Questions are transcribed from NTA's official question paper. Diagrams are described in words; open the official PDF to see them. Every answer was solved twice independently and is shown only when both agree. Where they did not, we say so.
Let an ellipse $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$, $a < b$, pass through the point $(4, 3)$ and have eccentricity $\frac{\sqrt{5}}{3}$. Then the length of its latus rectum is :
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Answer: $\frac{8\sqrt{5}}{3}$
Key idea: Ellipse with major axis along y-axis, eccentricity relation, and latus rectum formula
Since $a < b$, the major axis is along y-axis. Use $e = \frac{c}{b}$ where $c^2 = b^2 - a^2$
From $e = \frac{\sqrt{5}}{3}$: $c = \frac{\sqrt{5}b}{3}$, so $\frac{5b^2}{9} = b^2 - a^2$, giving $a^2 = \frac{4b^2}{9}$
Substitute point $(4,3)$ into ellipse equation: $\frac{16}{a^2} + \frac{9}{b^2} = 1$
Replace $a^2 = \frac{4b^2}{9}$: $\frac{36}{b^2} + \frac{9}{b^2} = 1$, so $b^2 = 45$ and $a^2 = 20$
Use latus rectum formula $L = \frac{2a^2}{b} = \frac{40}{\sqrt{45}} = \frac{40}{3\sqrt{5}} = \frac{8\sqrt{5}}{3}$
Let a circle C have its centre in the first quadrant, intersect the coordinate axes at exactly three points and cut off equal intercepts from the coordinate axes. If the length of the chord of C on the line $x+y=1$ is $\sqrt{14}$, then the square of the radius of C is _________.
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Answer: 8
Key idea: Circle passing through origin with equal intercepts on axes, chord length formula
Since the circle intersects axes at exactly 3 points with equal intercepts, it passes through $(0,0)$, $(a,0)$, and $(0,a)$ for some $a > 0$
The center equidistant from these three points is at $(\frac{a}{2}, \frac{a}{2})$ (can be verified by setting distances equal)
The radius squared is $r^2 = (\frac{a}{2})^2 + (\frac{a}{2})^2 = \frac{a^2}{2}$
Distance from center $(\frac{a}{2}, \frac{a}{2})$ to line $x+y=1$ is $d = \frac{|a-1|}{\sqrt{2}}$
Using chord length formula: $(\sqrt{14})^2 = 4(r^2 - d^2)$, we get $14 = 4(\frac{a^2}{2} - \frac{(a-1)^2}{2})$
Let O be the origin, and P and Q be two points on the rectangular hyperbola $xy = 12$ such that the mid point of the line segment PQ is $\left(\frac{1}{2}, -\frac{1}{2}\right)$. Then the area of the triangle OPQ equals :
(1) $\frac{3}{2}$
(2) $\frac{5}{2}$
(3) $\frac{7}{2}$
(4) $\frac{9}{2}$
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Answer: Option (3)
Key idea: Finding area of triangle formed by origin and two points on a hyperbola given their midpoint
Since $P(x_1, y_1)$ and $Q(x_2, y_2)$ lie on $xy = 12$, we have $x_1y_1 = 12$ and $x_2y_2 = 12$
From midpoint condition: $x_1 + x_2 = 1$ and $y_1 + y_2 = -1$
Substitute $y_1 = \frac{12}{x_1}$ and $y_2 = \frac{12}{x_2}$ into $y_1 + y_2 = -1$ to get $\frac{12(x_1+x_2)}{x_1x_2} = -1$
This gives $x_1x_2 = -12$. With $x_1 + x_2 = 1$, solve $t^2 - t - 12 = 0$ to get $x_1 = 4, x_2 = -3$
Find coordinates: $P = (4, 3)$ and $Q = (-3, -4)$
Area of $\triangle OPQ = \frac{1}{2}|x_1y_2 - x_2y_1| = \frac{1}{2}|4(-4) - (-3)(3)| = \frac{1}{2}|-7| = \frac{7}{2}$
Let the parabola $y = x^2 + px + q$ passing through the point $(1, -1)$ be such that the distance between its vertex and the $x$-axis is minimum. Then the value of $p^2 + q^2$ is :
(1) $2$
(2) $4$
(3) $5$
(4) $8$
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Answer: Option (2)
Key idea: Optimization of parabola vertex distance using constraint from a given point
Use the condition that parabola passes through $(1, -1)$: $-1 = 1 + p + q$, giving $q = -2 - p$
Find vertex coordinates: $x_v = -\frac{p}{2}$ and $y_v = -\frac{p^2}{4} + q = -\frac{p^2}{4} - p - 2$
Distance from vertex to $x$-axis is $|y_v| = \left|\frac{p^2}{4} + p + 2\right| = \frac{p^2 + 4p + 8}{4}$ (always positive)
Let A be the point $(3, 0)$ and circles with variable diameter AB touch the circle $x^2 + y^2 = 36$ internally. Let the curve C be the locus of the point B. If the eccentricity of C is e, then $72e^2$ is equal to __________.
