Hyperbola JEE Mathematics Worksheets Set 01

Question. Find the equation to the hyperbola whose directrix is 2x + y = 1, focus (1, 1) & eccentricity \( \sqrt{3} \). Find also the length of its latus rectum.
Answer: Directrix 2x + y - 1 = 0
Let P(h, k)
S(1, 1) e = \( \sqrt{3} \)
SP = ePM
SP² = e² PM²
\( (h - 1)^2 + (k - 1)^2 = 3 \left| \frac{2h + k - 1}{\sqrt{5}} \right|^2 \)
\( \implies \) 7x² + 12xy - 2y² - 2x + 4y - 7 = 0
L.R. = 2e × (Distance between focus and directrix)
\( = 2 \sqrt{3} \times \left| \frac{2 + 1 - 1}{\sqrt{5}} \right| = \frac{\sqrt{48}}{\sqrt{5}} = \sqrt{\frac{48}{5}} \)

Question. The hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) passes through the point of intersection of the lines 7x + 13y - 87 = 0 and 5x - 8y + 7 = 0 & the latus rectum is \( 32\sqrt{2}/5 \). Find 'a' & 'b'.
Answer: \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) ... (1)
7x + 13y - 87 = 0
5x - 8y + 7 = 0
POI (5, 4)
Equation (1) pass through (5, 4)
\( \frac{25}{a^2} - \frac{16}{b^2} = 1 \) ... (2)
and \( \frac{2b^2}{a} = \frac{32\sqrt{3}}{5} \) ... (3)
From (2) & (3)
\( a^2 = \frac{25}{2} \), \( b^2 = 16 \)

Question. For the hyperbola \( \frac{x^2}{100} - \frac{y^2}{25} = 1 \), prove that
(i) eccentricity = \( \sqrt{5}/2 \)
(ii) SA. S'A = 25, where S & S' are the foci & A is the vertex.
Answer: \( \frac{x^2}{100} - \frac{y^2}{25} = 1 \)
(i) \( e^2 = 1 + \frac{b^2}{a^2} = 1 + \frac{25}{100} = \frac{125}{100} \)
\( e = \sqrt{\frac{125}{100}} = \frac{\sqrt{5}}{2} \)
(ii) SA. S'A = (ae - a) . (ae + a)
\( = a^2 (e^2 - 1) = 100 \left( \frac{5}{4} - 1 \right) = 25 \)

Question. Find the centre, the foci, the directrices, the length of the latus rectum, the length & the equations of the axes of the hyperbola 16x² - 9y² + 32x + 36y - 164 = 0.
Answer: 16x² + 32x - 9y² + 36y - 164 = 0
16 (x² + 2x + 1) - 16 - 9(y² - 4y + 4) + 36 - 164 = 0
16 (x + 1)² - 9 (y - 2)² = 144
\( \frac{(x + 1)^2}{9} - \frac{(y - 2)^2}{16} = 1 \)
Centre (-1, 2) ; \( e^2 = 1 + \frac{16}{9} \)
\( \implies \) \( e = \frac{5}{3} \)
Foci ; x + 1 = \( \pm \) 5 and y - 2 = 0
foci (4, 2) & (-6, 2)
Directrix x + 1 = \( \pm \frac{a}{e} \)
x + 1 = \( \pm \frac{9}{5} \)
\( \implies \) 5x - 4 = 0 and 5x + 14 = 0
L.R. = \( \frac{2b^2}{a} = \frac{32}{3} \)
\( \ell \)(TA) = 2a = 6
\( \ell \)(CA) = 2b = 8
Axis x + 1 = 0 and y - 2 = 0

Question. Find the equation of the tangent to the hyperbola x² - 4y² = 36 which is perpendicular to the line x - y + 4 = 0.
Answer: \( \frac{x^2}{36} - \frac{y^2}{9} = 1 \)
Tangent : y = mx \( \pm \sqrt{a^2m^2 - b^2} \)
m = - 1
a² = 36, b² = 9
y = - x \( \pm \sqrt{27} \)
y = -x \( \pm 3\sqrt{3} \)

