Question 1

For the pair of parallel straight lines represented by 9x² - 6xy + y² + 18x - 6y + 8 = 0, what is the separation between them?

Accepted Answer

2/√10. The given equation represents a pair of parallel lines, and the separation between them can be calculated using the formula for the distance between two parallel lines derived from their general form. In this case, the correct calculation yields a separation of 2/√10.

Question 2

An ellipse has its two foci 10 units apart, and the length of its latus rectum is 15. If its axes are taken as the coordinate axes, which equation represents the ellipse?

Accepted Answer

3x² + 4y² = 300. 2c=10 -> c=5. Latus rectum 2b^2/a=15 -> b^2=7.5a. With a^2=b^2+25: a^2-7.5a-25=0 -> a=10, b^2=75. Ellipse x^2/100+y^2/75=1 -> 3x^2+4y^2=300.

Question 3

On the segment joining A(0, 0) and B(3a, 0), choose points P and Q so that AP = PQ = QB. Three circles are then constructed with AP, PQ, and QB as their respective diameters. If a point S is such that the sum of the squares of the tangents drawn from S to these three circles is b², then th

Accepted Answer

3(x² + y²) - 9ax + 8a² - b² = 0. With A(0,0),B(3a,0), P=(a,0),Q=(2a,0); the three circles have centres (a/2,0),(3a/2,0),(5a/2,0) and radius a/2. The sum of squared tangent lengths = sum of powers = 3(x^2+y^2) - 9ax + 8a^2. Setting this equal to b^2 gives 3(x^2+y^2) - 9ax + 8a^2 - b^2 = 0.

Question 4

Find the coordinates of the midpoint of the chord cut by the circle x² + y² + 4x - 2y - 3 = 0 on the line y = x + 2.

Accepted Answer

(-3/2, 1/2). The correct option is right because the midpoint of the chord can be found by substituting the line equation into the circle's equation, simplifying it, and then solving for the intersection points. The coordinates (-3/2, 1/2) accurately represent the midpoint of the chord formed by these intersections.

Question 5

A hyperbola has a transverse axis of length 2 sin θ and is confocal with the ellipse 3x² + 4y² = 12. Its equation is

Accepted Answer

x² cosec² θ - y² sec² θ = 1. 3x^2+4y^2=12 gives c^2=4-3=1, foci (+-1,0). Confocal hyperbola with transverse semi-axis sin(theta) has A=sin^2(theta), and c^2=A+B=1 gives B=cos^2(theta): x^2 cosec^2(theta) - y^2 sec^2(theta) = 1.

Question 6

Three points E, F and G are chosen on the parabola y² = 4ax such that their y-coordinates form a geometric progression. The point where the tangents at E and G meet lies on the

Accepted Answer

vertical line through F. For y^2=4ax with points at parameters t1,t2, tangents meet at (a*t1*t2, a(t1+t2)). y-coords in GP means t_E,t_F,t_G in GP, so t_E*t_G = t_F^2. The intersection x-coordinate a*t_E*t_G = a*t_F^2 equals F's x-coordinate, so the meeting point lies on the vertical line through F.

Question 7

In the parabola given by x² - 2x + y - 2 = 0, let AB be a focal chord and S be the focus. If AS = l1, then what is the value of BS?

Accepted Answer

l1/(4l1 - 1). (x-1)^2=-(y-3) has 4a=1 so a=1/4. For a focal chord 1/AS+1/BS=1/a=4, hence 1/BS=4-1/l1=(4l1-1)/l1, giving BS=l1/(4l1-1).

Question 8

For the point (⌊ P+1⌋, ⌊ P⌋), where ⌊ x⌋ denotes the greatest integer less than or equal to x, suppose it lies strictly inside the annular region enclosed by the circles x²+y²-2x-15=0 and x²+y²-2x-7=0. Then which of the following is true?

Accepted Answer

None of these. The point ((⌊ P+1⌋,⌊ P⌋) must lie strictly between the two circles defined by the equations, which restricts the possible values of P beyond the intervals provided in the other options. Therefore, none of the specified ranges accurately capture the valid values for P.

Question 9

A straight line passing through the point P(3, 11) intersects the circle x² + y² = 9 at the points A and B. The value of PA · PB is

Accepted Answer

121. The value of PA · PB is determined by the power of point P with respect to the circle, which is calculated as the square of the distance from P to the center of the circle minus the square of the radius. Here, the distance from P(3, 11) to the center (0, 0) is 3² + 11² = 130, and the radius squared is 9, leading to PA · PB = 130 - 9 = 121.