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Answer: 18
Key idea: Locus of points forming an ellipse using internal tangency condition
Let $B = (h, k)$. The variable circle has diameter $AB$, so its center is $\left(\frac{3+h}{2}, \frac{k}{2}\right)$ and radius is $\frac{|AB|}{2} = \frac{\sqrt{(h-3)^2 + k^2}}{2}$
For internal tangency with $x^2 + y^2 = 36$ (radius 6): Distance between centers = Difference of radii
Suppose that two chords, drawn from the point $(1, 2)$ on the circle $x^2 + y^2 + x - 3y = 0$ are bisected by the y-axis. If the other ends of these chords are R and S, and the mid point of the line segment RS is $(\alpha, \beta)$, then $6(\alpha + \beta)$ is equal to:
(1) 1
(2) 3
(3) 4
(4) 6
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Answer: Option (2)
Key idea: Chord properties and midpoint conditions in circles
Verify that point $P(1,2)$ lies on the circle $x^2 + y^2 + x - 3y = 0$
For a chord from $P(1,2)$ to point $(a,b)$ bisected by y-axis, the midpoint $\left(\frac{1+a}{2}, \frac{2+b}{2}\right)$ must have x-coordinate = 0, giving $a = -1$
Substitute $x = -1$ into the circle equation: $1 + y^2 - 1 - 3y = 0$, which gives $y^2 - 3y = 0$
Solve to get $y = 0$ or $y = 3$, so the two points are $R(-1, 0)$ and $S(-1, 3)$
Find the midpoint of $RS$: $(\alpha, \beta) = \left(-1, \frac{3}{2}\right)$
Consider the parabola P : $y^2 = 4ex$ and the ellipse E : $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$. Let the line segment joining the points of intersection of P and E, be their latus rectums. If the eccentricity of E is $e$, then $e^2 + 2\sqrt{2}$ is equal to ______.
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Answer: 3
Key idea: Intersection of parabola and ellipse on their latus rectums
Parabola $y^2 = 4ex$ has latus rectum at $x = e$ with points $(e, \pm 2e)$
Ellipse has latus rectum at $x = ae$ with semi-latus rectum $\frac{b^2}{a}$
For intersection on latus rectums: $x = e = ae$, giving $a = 1$
At intersection, $y$-coordinates match: $2e = 1 - e^2$ (using $b^2 = 1 - e^2$ for $a=1$)
Let $P(3\cos a, 2\sin a)$, $a \neq 0$, be a point on the ellipse $\frac{x^2}{9} + \frac{y^2}{4} = 1$. $Q$ be a point on the circle $x^2 + y^2 - 14x - 14y + 82 = 0$ and $R$ be a point on the line $x + y = 5$ such that the centroid of the triangle $PQR$ is $\left(2 + \cos a, 3 + \frac{2}{3}\sin a\right)$. Then the sum of the ordinates of all possible points $R$ is:
(1) 6
(2) 2
(3) 4
(4) 8
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Answer: Option (4)
Key idea: Centroid of triangle, coordinate geometry with ellipse, circle and line
Given $P(3\cos\alpha, 2\sin\alpha)$ on ellipse, $Q$ on circle, $R$ on line $x+y=5$, with centroid $G(2+\cos\alpha, 3+\frac{2}{3}\sin\alpha)$
Using centroid formula: $\frac{x_P+x_Q+x_R}{3} = 2+\cos\alpha$ and $\frac{y_P+y_Q+y_R}{3} = 3+\frac{2}{3}\sin\alpha$
Simplifying gives $x_Q + x_R = 6$ and $y_Q + y_R = 9$ (independent of $\alpha$)
Since $R$ is on line $x+y=5$, we get $x_R = 5-y_R$, so $Q = (1+y_R, 9-y_R)$
Substituting $Q$ into circle equation $x^2+y^2-14x-14y+82=0$ gives $2y_R^2-16y_R+24=0$
Solving: $y_R^2-8y_R+12=0 \Rightarrow y_R = 6$ or $y_R = 2$
Sum of ordinates of all possible points $R$ is $6+2=8$
Let $H: \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be a hyperbola such that the distance between its foci is $6$ and the distance between its directrices is $\frac{8}{3}$. If the line $x = a$ intersects the hyperbola $H$ at the points $A$ and $B$ such that the area of the triangle $AOB$ is $4\sqrt{15}$, where $O$ is the origin, then $a^2$ equals
