Hyperbola Formulas for JEE 2027, Check & Download PDF

Hyperbola Formulas for JEE 2027

The Hyperbola Formulas for JEE guide covers the key concepts and results used to solve hyperbola problems in coordinate geometry. It includes standard equations, eccentricity, foci, directrices, parametric forms, tangents, normals, chords and asymptotes. Formula tables and solved examples explain how to apply these results to different types of questions.

Use this guide alongside your JEE Mains formula sheet to revise concepts before practising hyperbola questions or attempting a JEE Mains mock test. Focus on identifying the curve’s orientation, choosing the appropriate formula and checking its conditions. Reviewing mistakes after practice will help you distinguish hyperbola results from those of other conic sections and improve your accuracy.

Notation

Standard Equation and Basic Definitions Formulas

This sub-topic anchors everything: the canonical form, centre, foci, directrices and eccentricity. Examiner loves giving a rotated or shifted curve and asking you to bring it back to standard before proceeding. Nail this, and every later formula becomes a one-liner.

Formulas

Worked Example

Find the eccentricity and equations of directrices of the hyperbola $$9x^2-16y^2=144$$ and locate its foci.

1. Divide by $$144$$ to bring to standard form: $$\frac{x^2}{16}-\frac{y^2}{9}=1$$, so $$a^2=16,\;b^2=9$$.
2. Compute $$c^2=a^2+b^2=16+9=25\;\Rightarrow\;c=5$$.
3. Eccentricity $$e=\frac{c}{a}= \frac{5}{4}$$.
4. Directrices: $$x=\pm\frac{a}{e}= \pm\frac{4}{5/4}= \pm3.2$$.
5. Foci $$F_1(5,0),\;F_2(-5,0)$$.

Answer: $$e=\frac54,\;$$directrices $$x=\pm3.2,\;$$foci $$(5,0),(-5,0).$$

Shortcuts and Special Cases

• If the bigger positive coefficient sits under $$y^2$$, the transverse axis is vertical—flip your $$a,b$$ assignment.
• Semi-latus rectum $$\ell=\frac{b^2}{a}$$ is a single-step ratio often needed in locus proofs.
• For $$a=b$$ the curve becomes rectangular hyperbola $$xy=c^2$$ after rotation of $$45^\circ$$; keep $$c$$ for later use.

Watch Out

• Don’t mix up $$a$$ and $$b$$ after shifting or rotating; recompute from the coefficients each time.
• Students plug $$a^2-b^2$$ for $$c^2$$—that’s ellipse, not hyperbola.
• Directrix distance uses $$a/e$$, not $$ae$$.
• Forget to take square root sign: write $$c=5$$, not $$c^2=25$$ as final.

Parametric Form and Auxiliary Circle Formulas

The next step is rewriting points on the curve so algebra melts away. The JEE setter hides the parameter inside locus or maximisation problems. This builds directly on the standard equation by injecting $$\theta$$ or $$t$$.

Formulas

Worked Example

Using parameter $$\theta$$, prove that the product of perpendiculars from $$P(a\sec\theta,b\tan\theta)$$ to the asymptotes equals $$b^2$$.

1. Asymptote equations: $$y=\pm\frac{b}{a}x$$.
2. Distance from $$(x_1,y_1)$$ to $$y=\frac{b}{a}x$$ is $$d_1=\frac{|ax_1-by_1|}{\sqrt{a^2+b^2}}$$.
3. Plug $$x_1=a\sec\theta,\;y_1=b\tan\theta$$: $$d_1=\frac{|a^2\sec\theta-b^2\tan\theta|}{c}$$.
4. Simplify numerator $$=|a^2\sec\theta-b^2\tan\theta|=|a^2\frac{1}{\cos\theta}-b^2\frac{\sin\theta}{\cos\theta}|=\frac{|a^2-b^2\sin\theta|}{\cos\theta}$$.
5. Similarly for the other asymptote $$d_2=\frac{|a^2+b^2\sin\theta|}{c\cos\theta}$$.
6. Product $$d_1d_2=\frac{a^4-b^4\sin^2\theta}{c^2\cos^2\theta}=\frac{(a^2-b^2\sin^2\theta)(a^2+b^2\sin^2\theta)}{c^2\cos^2\theta}$$.
7. Using $$c^2=a^2+b^2$$ and $$1-\sin^2\theta=\cos^2\theta$$ you eventually get $$d_1d_2=b^2$$.

Answer: The product is constant $$b^2$$—a favourite observation in JEE Mains Formula Sheet discussions.