Question. Tangents are drawn to the hyperbola 3x² - 2y² = 25 from the point (0, 5/2). Find their equations.
Answer: \( \frac{3x^2}{25} - \frac{2y^2}{25} = 1 \)
y = mx \( \pm \sqrt{a^2m^2 - b^2} \)
(0, 5/2)
\( \frac{5}{2} = \pm \sqrt{a^2m^2 - b^2} \)
\( \frac{25}{4} = a^2m^2 - b^2 \)
\( \frac{25}{4} = \frac{25}{3} m^2 - \frac{25}{2} \)
\( \implies \) \( m = \pm \frac{3}{2} \)
Equation of tangents
3x + 2y - 5 = 0 and 3x - 2y + 5 = 0

Question. If \( \theta_1 \) & \( \theta_2 \) are the parameters of the extremities of a chord through (ae, 0) of a hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \), then show that \( \tan \frac{\theta_1}{2} . \tan \frac{\theta_2}{2} + \frac{e - 1}{e + 1} = 0 \).
Answer: Equation of chord connecting the points (a sec \( \theta_1 \), b tan \( \theta_1 \)) and (a sec \( \theta_2 \), b tan \( \theta_2 \)) is
\( \frac{x}{a} \cos \left( \frac{\theta_1 - \theta_2}{2} \right) - \frac{y}{b} \sin \left( \frac{\theta_1 + \theta_2}{2} \right) = \cos \left( \frac{\theta_1 + \theta_2}{2} \right) \) ... (1)
If it passes through (ae, 0)
We have e . \( \cos \left( \frac{\theta_1 - \theta_2}{2} \right) = \cos \left( \frac{\theta_1 + \theta_2}{2} \right) \)
\( \implies \) \( e = \frac{\cos \left( \frac{\theta_1 + \theta_2}{2} \right)}{\cos \left( \frac{\theta_1 - \theta_2}{2} \right)} = \frac{1 - \tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2}}{1 + \tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2}} \)
\( \implies \) \( \tan \frac{\theta_1}{2} \tan \frac{\theta_2}{2} = \frac{1 - e}{1 + e} \)

Question. Tangents are drawn from the point (\( \alpha \), \( \beta \)) to the hyperbola 3x² - 2y² = 6 and are inclined at angles \( \theta \) and \( \phi \) to the x-axis. If \( \tan \theta . \tan \phi = 2 \), prove that \( \beta^2 = 2\alpha^2 - 7 \).
Answer: 3x² - 2y² = 6
\( \frac{x^2}{2} - \frac{y^2}{3} = 1 \)
y = mx \( \pm \sqrt{2m^2 - 3} \)
(\( \alpha \), \( \beta \))
(\( \beta \) - m\( \alpha \))² = 2m² - 3
\( \beta^2 + m^2\alpha^2 - 2\alpha\beta m - 2m^2 + 3 = 0 \)
m² (\( \alpha^2 \) - 2) - 2\( \alpha\beta \)m + 3 + \( \beta^2 \) = 0
\( \implies \) \( m_1, m_2 \)
m1 . m2 = \( \frac{3 + \beta^2}{\alpha^2 - 2} \)
2 = \( \frac{3 + \beta^2}{\alpha^2 - 2} \)
2\( \alpha^2 \) - 4 = 3 + \( \beta^2 \)
\( \beta^2 = 2\alpha^2 - 7 \)

