Sign in
12thPass logo
New chatPYQ LibraryDoubtsRank report
Sign in to see Recents

Your guest activity stays on this device

Sign in to save progress →
Sign in

Ellipse question

2025 · 28 Jan · Shift 2 · Q29
Guest · filters and generic practice availableBrowsing as a guest · PYQ filters and generic practice are available. Sign in only for personalised features and saved progress.
  1. PYQ Library
  2. /JEE Main
  3. /Mathematics
  4. /Ellipse
  5. /2025 · 28 Jan · Shift 2 · Q29

Ellipse question

2025 · 28 Jan · Shift 2 · Q29

JEE MainMathematicsEllipseMCQ+4 / −1
If the midpoint of a chord of the ellipse x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=19x2​+4y2​=1 is (2,4/3)(\sqrt{2}, 4 / 3)(2​,4/3), and the length of the chord is 2α3\frac{2 \sqrt{\alpha}}{3}32α​​, then α\alphaα is :
  1. A
    26
  2. B
    18
  3. C
    22
  4. D
    20
View written solutionFree

Correct answer: C

  1. Given ellipse

    x29+y24=1\frac{x^2}{9}+\frac{y^2}{4}=19x2​+4y2​=1

    Here, a2=9,b2=4a^2=9,\quad b^2=4a2=9,b2=4

  2. Chord with given midpoint

    Midpoint of the chord is (x1,y1)=(2,43)(x_1,y_1)=\left(\sqrt{2},\frac{4}{3}\right)(x1​,y1​)=(2​,34​)

    For the ellipse x2a2+y2b2=1,\frac{x^2}{a^2}+\frac{y^2}{b^2}=1,a2x2​+b2y2​=1, the chord whose midpoint is (x1,y1)(x_1,y_1)(x1​,y1​) is given by the midpoint form:

    xx1a2+yy1b2=x12a2+y12b2\frac{xx_1}{a^2}+\frac{yy_1}{b^2}=\frac{x_1^2}{a^2}+\frac{y_1^2}{b^2}a2xx1​​+b2yy1​​=a2x12​​+b2y12​​

    Substituting values:

    x29+y(4/3)4=(2)29+(4/3)24\frac{x\sqrt{2}}{9}+\frac{y(4/3)}{4}=\frac{(\sqrt{2})^2}{9}+\frac{(4/3)^2}{4}9x2​​+4y(4/3)​=9(2​)2​+4(4/3)2​

    2x9+y3=29+16/94\frac{\sqrt{2}x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{16/9}{4}92​x​+3y​=92​+416/9​

    2x9+y3=29+49=69=23\frac{\sqrt{2}x}{9}+\frac{y}{3}=\frac{2}{9}+\frac{4}{9}=\frac{6}{9}=\frac{2}{3}92​x​+3y​=92​+94​=96​=32​

    Multiply by 999:

    2x+3y=6\sqrt{2}x+3y=62​x+3y=6

  3. Find the points of intersection of this chord with the ellipse

    From the line,

    3y=6−2x  ⟹  y=2−23x3y=6-\sqrt{2}x \implies y=2-\frac{\sqrt{2}}{3}x3y=6−2​x⟹y=2−32​​x

    Substitute into ellipse:

    x29+14(2−23x)2=1\frac{x^2}{9}+\frac{1}{4}\left(2-\frac{\sqrt{2}}{3}x\right)^2=19x2​+41​(2−32​​x)2=1

    Expand:

    (2−23x)2=4−423x+29x2\left(2-\frac{\sqrt{2}}{3}x\right)^2=4-\frac{4\sqrt{2}}{3}x+\frac{2}{9}x^2(2−32​​x)2=4−342​​x+92​x2