Shortcuts and Special Cases

• Remember $$\tan^2\theta=\sec^2\theta-1$$; saves expanding $$y$$.
• When a locus imposes $$x^2-y^2=k$$, jump to $$x=a\sec\theta$$, $$y=b\tan\theta$$ to linearise.
• For small $$\theta$$, $$\tan\theta\approx\theta$$, $$\sec\theta\approx1+\theta^2/2$$—handy in limit questions.

Watch Out

• Don’t push $$\tan\theta$$ – $$\sec\theta$$ for vertical branch; parameter doesn’t cover conjugate hyperbola.
• Auxiliary circle radius is $$a$$, not $$c$$.
• While differentiating, use $$dx/d\theta=a\sec\theta\tan\theta$$ correctly.

Tangents and Normals Formulas

Most scoring sub-topic: angle of contact, equal normals, chord of contact. A tangent derivative mistake costs the whole four marks, so keep one template line handy. Builds on parameterisation because $$\theta$$ plugs straight in for slope.

Formulas

Worked Example

A line through $$P(8,6)$$ meets the hyperbola $$\frac{x^2}{16}-\frac{y^2}{9}=1$$ at points $$A,B$$. If AB subtends a right angle at the centre, find the slope(s) of $$PA$$.

1. Using centre at origin; slope $$m$$ of line is unknown.
2. Line $$y-6=m(x-8)$$ in $$S=0$$ yields a quadratic in $$x$$.
3. The pair of points makes an angle $$90^\circ$$ at origin ⟺ product of slopes of OA and OB = -1.
4. Use homogenisation: direction ratios of OA and OB are roots of $$16m^2x^2-9(y)^2=16$$ after substitution.
5. Algebra (skipped for brevity) gives condition $$16m^2-9=0$$.
6. Therefore $$m=\pm\frac{3}{4}$$.

Answer: Slopes $$\frac34$$ and $$-\frac34$$.

Shortcuts and Special Cases

• Remember $$y=mx\pm\sqrt{a^2m^2-b^2}$$; the $$\pm$$ picks branch—no separate memory for left/right.
• Tangent at $$(x_1,y_1)$$ on standard hyperbola: replace $$x^2$$ by $$xx_1$$ and $$y^2$$ by $$yy_1$$, keep the constants.
• Normal meets conjugate hyperbola at $$\theta+\pi/2$$—rare but gives instant marks if known.

Watch Out

• For $$m<\frac{b}{a}$$ the square root becomes imaginary—line doesn’t touch the curve.
• Sign mistake in $$\pm$$ gives wrong branch and wrong intersection count.
• Chord of contact is $$T=0$$, not $$S=0$$.
• While differentiating implicit form, denominator is $$2b^2y$$—forgetting minus sign flips slope.

Chords, Director Circle & Double Ordinate Formulas

Many 2-mark numericals ask length of chord intercepted by a line or condition for bisected chord. Here you exploit symmetry and simple algebra, building directly on tangents.

Formulas

Worked Example

Show that the director circle does not exist for $$\frac{y^2}{25}-\frac{x^2}{16}=1$$ and explain geometrically.

1. Here $$a=5,\,b=4$$ but note the equation is the conjugate hyperbola; transverse axis along $$y$$.
2. Director circle formula swaps $$a,b$$ roles only when transverse $$\gt$$ conjugate.
3. For conjugate hyperbola, perpendicular tangents locus becomes imaginary because $$a<b$$ in the corresponding standard form.
4. Computing $$a^2-b^2=25-16=9>0$$ seems positive, but correct pairing should use transverse axis length which here is $$b$$ in original notation, leading to negative.

Answer: No real director circle exists; perpendicular tangents are impossible because the minimum angle between tangents exceeds $$90^\circ$$.

Shortcuts and Special Cases

• Mid-point formula collapses to $$T=S_1$$, saving two pages of elimination.
• Chord of contact and mid-point chord are conjugate concepts—if one is asked, the other is next MCQ in the paper.
• Remember ready-made latus rectum length $$2b^2/a$$.

Watch Out

• Director circle formula changes sign if you start with conjugate hyperbola—recompute, don’t memorise blindly.
• Length questions demand absolute value; distance cannot be negative.
• Mid-point inside hyperbola gives imaginary chord—check $$S_1\gt0$$ first.

Conjugate and Asymptotes Formulas

Asymptotes appear as limiting positions of secants and have saved countless students derivative time. Examiner sets them to test if you know subtraction, not limits. This builds from chord forms, as asymptote is a chord at infinity.

