) Step 1: Establish the Equations for the Hyperbola
The given hyperbola \(H\) is \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\). The point \(P(4,2 \sqrt{3})\) lies on \(H\), so it satisfies the equation:
\(
\frac{4^2}{a^2}-\frac{(2 \sqrt{3})^2}{b^2}=1 \Longrightarrow \frac{16}{a^2}-\frac{12}{b^2}=1
\)
For a hyperbola, the focal distances of a point \(\boldsymbol{P ( x , y ) \text { are } | e x – a | \text { and } | e x + a | \text { , where }} e\) is the eccentricity. The product of these distances is:
\(
(e x-a)(e x+a)=e^2 x^2-a^2=32
\)
Substituting \(x=4\) into the product equation:
\(
16 e^2-a^2=32
\)
Step 2: Solve for Hyperbola Parameters
We know that \(b^2=a^2\left(e^2-1\right)\), which implies \(a^2 e^2=a^2+b^2\). Substituting this into the product equation:
\(
16 e^2-a^2=32 \Longrightarrow 16\left(\frac{a^2+b^2}{a^2}\right)-a^2=32
\)
\(
16+\frac{16 b^2}{a^2}-a^2=32 \Longrightarrow \frac{16 b^2}{a^2}-a^2=16 \Longrightarrow \frac{16 b^2-a^4}{a^2}=16
\)
From the point \(P\) equation, we have \(\frac{16 b^2-12 a^2}{a^2 b^2}=1 \Longrightarrow 16 b^2-12 a^2=a^2 b^2\).
Solving these simultaneous equations for \(a^2\) and \(b^2\) :
\(a^2=8\)
\(b^2=12\)
Step 3: Calculate \(p^2+q^2\)
The length of the conjugate axis \(p\) is \(2 b\).
\(
p=2 \sqrt{12} \Longrightarrow p^2=4(12)=48
\)
The length of the latus rectum \(q\) is \(\frac{2 b^2}{a}\).
\(
\begin{gathered}
q=\frac{2(12)}{\sqrt{8}}=\frac{24}{2 \sqrt{2}}=\frac{12}{\sqrt{2}}=6 \sqrt{2} \\
q^2=36(2)=72
\end{gathered}
\)
Summing these values:
\(
p^2+q^2=48+72=120
\)

Hint

(d) Step 1: Analyze the first hyperbola \(\boldsymbol{H}_{\mathbf{1}}\)
The hyperbola \(H_1\) is given by \(\frac{x^2}{a^2}-\frac{y^2}{b^2}=1\).
Eccentricity: \(e_1=\sqrt{1+\frac{b^2}{a^2}}=\sqrt{\frac{5}{2}} \Longrightarrow \frac{b^2}{a^2}=\frac{5}{2}-1=\frac{3}{2}\).
Latus Rectum \(\left(L R_1\right): L R_1=\frac{2 b^2}{a}=15 \sqrt{2}\).
Substituting \(b^2=\frac{3}{2} a^2\) :
\(
\frac{2\left(\frac{3}{2} a^2\right)}{a}=3 a=15 \sqrt{2} \Longrightarrow a=5 \sqrt{2}
\)
Step 2: Determine dimensions of the second hyperbola \(\mathrm{H}_2\)
The hyperbola \(H_2\) is given by \(-\frac{x^2}{A^2}+\frac{y^2}{B^2}=1\). This is a vertical hyperbola.
Transverse Axis Length: The length is 2B.
Product of Transverse Axes: \((2 a)(2 B)=100 \sqrt{10}\).
Substituting \(a=5 \sqrt{2}\) :
\(
4(5 \sqrt{2}) B=100 \sqrt{10} \Longrightarrow 20 \sqrt{2} B=100 \sqrt{10} \Longrightarrow B=\frac{5 \sqrt{10}}{\sqrt{2}}=5 \sqrt{5} \text {. }
\)
Step 3: Calculate \(e_2\) and the final expression
Latus Rectum \(\left(L R_2\right): L R_2=\frac{2 A^2}{B}=12 \sqrt{5}\).
Substituting \(B=5 \sqrt{5}\) :
\(
\frac{2 A^2}{5 \sqrt{5}}=12 \sqrt{5} \Longrightarrow 2 A^2=12 \times 25 \Longrightarrow A^2=150.
\)
Eccentricity (\(e_2\)): For a vertical hyperbola, \(e_2^2=1+\frac{\boldsymbol{A}^2}{\boldsymbol{B}^2}\).
\(
e_2^2=1+\frac{150}{(5 \sqrt{5})^2}=1+\frac{150}{125}=1+\frac{6}{5}=\frac{11}{5}
\)
Final Value: \(25 e_2^2=25 \times \frac{11}{5}=5 \times 11=55\).