    So,

    x29+14(4−423x+29x2)=1\frac{x^2}{9}+\frac{1}{4}\left(4-\frac{4\sqrt{2}}{3}x+\frac{2}{9}x^2\right)=19x2​+41​(4−342​​x+92​x2)=1

    x29+1−23x+118x2=1\frac{x^2}{9}+1-\frac{\sqrt{2}}{3}x+\frac{1}{18}x^2=19x2​+1−32​​x+181​x2=1

    x29+x218−23x=0\frac{x^2}{9}+\frac{x^2}{18}-\frac{\sqrt{2}}{3}x=09x2​+18x2​−32​​x=0

    x26−23x=0\frac{x^2}{6}-\frac{\sqrt{2}}{3}x=06x2​−32​​x=0

    x(x6−23)=0x\left(\frac{x}{6}-\frac{\sqrt{2}}{3}\right)=0x(6x​−32​​)=0

    Hence,

    x=0orx=22x=0 \quad \text{or} \quad x=2\sqrt{2}x=0orx=22​

    Corresponding yyy values:

    • If x=0x=0x=0, then y=2y=2y=2
    • If x=22x=2\sqrt{2}x=22​, then y=2−23(22)=2−43=23y=2-\frac{\sqrt{2}}{3}(2\sqrt{2})=2-\frac{4}{3}=\frac{2}{3}y=2−32​​(22​)=2−34​=32​

    So the endpoints of the chord are:

    (0,2),(22,23)(0,2),\quad (2\sqrt{2},\tfrac{2}{3})(0,2),(22​,32​)

  4. Find the length of the chord

    L=(22−0)2+(23−2)2L=\sqrt{(2\sqrt{2}-0)^2+\left(\frac{2}{3}-2\right)^2}L=(22​−0)2+(32​−2)2​

    =8+(−43)2=\sqrt{8+\left(-\frac{4}{3}\right)^2}=8+(−34​)2​

    =8+169=\sqrt{8+\frac{16}{9}}=8+916​​

    =72+169=889=2223=\sqrt{\frac{72+16}{9}}=\sqrt{\frac{88}{9}}=\frac{2\sqrt{22}}{3}=972+16​​=988​​=3222​​

  5. Compare with given form

    Given chord length is

    2α3\frac{2\sqrt{\alpha}}{3}32α​​

    Therefore,

    2α3=2223\frac{2\sqrt{\alpha}}{3}=\frac{2\sqrt{22}}{3}32α​​=3222​​

    So,

    α=22\alpha=22α=22

  6. Option check

    • A: 262626 ❌
    • B: 181818 ❌
    • C: 222222 ✅
    • D: 202020 ❌

Thus, the correct answer is Option C.

PreviousNext

More from Ellipse

  • Let the ellipse E1​:a2x2​+b2y2​=1, a>b and E2​:A2x2​+B2y2​=1, A<B have same eccentricity 3​1​. Let the product of their lengths of latus rectums be 3​32​…2025 · MCQ
  • If αx+βy=109 is the equation of the chord of the ellipse 9x2​+4y2​=1, whose mid point is (25​,21​). then α+β is equal to :2025 · MCQ
  • Let a2x2​+b2y2​=1,a>b be an ellipse, whose eccentricity is 2​1​ and the length of the latusrectum is 14​. Then the square of the eccentricity of a2x2​−b2y2​=1…2024 · MCQ
  • Let P be a point on the ellipse 9x2​+4y2​=1. Let the line passing through P and parallel to y-axis meet the circle x2+y2=9 at point Q such that P and Q are on…2024 · MCQ
  • Let the line 2x+3y−k=0,k>0, intersect the x-axis and y-axis at the points A and B, respectively. If the equation of the circle having the line segment AB as a diameter is x2+y2−3x−2y=0…2024 · MCQ
  • Let f(x)=x2+9,g(x)=x−9x​ and a=f∘g(10),b=g∘f(3). If e and l denote the eccentricity and the length of the latus rectum of the ellipse ax2​+ by2​=1…2024 · MCQ
  • The length of the chord of the ellipse 25x2​+16y2​=1, whose mid point is (1,52​), is equal to :2024 · MCQ
  • If the length of the minor axis of an ellipse is equal to half of the distance between the foci, then the eccentricity of the ellipse is :2024 · MCQ