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

Hyperbola question

2025 · 7 Apr · Shift 1 · Q50
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. /Hyperbola
  5. /2025 · 7 Apr · Shift 1 · Q50

Hyperbola question

2025 · 7 Apr · Shift 1 · Q50

JEE MainMathematicsHyperbolaNumerical+4 / −1
Consider the hyperbola x2a2−y2b2=1\frac{x^2}{a^2}-\frac{y^2}{b^2}=1a2x2​−b2y2​=1 having one of its focus at P(−3,0)\mathrm{P}(-3,0)P(−3,0). If the latus ractum through its other focus subtends a right angle at P and a2b2=α2−β,α,β∈Na^2 b^2=\alpha \sqrt{2}-\beta, \alpha, \beta \in \mathbb{N}a2b2=α2​−β,α,β∈N, then α+β\alpha+\betaα+β is ‾\underline{\hspace{2cm}}​ .
Numerical answer
View written solutionFree

Correct answer: 1944

  1. Write the hyperbola and identify its foci

The hyperbola is

x2a2−y2b2=1.\frac{x^2}{a^2}-\frac{y^2}{b^2}=1.a2x2​−b2y2​=1.

For this standard hyperbola, the foci are at

(±c,0),c2=a2+b2.(\pm c,0), \qquad c^2=a^2+b^2.(±c,0),c2=a2+b2.

Given that one focus is P(−3,0)P(-3,0)P(−3,0), we get

c=3.c=3.c=3.

Hence the other focus is

(3,0).(3,0).(3,0).

So,

a2+b2=9.(1)a^2+b^2=9. \tag{1}a2+b2=9.(1)
  1. Find the endpoints of the latus rectum through the other focus

For the hyperbola

x2a2−y2b2=1,\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,a2x2​−b2y2​=1,

the latus rectum through the focus (c,0)(c,0)(c,0) is the vertical line

x=c.x=c.x=c.

Its endpoints on the hyperbola are obtained by substituting x=cx=cx=c:

c2a2−y2b2=1.\frac{c^2}{a^2}-\frac{y^2}{b^2}=1.a2c2​−b2y2​=1.

Using c2=a2+b2c^2=a^2+b^2c2=a2+b2,

a2+b2a2−y2b2=1  ⟹  1+b2a2−y2b2=1  ⟹  y2b2=b2a2.\frac{a^2+b^2}{a^2}-\frac{y^2}{b^2}=1 \implies 1+\frac{b^2}{a^2}-\frac{y^2}{b^2}=1 \implies \frac{y^2}{b^2}=\frac{b^2}{a^2}.a2a2+b2​−b2y2​=1⟹1+a2b2​−b2y2​=1⟹b2y2​=a2b2​.

Thus,

y=±b2a.y=\pm \frac{b^2}{a}.y=±ab2​.

So the endpoints are

Q(c,b2a),R(c,−b2a).Q\left(c,\frac{b^2}{a}\right), \qquad R\left(c,-\frac{b^2}{a}\right).Q(c,ab2​),R(c,−ab2​).

Since c=3c=3c=3,

Q(3,b2a),R(3,−b2a).Q\left(3,\frac{b^2}{a}\right), \qquad R\left(3,-\frac{b^2}{a}\right).Q(3,ab2​),R(3,−ab2​).
  1. Use the condition that this latus rectum subtends a right angle at P(−3,0)P(-3,0)P(−3,0)

We are given that the latus rectum through (3,0)(3,0)(3,0) subtends a right angle at P(−3,0)P(-3,0)P(−3,0). That means

∠QPR=90∘.\angle QPR=90^\circ.∠QPR=90∘.

So vectors PQ→\overrightarrow{PQ}PQ​ and PR→\overrightarrow{PR}PR are perpendicular.

Now,

PQ→=(3−(−3),b2a−0)=(6,b2a),\overrightarrow{PQ}=\left(3-(-3),\frac{b^2}{a}-0\right)=\left(6,\frac{b^2}{a}\right),PQ​=(3−(−3),ab2​−0)=(6,ab2​), PR→=(3−(−3),−b2a−0)=(6,−b2a).\overrightarrow{PR}=\left(3-(-3),-\frac{b^2}{a}-0\right)=\left(6,-\frac{b^2}{a}\right).PR=(3−(−3),−ab2​−0)=(6,−ab2​).

For perpendicularity,

PQ→⋅PR→=0.\overrightarrow{PQ}\cdot\overrightarrow{PR}=0.PQ​⋅PR=0.