Question 10

The curve given by the equation √(ax) + √(by) = 1 is a

Accepted Answer

parabola. The equation ( rac{ ext{sqrt}(ax)}{1} + rac{ ext{sqrt}(by)}{1} = 1 ext{,} ext{ which can be manipulated to resemble the standard form of a parabola, indicates that it represents a parabolic curve. This is due to the presence of square root terms that define the relationship between x and y in a way characteristic of parabolas.

Question 11

For the ellipse x²/a² + y²/b² = 1, what is the maximum area of a rectangle that can be drawn inside it?

Accepted Answer

2ab. A rectangle inscribed in x^2/a^2+y^2/b^2=1 with corner (a cosT, b sinT) has area 4*a cosT*b sinT = 2ab sin2T, maximized when 2T=90 deg giving maximum area 2ab.

Question 12

The straight line 3x + 2y + 1 = 0 intersects the hyperbola 4x² - y² = 4a² at points P and Q. The point where the tangents drawn to the hyperbola at P and Q intersect is

Accepted Answer

(-3a², 8a²). For x^2/a^2 - y^2/4a^2 = 1, the chord of contact from (x1,y1) is 4x1 x - y1 y = 4a^2. Matching with 3x+2y+1=0 gives x1=-3a^2, y1=8a^2, so the point is (-3a^2, 8a^2).

Question 13

From any point on the circle 15x² + 15y² - 48x + 64y = 0, tangents are drawn to the circles 5x² + 5y² - 24x + 32y + 75 = 0 and 5x² + 5y² - 48x + 64y + 300 = 0. The lengths of these tangents are in the ratio

Accepted Answer

1: 2. Reduce all circles to monic form. For a point on the given circle, substitute S = x^2+y^2 from its equation into the tangent-length formulas: L1^2 = -(8/5)x+(32/15)y+15 and L2^2 = -(32/5)x+(128/15)y+60 = 4*L1^2. Hence L1:L2 = 1:2.

Question 14

Find the equation of the parabola if its focus is at the origin and the tangent at its vertex is given by x - y + 1 = 0.

Accepted Answer

x² + y² + 2xy - 4x + 4y - 4 = 0. The correct option represents a parabola with its focus at the origin and aligns with the given tangent line at the vertex, confirming that it satisfies the conditions of the problem.

Question 15

The parametric equations x = 2 - 3 sec t and y = 1 + 4 tan t describe which of the following curves?

Accepted Answer

A hyperbola with centre at (2, 1) and eccentricity 5/3. From x = 2-3sec t, y = 1+4tan t: (x-2)^2/9 - (y-1)^2/16 = 1, a hyperbola centred at (2,1) with a^2=9, b^2=16. Eccentricity e = sqrt(1 + 16/9) = 5/3.

Question 16

A circle goes through the point (a,b) and intersects the circle x²+y²=4 at right angles. The set of all possible centres of such circles is

Accepted Answer

2ax + 2by - (a² + b² + 4) = 0. Let the circle be x^2+y^2+2gx+2fy+c=0. Orthogonality with x^2+y^2=4 gives c=4. Passing through (a,b): a^2+b^2+2ga+2fb+4=0. With center (x,y)=(-g,-f), the locus of centers is 2ax+2by-(a^2+b^2+4)=0.

Question 17

A point P moving on an ellipse of eccentricity e is connected to the two foci S1 and S2. The incenter of triangle PS1S2 is located on

Accepted Answer

none of these. The incenter I=(c cos t, c*sqrt(a^2-c^2) sin t/(a+c)) traces an ellipse with semi-axes c and c*sqrt(a^2-c^2)/(a+c); its eccentricity satisfies e_I^2 = 2e/(1+e), which does not match sqrt((3+e^2)/4). So none of the given options is correct.

Question 18

Four points lying on the rectangular hyperbola xy = c² are concyclic, and their coordinates are given by (c t1, c/t1), (c t2, c/t2), (c t3, c/t3), and (c t4, c/t4). Then

Accepted Answer

t1 t2 t3 t4 = 1. Substituting (ct, c/t) into a general circle and multiplying by t^2 gives c^2 t^4 + 2gc t^3 + k t^2 + 2fc t + c^2 = 0. The product of the four roots is c^2/c^2 = 1, so t1 t2 t3 t4 = 1.

Question 19

A family of chords is drawn from the origin to the circle (x-1)²+y²=1. The equation of the locus of the midpoints of these chords is:

Accepted Answer

x² + y² = x. The locus of the midpoints of the chords drawn from the origin to the circle can be derived using the midpoint formula and the properties of circles. The resulting equation, x² + y² = x, represents a circle that is centered at (0.5, 0) with a radius of 0.5, which accurately describes the midpoints of the chords.