(1) 12
(2) 9
(3) 15
(4) 18
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Let A, B and C be the vertices of a variable right angled triangle inscribed in the parabola $y^2 = 16x$. Let the vertex B containing the right angle be $(4, 8)$ and the locus of the centroid of $\triangle ABC$ be a conic $C_0$. Then three times the length of latus rectum of $C_0$ is _________
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Answer: 16
Key idea: Locus of centroid of right-angled triangle inscribed in a parabola
Parametrize points on parabola $y^2 = 16x$ as $A(t_1^2, 4t_1)$, $B(4, 8)$, $C(t_3^2, 4t_3)$
Apply perpendicularity condition $\vec{BA} \cdot \vec{BC} = 0$ to get $t_1t_3 + 2(t_1 + t_3) + 20 = 0$
Let a focus of the ellipse E : $\frac{x^2}{a^2} + \frac{y^2}{b^2} = 1$ be S(4, 0) and its eccentricity be $\frac{4}{5}$. If the point P(3, $\alpha$) lies on E and O is the origin, then the area of $\Delta$POS is equal to:
(1) 12/5
(2) 14/5
(3) 24/5
(4) 48/5
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Answer: Option (3)
Key idea: Ellipse properties: finding parameters from focus and eccentricity, and calculating area of triangle
Given focus $S(4,0)$ and eccentricity $e = \frac{4}{5}$. Since focus is at $(ae, 0)$, we have $ae = 4$, so $a = 5$
Find $b$ using $b^2 = a^2(1-e^2) = 25(1-\frac{16}{25}) = 9$, so $b = 3$
The ellipse equation is $\frac{x^2}{25} + \frac{y^2}{9} = 1$
Substitute $P(3, \alpha)$ into the ellipse: $\frac{9}{25} + \frac{\alpha^2}{9} = 1$, giving $\alpha^2 = \frac{144}{25}$, so $\alpha = \frac{12}{5}$
Since $O(0,0)$ and $S(4,0)$ lie on the x-axis, base $OS = 4$ and height $= |\alpha| = \frac{12}{5}$
Area of $\triangle POS = \frac{1}{2} \times 4 \times \frac{12}{5} = \frac{24}{5}$
Let P be a moving point on the circle $x^2 + y^2 - 6x - 8y + 21 = 0$. Then, the maximum distance of P from the vertex of the parabola $x^2 + 6x + y + 13 = 0$ is equal to:
(1) 8
(2) 10
(3) 12
(4) 9
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Answer: Option (3)
Key idea: Maximum distance from a point on a circle to a fixed point
Convert circle $x^2 + y^2 - 6x - 8y + 21 = 0$ to standard form: $(x-3)^2 + (y-4)^2 = 4$. Center $C(3,4)$, radius $r=2$
Rewrite parabola $x^2 + 6x + y + 13 = 0$ as $y = -(x+3)^2 - 4$. Vertex is $V(-3,-4)$
Calculate distance from center $C(3,4)$ to vertex $V(-3,-4)$: $d = \sqrt{(6)^2 + (8)^2} = \sqrt{100} = 10$
Maximum distance from point $P$ on circle to vertex $V$ is: $d_{max} = d + r = 10 + 2 = 12$
Let the point P be the vertex of the parabola $y = x^2 - 6x + 12$. If a line passing through the point P intersects the circle $x^2 + y^2 - 2x - 4y + 3 = 0$ at the points R and S, then the maximum value of $(PR + PS)^2$ is :
(1) 10
(2) 20
(3) 25
(4) 5
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Answer: Option (2)
Key idea: Parabola vertex, circle geometry, chord properties, and optimization
Find vertex of parabola $y = x^2 - 6x + 12$ by completing square: $y = (x-3)^2 + 3$, so $P = (3, 3)$
Rewrite circle $x^2 + y^2 - 2x - 4y + 3 = 0$ in standard form: $(x-1)^2 + (y-2)^2 = 2$, center $C = (1, 2)$, radius $r = \sqrt{2}$
Calculate distance $PC = \sqrt{(3-1)^2 + (3-2)^2} = \sqrt{5}$. Since $\sqrt{5} > \sqrt{2}$, point $P$ is outside the circle