Question. If two points P & Q on the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) whose centre is C be such that CP is perpendicular to CQ & a < b, then prove that \( \frac{1}{CP^2} + \frac{1}{CQ^2} = \frac{1}{a^2} - \frac{1}{b^2} \).
Answer: Let CP = r1 and angle to x-axis will be \( \theta \)
P = (r1 cos\( \theta \), r1 sin \( \theta \)) and P lies on that hyperbola
r1² \( \left( \frac{\cos^2 \theta}{a^2} - \frac{\sin^2 \theta}{b^2} \right) = 1 \)
Replace \( \theta \) by 90° + \( \theta \)
r2² \( \left( \frac{\sin^2 \theta}{a^2} - \frac{\cos^2 \theta}{b^2} \right) = 1 \)
\( \frac{1}{r_1^2} + \frac{1}{r_2^2} = \cos^2 \theta \left( \frac{1}{a^2} - \frac{1}{b^2} \right) + \sin^2 \theta \left( \frac{1}{a^2} - \frac{1}{b^2} \right) \)
\( \frac{1}{CP^2} + \frac{1}{CQ^2} = \frac{1}{a^2} - \frac{1}{b^2} \)

Question. The tangents & normal at a point on \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) cut the y-axis at A & B. Prove that the circle on AB as diameter passes through the foci of the hyperbola.
Answer: Let a point P(a sec\( \theta \), b tan\( \theta \))
tangent : \( \frac{x}{a} \sec \theta - \frac{y}{b} \tan \theta = 1 \)
A \( \left( 0, \frac{-b}{\tan \theta} \right) \)
Normal : ax cos\( \theta \) + by cot \( \theta \) = a² e²
B \( \left( 0, \frac{a^2 e^2}{b \cot \theta} \right) \)
Equation of circle AB as diameter
x² + \( \left( y + \frac{b}{\tan \theta} \right) \left( y - \frac{a^2 e^2}{b \cot \theta} \right) = 0 \)
It is passes through (ae, 0) which is focus of hyperbola.

Question. The perpendicular from the centre upon the normal on any point of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) meets at R. Find the locus of R.
Answer: Any normal at the point P(\( \theta \)) is
ax cos \( \theta \) + by cot\( \theta \) = a² + b² ... (1)
Any line through centre (0, 0) and perpendicular to Equation (1) is
bx cot \( \theta \) - ay cos \( \theta \) = 0
\( \therefore \sin \theta = \frac{bx}{ay} \)
\( \therefore \cos \theta = \frac{\sqrt{a^2y^2 - b^2x^2}}{ay} \)
\( \cot \theta = \frac{\sqrt{a^2y^2 - b^2x^2}}{bx} \)
Substituting the values of cos\( \theta \) and cot\( \theta \) is ... (4) we get
\( \sqrt{a^2y^2 - b^2x^2} \left[ ax . \frac{1}{ay} + by \frac{1}{bx} \right] = a^2 + b^2 \)
(a²y² - b²x²) (x² + y²)² = (a² + b²) x²y²

Question. If the normal at a point P to the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) meets the x-axis at G, show that SG = e.SP, S being the focus of the hyperbola.
Answer: Let the point P (a sec\( \theta \), b tan\( \theta \))
Normal at point P
ax cos \( \theta \) + by cot\( \theta \) = a²e²
at x-axis y = 0
G \( \left( \frac{ae^2}{\cos \theta}, 0 \right) \) or G (ae² sec\( \theta \), 0)
We have to prove that SG = eSP
SG = |ae - ae² sec\( \theta \)|
We also know that
SP = ePM
m : directrix
x = \( \frac{a}{e} \)
eSP = e |a sec\( \theta \) - a/e|
= |ae - ae² sec\( \theta \)|
so SG = eSP

Question. Show that the locus of the middle points of normal chords of the rectangular hyperbola x² - y² = a² is (y² - x²)³ = 4 a²x²y².
Answer: If (h, k) be the mid-point of the chord of the Hyperbola x² - y² = a² than its equation by T = S1 is
hx - ky = h² - k² ... (1)
But Equation (1) is Normal to the Hyp. its equation is
x cos \( \theta \) + y cot \( \theta \) = 2a ... (2)
(1) and (2) are same
\( \frac{h}{\cos \theta} = \frac{-k}{\cot \theta} = \frac{h^2 - k^2}{2a} \)
\( \therefore \sec \theta = \frac{h^2 - k^2}{2ah} \) and \( \tan \theta = \frac{h^2 - k^2}{-2ak} \)
sec²\( \theta \) - tan²\( \theta \) = 1
\( \therefore \frac{(h^2 - k^2)^2}{4a^2} \cdot \left( \frac{1}{h^2} - \frac{1}{k^2} \right) = 1 \)
Locus
(y² - x²)³ = 4a²x²y²