Question 20

For a point P = (x, y), let the foci be F1 = (3, 0) and F2 = (-3, 0). If the point satisfies 16x² + 25y² = 400, then the value of PF1 + PF2 is

Accepted Answer

10. Dividing by 400 gives x^2/25 + y^2/16 = 1 with a=5, b=4, c=3, so foci are (+-3,0) as given. For any point on the ellipse PF1+PF2 = 2a = 10.

Question 21

For the hyperbola given by 2x² + 5xy + 2y² + 4x + 5y = 0, the equation representing the pair of its asymptotes is

Accepted Answer

2x² + 5xy + 2y² + 4x + 5y + 2 = 0. Asymptotes have the same second-degree part plus a constant: 2x^2+5xy+2y^2+4x+5y+c = 0 must represent two lines, so its 3x3 determinant vanishes. Solving gives c = 2, so the asymptote pair is 2x^2+5xy+2y^2+4x+5y+2 = 0.

Question 22

For the circles x² + y² - 4x - 4y = 0 and x² + y² = 16, what angle is subtended at the origin by their common chord?

Accepted Answer

π/2. Subtracting the circle equations gives the common chord x+y=4, which meets x^2+y^2=16 at (4,0) and (0,4). These two points subtend a right angle (pi/2) at the origin.

Question 23

Which of the following is an equation of a common tangent to the parabola y² = 8x and the circle x² + y² - 12x + 4 = 0?

Accepted Answer

y = x + 2. For y^2=8x, tangents are y = mx + 2/m. The circle is centre (6,0), r = 4sqrt(2). Distance condition gives m = +-1, so the common tangents are y = x + 2 and y = -x - 2; among the options the valid one is y = x + 2. (y = x - 2 is not tangent to the circle.)

Question 24

An ellipse has its axes along the coordinate axes. If it passes through the point (4, -1) and is tangent to the line x + 4y - 10 = 0, then its equation is

Accepted Answer

Both (a) and (b). Both options (a) and (b) represent valid equations of ellipses that satisfy the given conditions of passing through the point (4, -1) and being tangent to the specified line, confirming that they are both correct.

Question 25

In the parabola y² = 4ax, a double ordinate has length 8a. The angle it subtends at the vertex is

Accepted Answer

π/2. A double ordinate of length 8a on y^2=4ax has endpoints (4a, 4a) and (4a, -4a). Their slopes from the vertex (0,0) are +1 and -1, so the angle subtended is 90 degrees = pi/2.

Question 26

A circle is given by the equation x² + y² + 16x - 24y + 183 = 0. If this circle is reflected in the line 4x + 7y + 13 = 0, what is the equation of the resulting image circle?

Accepted Answer

x² + y² + 32x + 4y + 235 = 0. The circle has centre (-8,12), r=5. Reflecting (-8,12) in 4x+7y+13=0 gives the image centre (-16,-2), radius still 5. The image circle is (x+16)^2+(y+2)^2=25, i.e. x^2+y^2+32x+4y+235=0.

Question 27

Determine the point on the parabola y = x² + 7x + 2 that lies nearest to the line y = 3x - 3.

Accepted Answer

(-2,-8). The nearest point has tangent parallel to y=3x-3. dy/dx=2x+7=3 => x=-2, y=(-2)^2+7(-2)+2=-8. The point is (-2,-8).

Question 28

For which values of n does the line x/a + y/b = 2 act as a tangent to the curve (x/a)ⁿ + (y/b)ⁿ = 2 at the point (a,b)?

Accepted Answer

n can be any real number. Differentiating (x/a)^n+(y/b)^n=2 gives n(x/a)^(n-1)/a + n(y/b)^(n-1) y'/b = 0. At (a,b) this is 1/a + y'/b = 0, so y'=-b/a, and the tangent is x/a+y/b=2 independent of n. Thus n can be any real number.

Question 29

The curves x²/α + y²/4 = 1 and y³ = 16x meet orthogonally. Then one possible value of α is:

Accepted Answer

4/3. The curves meet orthogonally when the product of their slopes at the intersection points equals -1. By calculating the derivatives of both curves and setting up the condition for orthogonality, we find that α must equal 4/3 to satisfy this relationship.

Question 30

For the curve xy = c², suppose the normal drawn at the point corresponding to parameter t1 intersects the curve again at parameter t2. Then which relation is true?

Accepted Answer

t1³ t2 = −1. Parametrize xy=c^2 as (ct, c/t). Slope of curve is -1/t^2, so the normal at t1 has slope t1^2. Setting the normal line equal to the curve again gives t2=-1/t1^3, hence t1^3 t2 = -1.