For chord $RS$ with midpoint $M$, we have $PR + PS = 2PM$ where $PM = \sqrt{PC^2 - CM^2} = \sqrt{5 - CM^2}$
To maximize $PR + PS$, minimize $CM$. Minimum occurs when line passes through center $C$, giving $CM = 0$
Maximum value: $PR + PS = 2\sqrt{5}$, so $(PR + PS)^2 = 4 \times 5 = 20$
Let the directrix of the parabola P : $y^2 = 8x$, cut x-axis at the point A. Let B $(\alpha, \beta)$, $\alpha > 1$, be a point on P such that the slope of AB is $3/5$. If BC is a focal chord of P, then six times the area of $\triangle ABC$ is :
Let the eccentricity e of a hyperbola satisfy the equation $6e^2 - 11e + 3 = 0$. If the foci of the hyperbola are $(3, 5)$ and $(3, -4)$, then the length of its latus rectum is :
(1) $\frac{11}{\sqrt{3}}$
(2) $\frac{17}{\sqrt{3}}$
(3)
(4)
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Let $e_1$ and $e_2$ be two distinct roots of the equation $x^2 - \alpha x + 2 = 0$. Let the sets $\{a \in \mathbb{R} : e_1$ and $e_2$ are the eccentricities of hyperbolas$\} = (\alpha, \beta)$, and $\{a \in \mathbb{R} : e_1$ and $e_2$ are the eccentricities of an ellipse and a hyperbola, respectively$\} = (\gamma, \infty)$. Then $\alpha^2 + \beta^2 + \gamma^2$ is equal to:
(1) 18
(2) 22
(3) 26
(4) 34
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Answer: Option (3)
Key idea: Eccentricity constraints for conics, Vieta's formulas, and interval analysis
Use Vieta's formulas: $e_1 + e_2 = \alpha$ and $e_1 \cdot e_2 = 2$. For distinct roots: $\alpha^2 > 8$.
For both hyperbolas ($e_1, e_2 > 1$): Need $(e_1-1)(e_2-1) > 0 \Rightarrow 3 - \alpha > 0$. Combined with $\alpha > 2\sqrt{2}$, get $(\alpha, \beta) = (2\sqrt{2}, 3)$.
For ellipse and hyperbola: Let $0 < e_1 < 1$ and $e_2 > 1$. Since $e_1 e_2 = 2$ and $e_1 < 1$, we get $e_2 > 2$.
With $e_1 = \frac{2}{e_2}$, we have $\alpha = \frac{2}{e_2} + e_2$. For $e_2 > 2$, the function $f(x) = \frac{2}{x} + x$ is increasing.
At $e_2 = 2$: $\alpha = 3$. As $e_2 \to \infty$: $\alpha \to \infty$. So $(\gamma, \infty) = (3, \infty)$, giving $\gamma = 3$.
If the eccentricity $e$ of the hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, passing through $(6, 4\sqrt{3})$, satisfies $15(e^2 + 1) = 34e$, then the length of the latus rectum of the hyperbola $\frac{x^2}{b^2} - \frac{y^2}{2(a^2+1)} = 1$ is:
(1) $10$
(2) $20$
(3) $25$
(4) $30$
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Answer: Option (1)
Key idea: Hyperbola eccentricity, latus rectum, and point on conic
Solve $15(e^2 + 1) = 34e$ to get $15e^2 - 34e + 15 = 0$. Using quadratic formula: $e = \frac{5}{3}$ (since $e > 1$ for hyperbola)
For hyperbola $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$, use $e^2 = 1 + \frac{b^2}{a^2}$. So $\frac{25}{9} = 1 + \frac{b^2}{a^2}$, giving $b^2 = \frac{16a^2}{9}$
Substitute point $(6, 4\sqrt{3})$ into hyperbola: $\frac{36}{a^2} - \frac{48}{b^2} = 1$
Using $b^2 = \frac{16a^2}{9}$: $\frac{36}{a^2} - \frac{27}{a^2} = 1$, which gives $a^2 = 9$ and $b^2 = 16$
Let chord PQ of length $3\sqrt{13}$ of the parabola $y^2 = 12x$ be such that the ordinates of points P and Q are in the ratio 1:2. If the chord PQ subtends an angle $\alpha$ at the focus of the parabola, then $\sin \alpha$ is equal to:
(1) $\frac{3}{5}$
(2) $\frac{4}{5}$
(3) $\frac{5}{13}$
(4) $\frac{12}{13}$
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Answer: Option (1)
Key idea: Parabola focal chord properties and angle subtended at focus using cosine rule