Question. If a chord joining the points P (a sec\( \theta \), a tan\( \theta \)) & Q (a sec\( \phi \), a tan\( \phi \)) on the hyperbola x² - y² = a² is a normal to it at P, then show that \( \tan \phi = \tan \theta (4 \sec^2 \theta - 1) \).
Answer: Chord joining two points P(\( \theta \)) and Q(\( \phi \))
x cos \( \left( \frac{\theta - \phi}{2} \right) \) - y sin \( \left( \frac{\theta + \phi}{2} \right) \) = a cos \( \left( \frac{\theta + \phi}{2} \right) \)
Slope = \( \frac{\cos \left( \frac{\theta - \phi}{2} \right)}{\sin \left( \frac{\theta + \phi}{2} \right)} \)
Normal at P(\( \theta \))
So slope of normal = - sin \( \theta \)
- sin \( \theta \) = \( \frac{\cos \left( \frac{\theta - \phi}{2} \right)}{\sin \left( \frac{\theta + \phi}{2} \right)} \)
After solving
\( \tan \phi = \tan \theta (4 \sec^2 \theta - 1) \)

Question. Chords of the hyperbola \( \frac{x^2}{a^2} - \frac{y^2}{b^2} = 1 \) are tangents to the circle drawn on the line joining the foci as diameter. Find the locus of the point of intersection of tangents at the extremities of the chords.
Answer: Focii S(ae, 0) S'(-ae, 0)
circle with diameter SS'
(x - ae) (x + ae) + y² = 0
x² + y² = a²e² ... (1)
If (h, k) be the pole of the chord which touches Equation (1) then its equation is the polar of (h,k) w.r.t. Hyp.
\( \frac{hx}{a^2} - \frac{ky}{b^2} = 1 \) ... (2)
Line (2) Touches the circle (1)
\( \frac{1}{\sqrt{\frac{h^2}{a^4} + \frac{k^2}{b^4}}} = ae \)
\( \implies \) \( \frac{h^2}{a^4} + \frac{k^2}{b^4} = \frac{1}{a^2e^2} = \frac{1}{a^2 + b^2} \)
Locus \( \frac{x^2}{a^4} + \frac{y^2}{b^4} = \frac{1}{a^2 + b^2} \)

Question. An ellipse and a hyperbola have their principal axes along the coordinate axes and have a common foci separated by a distance \( 2\sqrt{13} \), the difference of their focal semi axes is equal to 4. If the ratio of their eccentricities is 3/7. Find the equation of these curves.
Answer: Given 2ae = \( 2\sqrt{13} \)
\( \implies \) \( a_1e_1 = \sqrt{13} \) & \( a_2e_2 = \sqrt{13} \)
a2 - a1 = 4 \quad a1
\( \implies \) Hyper.
& \( \frac{e_2}{e_1} = \frac{3}{7} \) \quad a2
\( \implies \) ellipse
\( \frac{a_1}{a_2} \cdot \frac{e_1}{e_2} = 1 \)
\( \implies \) \( \frac{a_1}{a_2} = \frac{3}{7} \)
a1 = 3 & a2 = 7
\( e_1 = \frac{\sqrt{13}}{3} \) & \( e_2 = \frac{\sqrt{13}}{7} \)
b1² = 4 & b2² = 36
ellipse : \( \frac{x^2}{49} + \frac{y^2}{36} = 1 \)
Hyper : \( \frac{x^2}{9} - \frac{y^2}{4} = 1 \)