Question 31

For the point (a,0), the greatest possible distance from the curve 2x²+y²-2x=0 is

Accepted Answer

√(1-2a+2a²). From 2x^2+y^2-2x=0, y^2=2x-2x^2 with 0<=x<=1. Distance^2 from (a,0): (x-a)^2+y^2 = -x^2+(2-2a)x+a^2, maximized at x=1-a giving (1-a)^2+a^2 = 1-2a+2a^2. So greatest distance = sqrt(1-2a+2a^2).

Question 32

For the parabola y² = 36x, let LL' denote its latus rectum and let PP' be a double ordinate drawn at a point between the vertex and the latus rectum. The area of the trapezium PP'LL' is greatest when the distance of PP' from the vertex is

Accepted Answer

1. The area of the trapezium PP'LL' is maximized when the distance from the vertex to the double ordinate PP' is equal to the distance from the vertex to the latus rectum, which is 1 unit for the given parabola. This is because the area depends on the height and the width of the trapezium, and at this specific distance, the dimensions yield the largest area.

Question 33

A tangent to the ellipse x²/27 + y² = 1 is taken at the point (√3 cos θ, sin θ), where θ lies in (0, π/2). For which value of θ will the sum of the intercepts cut off by this tangent on the coordinate axes be the least?

Accepted Answer

π/6. Tangent at (3sqrt3 cos th, sin th) has x-intercept 27/(3sqrt3 cos th) and y-intercept 1/sin th, so the sum S=3sqrt3/cos th + 1/sin th. dS/dth=0 gives tan^3 th=1/(3sqrt3), i.e. tan th=1/sqrt3, so th=pi/6 (minimum S=8).

Question 34

For the pair of curves ax² + by² = 1 and a₁x² + b₁y² = 1 to intersect at right angles, which relation must hold?

Accepted Answer

(a-a₁)/(aa₁) = (b-b₁)/(bb₁). At an intersection, slopes give (ax/by)(a1x/b1y)=-1, and subtracting the conics gives x^2/y^2 = -(b-b1)/(a-a1). Combining yields aa1(b-b1)=bb1(a-a1), i.e. (a-a1)/(aa1) = (b-b1)/(bb1).

Question 35

Find the area, in square units, of the part of the first quadrant enclosed by the parabola y = 9x² and the lines x = 0, y = 1, and y = 4.

Accepted Answer

14/9. The area is determined by integrating the difference between the upper line (y = 4) and the parabola (y = 9x²) from the intersection points of the parabola with y = 1 and y = 4. This results in a calculated area of 14/9 square units, which represents the region in the first quadrant bounded by the specified lines and the parabola.

Question 36

The region bounded by the parabola y² = 4ax and the line y = ax has area 1/3 square unit. For the same parabola, what is the area enclosed by the line y = 4x?

Accepted Answer

8/3 square units. Area between y^2=4ax and y=ax is 8/(3a)=1/3, so a=8 and the parabola is y^2=32x. With y=4x: intersection at x=2, area = integral 0->2 (sqrt(32x)-4x) dx = 32/3 - 8 = 8/3 square units.

Question 37

For the two ellipses x²/a² + y²/b² = 1 and x²/b² + y²/a² = 1, where 0<b<a, what is the area of their overlapping region?

Accepted Answer

4abtan⁻¹(b/a). The two ellipses intersect on the lines y=+-x. Integrating the common interior region gives the standard result area = 4ab*arctan(b/a); numerical integration for a=2,b=1 gives 3.709, matching 4ab*arctan(b/a)=3.709.

Question 38

Find the area enclosed by the curves y - 1 = |x|, y = 0, and |x| = 1/2.

Accepted Answer

5/4. Region under y=1+|x| above y=0, between x=-1/2 and 1/2. Area = Integral(1+|x|)dx = 2*Integral_0^{1/2}(1+x)dx = 2*(0.5+0.125) = 5/4.

Question 39

Determine the area enclosed by the curve x² = 2 - y - y² and the y-axis.

Accepted Answer

9/2. The curve x = 2 - y - y^2 crosses the y-axis (x=0) at y = -2 and y = 1. Area = integral_{-2}^{1} (2 - y - y^2) dy = [2y - y^2/2 - y^3/3]_{-2}^{1} = 7/6 + 10/3 = 9/2.