Let ordinates be $y_1 = t$ and $y_2 = 2t$ (ratio 1:2). Using $y^2 = 12x$, get $P = \left(\frac{t^2}{12}, t\right)$ and $Q = \left(\frac{t^2}{3}, 2t\right)$
Apply length condition $|PQ| = 3\sqrt{13}$ to get $t^2(t^2 + 16) = 1872$, solving gives $t^2 = 36$, so $t = 6$
Let the centre of the circle $x^2 + y^2 + 2gx + 2fy + 25 = 0$ be in the first quadrant and lie on the line $2x - y = 4$. Let the area of an equilateral triangle inscribed in the circle be $27\sqrt{3}$. Then the square of the length of the chord of the circle on the line $x = 1$ is _______
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Answer: 80
Key idea: Circle properties, equilateral triangle inscribed in circle, chord length formula
Center of circle $x^2 + y^2 + 2gx + 2fy + 25 = 0$ is $(-g, -f)$ with radius $r = \sqrt{g^2 + f^2 - 25}$. Since center lies on $2x - y = 4$: $-2g + f = 4$, so $f = 2g + 4$
For equilateral triangle inscribed in circle with radius $r$, area $= \frac{3\sqrt{3}r^2}{4}$. Given area $= 27\sqrt{3}$, we get $r^2 = 36$, so $r = 6$
From $r^2 = g^2 + f^2 - 25 = 36$, we have $g^2 + f^2 = 61$. Substituting $f = 2g + 4$: $5g^2 + 16g - 45 = 0$
Solving: $g = \frac{9}{5}$ or $g = -5$. Since center must be in first quadrant, $g = -5$ and $f = -6$ giving center $(5, 6)$
Distance from center $(5, 6)$ to line $x = 1$ is $d = 4$. Chord length $= 2\sqrt{r^2 - d^2} = 2\sqrt{36 - 16} = 4\sqrt{5}$
Let $C$ be a circle having centre in the first quadrant and touching the $x$-axis at a distance of $3$ units from the origin. If the circle $C$ has an intercept of length $6\sqrt{3}$ on $y$-axis, then the length of the chord of the circle $C$ on the line $x - y = 3$ is :
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Answer: 6√2
Key idea: Circle touching x-axis, y-intercept, and chord length using perpendicular distance formula
Since the circle touches the x-axis at $(3, 0)$ and center is in first quadrant, center is at $(3, r)$ where $r$ is the radius
For y-axis intercept, substitute $x = 0$ in $(x-3)^2 + (y-r)^2 = r^2$ to get intercept length $= 2\sqrt{r^2 - 9}$
Given intercept length $= 6\sqrt{3}$, so $2\sqrt{r^2 - 9} = 6\sqrt{3}$ which gives $r^2 = 36$, thus $r = 6$
Center is $(3, 6)$ with radius $6$. Find perpendicular distance from center to line $x - y - 3 = 0$: $d = \frac{|3-6-3|}{\sqrt{2}} = 3\sqrt{2}$
The eccentricity of an ellipse E with centre at the origin O is $\frac{\sqrt{3}}{2}$ and its directrices are $x = \pm \frac{4\sqrt{6}}{3}$. Let H: $\frac{x^2}{a^2} - \frac{y^2}{b^2} = 1$ be a hyperbola whose eccentricity is equal to the length of semi-major axis of E, and whose length of latus rectum is equal to the length of minor axis of E. Then the distance between the foci of H is:
(1) $\frac{4\sqrt{2}}{\sqrt{7}}$
(2) $\frac{4\sqrt{2}}{7}$
(3) $\frac{4}{\sqrt{7}}$
(4) $\frac{8}{7}$
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Answer: Option (4)
Key idea: Ellipse and Hyperbola Properties: Eccentricity, Directrix, Latus Rectum, and Foci
Step 1: For ellipse $E$, use directrix formula $\frac{A}{e} = \frac{4\sqrt{6}}{3}$ with $e = \frac{\sqrt{3}}{2}$ to get semi-major axis $A = 2\sqrt{2}$
Step 2: Use $e^2 = 1 - \frac{B^2}{A^2}$ to find semi-minor axis $B = \sqrt{2}$, so minor axis length $= 2\sqrt{2}$
Step 3: For hyperbola $H$, eccentricity $e_H = 2\sqrt{2}$ (equals $A$ of ellipse). Use $e_H^2 = 1 + \frac{b^2}{a^2}$ to get $\frac{b^2}{a^2} = 7$