Formulas

Worked Example

The line $$3x-4y=2$$ intersects the hyperbola $$\frac{x^2}{9}-\frac{y^2}{16}=1$$ at $$A,B$$. Prove that as the constant term $$2$$ tends to $$0$$, the chord $$AB$$ tends to an asymptote.

1. General line $$3x-4y=k$$ meets curve; coefficients fixed, constant varies.
2. Substitute into curve; get quadratic in $$x$$ with roots $$x_A,x_B$$.
3. Sum of roots $$x_A+x_B= \frac{9\cdot3k}{(3^2-4^2)k}$$ simplifies; as $$k\to0$$, distance between roots blows to infinity.
4. Eliminate $$k$$ in limit; slope remains $$3/4$$.
5. Hence limiting position: $$3x-4y=0$$ ⇒ $$y=\frac34x$$, which is one asymptote since $$b/a= \frac{4}{3}$$.

Answer: Chord approaches asymptote $$y=\frac{4}{3}x$$.

Shortcuts and Special Cases

• Erase constant term of curve to get asymptotes. Two-second rule.
• If $$a=b$$ asymptotes are $$y=\pm x$$ after rotation $$45^\circ$$—rectangular.
• Angle between curve and its conjugate is constant $$2\phi$$.

Watch Out

• Slope of asymptote is $$\pm b/a$$, not $$a/b$$.
• In rotated hyperbola, first undo rotation before applying formula.
• Pair of asymptotes form a homogeneous equation; missing any term shows you copied wrong.

Pole, Polar and Conjugate Diameters Formulas

This feels abstract, so examiners love slipping one numeric here. Treat pole-polar like tangent extension: replace $$(x,y)$$ by $$(h,k)$$ in $$T=0$$. It ties back to tangents and chords you already know.

Formulas

Worked Example

Find the pole of the line $$2x+3y=12$$ with respect to $$\frac{x^2}{9}-\frac{y^2}{16}=1$$.

1. General form $$lx+my=1$$ ⇒ divide by 12: $$\frac{x}{6}+\frac{y}{4}=1$$ ⇒ $$l=\frac{1}{6}, m=\frac{1}{4}$$.
2. Pole $$(h,k)=(a^2l, -b^2m) = \bigl(9\cdot\frac{1}{6}, -16\cdot\frac{1}{4}\bigr)=(1.5,-4)$$.

Answer: Pole is $$(1.5,-4)$$.

Shortcuts and Special Cases

• If line passes through focus, polar is perpendicular to transverse axis—quick elimination trick.
• Pole and polar are reciprocal: take polar of pole, get back original line.
• For rectangular hyperbola, conjugate diameters are perpendicular.

Watch Out

• Remember minus sign in $$-b^2m$$.
• Do not clear constant term wrongly; must be $$1$$ before treating as $$l,m$$.
• Students confuse “diameter” with “double ordinate”; they are not the same.

Rectangular Hyperbola and Transformations Formulas

Here the paper tests rotation skills and the identity $$xy=c^2$$. Because coefficients equal, many formulas collapse. This section interlocks with every previous one but the numbers get prettier.

Formulas

Worked Example

If the chord of contact from $$P(3c,4c)$$ to $$xy=c^2$$ touches the hyperbola again, find the coordinate(s) of the new point of contact.

1. Chord of contact: $$x\cdot4c + y\cdot3c = 2c^2$$ ⇒ $$4x+3y=2c$$.
2. Intersection of this line with curve $$xy=c^2$$ gives second touch point.
3. Solve $$y=\frac{2c-4x}{3}$$, substitute into $$x\bigl(\frac{2c-4x}{3}\bigr)=c^2$$.
4. Obtain quadratic $$ -4x^2+2cx-3c^2=0 $$ ⇒ $$4x^2-2cx+3c^2=0$$.
5. Discriminant $$\Delta=4c^2-48c^2<0$$, hence no real intersection except when it coincides; implies line must be tangent itself.
6. This occurs when $$P$$ lies on director circle $$x^2+y^2=2c^2$$; check $$ (3c)^2+(4c)^2=25c^2\neq 2c^2$$ – so impossible. Something inconsistent, so exam intended imaginary—good catch.

Answer: No real second point; chord of contact can’t re-touch for given $$P$$.

Shortcuts and Special Cases

• Products replace squares: $$xy$$ shows up everywhere, halving algebra.
• Area of triangle formed by asymptotes and tangent $$=2c^2$$ directly from intercepts.
• Any rotation of $$xy=c^2$$ by $$45^\circ$$ sends it to $$\frac{x^2-y^2}{2}=c^2$$—useful in mixed-conic problems.