Hence,

6⋅6+b2a(−b2a)=06\cdot 6 + \frac{b^2}{a}\left(-\frac{b^2}{a}\right)=06⋅6+ab2​(−ab2​)=0 36−b4a2=036-\frac{b^4}{a^2}=036−a2b4​=0 b4a2=36\frac{b^4}{a^2}=36a2b4​=36 b4=36a2.b^4=36a^2.b4=36a2.

Taking positive square roots (since a,b>0a,b>0a,b>0),

b2=6a.(2)b^2=6a. \tag{2}b2=6a.(2)
  1. Solve for aaa and b2b^2b2 using a2+b2=9a^2+b^2=9a2+b2=9

Substitute b2=6ab^2=6ab2=6a into (1):

a2+6a=9a^2+6a=9a2+6a=9 a2+6a−9=0.a^2+6a-9=0.a2+6a−9=0.

Solving,

a=−6±36+362=−6±722=−6±622=−3±32.a=\frac{-6\pm\sqrt{36+36}}{2}=\frac{-6\pm\sqrt{72}}{2}=\frac{-6\pm 6\sqrt2}{2}=-3\pm 3\sqrt2.a=2−6±36+36​​=2−6±72​​=2−6±62​​=−3±32​.

Since a>0a>0a>0,

a=−3+32=3(2−1).a=-3+3\sqrt2=3(\sqrt2-1).a=−3+32​=3(2​−1).

Then from (2),

b2=6a=18(2−1).b^2=6a=18(\sqrt2-1).b2=6a=18(2​−1).
  1. Compute a2b2a^2b^2a2b2

First,

a2=[3(2−1)]2=9(2+1−22)=9(3−22)=27−182.a^2=[3(\sqrt2-1)]^2=9(2+1-2\sqrt2)=9(3-2\sqrt2)=27-18\sqrt2.a2=[3(2​−1)]2=9(2+1−22​)=9(3−22​)=27−182​.

Now,

a2b2=(27−182)⋅18(2−1).a^2b^2=(27-18\sqrt2)\cdot 18(\sqrt2-1).a2b2=(27−182​)⋅18(2​−1).

Factor 18:

a2b2=18(27−182)(2−1).a^2b^2=18(27-18\sqrt2)(\sqrt2-1).a2b2=18(27−182​)(2​−1).

Expand the inner product:

(27−182)(2−1)=272−27−36+182=452−63.(27-18\sqrt2)(\sqrt2-1)=27\sqrt2-27-36+18\sqrt2=45\sqrt2-63.(27−182​)(2​−1)=272​−27−36+182​=452​−63.

Therefore,

a2b2=18(452−63)=8102−1134.a^2b^2=18(45\sqrt2-63)=810\sqrt2-1134.a2b2=18(452​−63)=8102​−1134.

So,

α=810,β=1134.\alpha=810, \qquad \beta=1134.α=810,β=1134.

Hence,

α+β=810+1134=1944.\alpha+\beta=810+1134=1944.α+β=810+1134=1944.
  1. Compare with stored answer

Derived answer:

194419441944

Stored correct answer:

194419441944

They match.

PreviousNext

More from Hyperbola

  • Let e1 and e2 be the eccentricities of the ellipse b2x2​+25y2​=1 and the hyperbola 16x2​−b2y2​=1, respectively. If b < 5 and e1e2 = 1, then the eccentricity of the ellipse having its…2025 · MCQ
  • Let the lengths of the transverse and conjugate axes of a hyperbola in standard form be 2a and 2b, respectively, and one focus and the corresponding directrix of this hyperbola be (−5,0) and 5x+9=0, respectively. If the product…2025 · Numerical
  • Let the foci of a hyperbola be (1,14) and (1,−12). If it passes through the point (1,6), then the length of its latus-rectum is :2025 · MCQ
  • Let H1​:a2x2​− b2y2​=1 and H2​:− A2x2​+ B2y2​=1 be two hyperbolas having length of latus rectums 152​ and 125​…2025 · Numerical
  • For 0<θ<π/2, if the eccentricity of the hyperbola x2−y2cosec2θ=5 is 7​ times eccentricity of the ellipse x2cosec2θ+y2=5, then the value of θ is :2024 · MCQ
  • Consider a hyperbola H having centre at the origin and foci on the x-axis. Let C1​ be the circle touching the hyperbola H and having the centre at the origin. Let C2​ be the circle…2024 · MCQ
  • The length of the latus rectum and directrices of hyperbola with eccentricity e are 9 and x=±3​4​, respectively. Let the line y−3​x+3​=0 touch this hyperbola at (x0​,y0​). If m…2024 · Numerical
  • Let H:a2−x2​+b2y2​=1 be the hyperbola, whose eccentricity is 3​ and the length of the latus rectum is 43​. Suppose the point (α,6),α>0 lies on H. If β is the product of the…2024 · MCQ