Step 4: Latus rectum of $H$ equals minor axis of $E$: $\frac{2b^2}{a} = 2\sqrt{2}$, giving $b^2 = a\sqrt{2}$
Step 5: Solve $b^2 = 7a^2$ and $b^2 = a\sqrt{2}$ simultaneously to get $a = \frac{\sqrt{2}}{7}$
Let $x = 9$ be a directrix of an ellipse E, whose centre is at the origin and eccentricity is $\frac{1}{3}$. Let P $(\alpha, 0)$, $\alpha > 0$, be a focus of E and AB be a chord passing through P. Then the locus of the mid point of AB is:
(1) $9y^2 = 8x(1-x)$
(2) $3y^2 = 4x(1-x)$
(3)
(4)
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Answer: Option (1)
Key idea: Locus of midpoint of focal chord of ellipse
Find semi-major axis: Using directrix $x = \frac{a}{e} = 9$ and $e = \frac{1}{3}$, we get $a = 3$
Let O be the vertex of the parabola $y^2 = 4x$ and its chords OP and OQ are perpendicular to each other. If the locus of the mid-point of the line segment PQ is a conic C, then the length of its latus rectum is:
(1) $1$
(2) $2$
(3) $4$
(4) $8$
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Answer: Option (2)
Key idea: Locus of midpoint of chord, perpendicular chords from vertex of parabola, parametric form of parabola
Step 1: Use parametric form: Let $P = (t_1^2, 2t_1)$ and $Q = (t_2^2, 2t_2)$ on parabola $y^2 = 4x$
Step 2: Apply perpendicularity: Slopes of OP and OQ are $\frac{2}{t_1}$ and $\frac{2}{t_2}$. Since perpendicular: $\frac{2}{t_1} \cdot \frac{2}{t_2} = -1 \Rightarrow t_1 t_2 = -4$
Step 3: Find midpoint: If $(h,k)$ is midpoint of PQ, then $h = \frac{t_1^2 + t_2^2}{2}$ and $k = t_1 + t_2$
Step 4: Derive locus: From $k^2 = t_1^2 + t_2^2 + 2t_1t_2 = t_1^2 + t_2^2 - 8$, we get $2h = k^2 + 8$, so $y^2 = 2(x-4)$
Step 5: Identify parameters: Comparing $y^2 = 2(x-4)$ with $y^2 = 4a(x-h)$, we get $4a = 2$, so $a = \frac{1}{2}$
Let $\frac{x^2}{f(a^2+7a+3)} + \frac{y^2}{f(3a+15)} = 1$ represent an ellipse with major axis along $y$-axis, where $f$ is a strictly decreasing positive function on $\mathbb{R}$. If the set of all possible values of $a$ is $\mathbb{R} - [\alpha, \beta]$, then $\alpha^2 + \beta^2$ is equal to :
(1) 28
(2) 40
(3) 61
(4) 24
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Answer: Option (2)
Key idea: Ellipse with major axis condition and properties of strictly decreasing functions
For ellipse $\frac{x^2}{A} + \frac{y^2}{B} = 1$ with major axis along $y$-axis, we need $B > A$, i.e., $f(3a+15) > f(a^2+7a+3)$
Since $f$ is strictly decreasing, $f(3a+15) > f(a^2+7a+3)$ implies $3a+15 < a^2+7a+3$
Consider the circle $C: x^2 + y^2 - 6x - 8y - 11 = 0$. Let a variable chord AB of the circle C subtend a right angle at the origin. If the locus of the foot of the perpendicular drawn from the origin on the chord AB is the circle $x^2 + y^2 - \alpha x - \beta y - \gamma = 0$, then $\alpha + \beta + 2\gamma$ is equal to _______.
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Answer under review. Our two independent solutions did not agree, so we are not showing an answer yet. Check it against NTA's final answer key.
Key idea: Locus of foot of perpendicular from origin to chord subtending right angle at origin
FAQs
Frequently asked questions
How many questions come from Conic Sections (Parabola, Ellipse, Hyperbola) in JEE Main?
In the 9 shifts of JEE Main 2026 Session 2 we found 25 questions on it, about 2.8 per shift.
Where can I practise Conic Sections (Parabola, Ellipse, Hyperbola) PYQs for free?
On this page, with solutions. For more, use NTA Abhyas and our free mock tests.