Question 40

Find the area enclosed by the parabola y² = 4a²(x - 1), the vertical line x = 1, and the horizontal line y = 4a:

Accepted Answer

16a/3 sq unit. On the parabola x-1=y^2/(4a^2). Area = Integral_0^{4a} y^2/(4a^2) dy = (1/(4a^2))*(4a)^3/3 = 16a/3 sq unit.

Question 41

Find the area of the triangle enclosed by the tangent and the normal drawn at the point (1, √3) on the circle x² + y² = 4, together with the x-axis.

Accepted Answer

2√3 sq. units. At (1,sqrt3) on x^2+y^2=4 the tangent has slope -1/sqrt3 and meets the x-axis at (4,0); the normal passes through the centre (0,0). The triangle has vertices (0,0),(4,0),(1,sqrt3), so area = (1/2)*4*sqrt3 = 2*sqrt3 sq. units.

Question 42

Two spheres with radii 3 and 4 intersect at right angles. What is the radius of their common circle of intersection?

Accepted Answer

12/5. For spheres intersecting at right angles, d^2=r1^2+r2^2=9+16=25 so d=5. The radius of the common circle is r1*r2/d = 3*4/5 = 12/5.

Question 43

Let L1 be a line through the origin, and let L2 be the line x + y = 1. If the chord cut off by the circle x² + y² − x + 3y = 0 on L1 has the same length as the chord cut off on L2, then which equation can describe L1?

Accepted Answer

Both (a) and (b). Both options (a) and (b) represent lines through the origin that can yield chords of equal length when intersected with the given circle. The specific slopes of these lines allow for the necessary geometric conditions to be met, resulting in equal chord lengths.

Question 44

Find the area of the region bounded by the parabola y² = 12x and its latus rectum.

Accepted Answer

24. The area of the region bounded by the parabola y² = 12x and its latus rectum can be calculated using the formula for the area of a parabola segment. The latus rectum of this parabola is a vertical line segment that intersects the parabola at points where y = ±6, leading to an area of 24 square units.

Question 45

If the three different points P(3u², 2u³), Q(3v², 2v³), and R(3w², 2w³) lie on one straight line, then which relation must hold?

Accepted Answer

uv + vw + wu = 0. The points P, Q, and R are collinear if the area of the triangle they form is zero, which leads to the condition that the determinant formed by their coordinates equals zero. This results in the relation uv + vw + wu = 0, indicating that the weighted sum of the products of the coordinates must balance out to maintain collinearity.

Question 46

Find the area of the triangle enclosed by the common tangents to the circles x² + y² - 6x = 0 and x² + y² + 2x = 0.

Accepted Answer

3√3. The area of the triangle formed by the common tangents to the two circles can be calculated using the formula for the area of a triangle given the lengths of the tangents. The specific configuration of the circles leads to an area of 3√3, which is derived from the geometric properties of the tangents and the distance between the centers of the circles.

Question 47

Find the equation of the locus of the centres of all circles that intersect each of the circles x² + y² + 4x - 6y + 9 = 0 and x² + y² - 5x + 4y - 2 = 0 at right angles.

Accepted Answer

9x - 10y + 11 = 0. The correct option represents the locus of centers of circles that intersect the given circles at right angles, which is derived from the condition that the power of the point (the center of the new circle) with respect to each of the original circles must be equal to the square of the radius of the new circle.

Question 48

For a point P lying on the parabola y² = 4ax, let the line through P and the vertex meet the perpendicular drawn from the focus to the tangent at P at a point R. The locus of R is given by:

Accepted Answer

2x² + y² - 2ax = 0. The correct option describes the locus of point R derived from the geometric properties of the parabola and the relationships between the focus, vertex, and tangent line. This equation captures the necessary conditions that arise from the intersection of the line through point P and the perpendicular from the focus, leading to the specified quadratic form.

Question 49

Consider the curves given by x³ - 3xy² + 2 = 0 and 3x²y - y³ = 2. How do these two curves intersect?

Accepted Answer

They intersect orthogonally. The curves intersect orthogonally when their tangent lines at the points of intersection are perpendicular to each other, which occurs when the product of their slopes is -1. This condition is satisfied for the given curves, indicating that they cross at right angles.

Question 50

The pair of straight lines given by ax² + 2(a + b)xy + by² = 0 are diameters of a circle and split it into four sectors. If the area of one sector is three times the area of another sector, then which relation between a and b must hold?

Accepted Answer

3a² + 2ab + 3b² = 0. The correct option indicates a specific relationship between the coefficients a and b that ensures the areas of the sectors created by the diameters of the circle are in the ratio of 3:1. This relationship arises from the geometric properties of the circle and the angles formed by the intersecting lines.

JEE Main Maths: Conic Sections questions with solutions

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