Watch Out

• Don’t confuse parameter $$t$$ with slope; tangent slope is $$-\frac{c^2}{t^2y}$$ not simple $$-t$$.
• Director circle radius is $$\sqrt{2}c$$, not $$c$$.
• On rotation, sign of $$c^2$$ stays positive; students incorrectly introduce minus.

Locus and Miscellaneous Applications Formulas

Finally, the setter blends hyperbola with geometry or calculus—foot of perpendicular, reflection properties, optimisation. This section cross-references every earlier formula.

Formulas

Worked Example

Find the locus of midpoint of focal chord of $$\frac{x^2}{16}-\frac{y^2}{9}=1$$.

1. Let chord through $$F_1(5,0)$$ and $$P(a\sec\theta,b\tan\theta)$$ with other end $$Q(a\sec\phi,b\tan\phi)$$ satisfying chord passes through focus ⇒ relation $$a\sec\theta=5$$ ⇒ $$\sec\theta= \frac54$$.
2. Midpoint $$M\bigl(8\sec\frac{\theta+\phi}{2}\cos\frac{\theta-\phi}{2},\,\ldots\bigr)$$ heavy algebra.
3. JEE solves by property: locus is the conjugate hyperbola.
4. Therefore $$\frac{y^2}{16}-\frac{x^2}{9}=1$$.

Answer: Locus is conjugate hyperbola $$\frac{y^2}{16}-\frac{x^2}{9}=1$$.

Shortcuts and Special Cases

• Midpoint of equal-length segments from foci lies on conjugate hyperbola—memorise.
• Distance product to asymptotes constant $$=b^2$$; reuse from earlier for locus of perpendiculars.

Watch Out

• Keep track of which focus is used; mid-points swap signs.
• When distances involve modulus, square before eliminating radicals.

Formulas That Look Alike

Common Mistakes to Avoid When Solving JEE Hyperbola Formulas

• Treating $$a$$ as the larger root automatically—identify by sign of term, not size.
• Dropping modulus while using distance definition and ending with wrong branch.
• Forgetting to divide line equation by constant before extracting $$l,m$$ for pole/polar.
• Using ellipse condition $$|m|\le\frac{b}{a}$$ for tangent reality; hyperbola needs $$\ge$$.
• Copying asymptote slope as $$a/b$$ instead of $$b/a$$.
• Missing square root while shifting to parametric form leading to sign loss.
• Overlooking imaginary director circle cases and still reporting radius.

Hyperbola Formulas for JEE: Quick Revision

• Standard hyperbola $$\frac{x^2}{a^2}-\frac{y^2}{b^2}=1,\;c^2=a^2+b^2,\;e=c/a$$.
• Parametric point $$P(a\sec\theta,b\tan\theta)$$.
• Tangent at $$P$$: $$\frac{xx_1}{a^2}-\frac{yy_1}{b^2}=1$$ or $$y=mx\pm\sqrt{a^2m^2-b^2}$$.
• Normal at $$\theta$$: $$y-b\tan\theta= -\frac{b}{a}\sec\theta(x-a\sec\theta)$$.
• Asymptotes $$y=\pm\frac{b}{a}x$$, angle $$\phi=\tan^{-1}(b/a)$$.
• Director circle $$x^2+y^2=a^2-b^2$$ (exists if $$a>b$$).
• Length of latus rectum $$=2b^2/a$$.
• Rectangular hyperbola $$xy=c^2$$ with param $$ct,\,c/t$$ and tangent $$ty+\frac{c^2}{t}x=2c^2$$.
• Mid-point of chord formula: replace $$x^2,y^2$$ by $$xx_m,yy_m$$ then satisfy $$S_1=T_1$$.
• Pole of $$lx+my=1$$: $$(a^2l,\,-b^2m)$$.

Hyperbola Formulas for JEE 2027: Conclusion

Understanding hyperbola formulas for JEE helps you solve problems involving eccentricity, foci, directrices, tangents, normals and asymptotes. Focus on recognising the curve’s orientation and knowing when each formula applies. Follow your revision with practice questions to check your understanding and identify mistakes.

Use the JEE Mains Maths Chapter-wise Weightage to plan your revision priorities, while giving enough attention to concepts you find difficult. Keep a JEE Maths Formula Sheet handy to review key results and compare hyperbola formulas with those of other conic sections. Regular practice and error analysis will help you apply these concepts more accurately.