The chord of contact of the tangents through P to the hyperbola x 2 a 2 - y 2 b 2 = 1 subtends a right angle at the centre. Prove that the locus of P is the ellipse b 4 x 2 + a 4 y 2 = a 2 b 2 b 2 - a 2 .

Let, P h , k . The equation of chord of contact of tangents drawn from the point P h , k to the hyperbola x 2 a 2 - y 2 b 2 = 1 is h x a 2 - k y b 2 - 1 = 0 . Solve h x a 2 - k y b 2 - 1 = 0 & x 2 a 2 - y 2 b 2 = 1 = x 2 a 2 - 1 b 2 h b 2 x k a 2 - b 2 k 2 = 1 = 1 a 2 - b 2 k 2 × h 2 a 4 x 2 + 2 h a 2 × b 2 k 2 x - b 2 k 2 - 1 = 0 Let, x 1 & x 2 are the roots of the above equation. x 1 + x 2 = - 2 h a 2 × b 2 k 2 1 a 2 - b 2 k 2 × h 2 a 4 & x 1 x 2 = - b 2 k 2 - 1 1 a 2 - b 2 k 2 × h 2 a 4 Let, A x 1 , b 2 k h x 1 a 2 - 1 & B x 2 , b 2 k h x 2 a 2 - 1 is the point of intersection of hyperbola of chord of contact. If A B subtends a right-angle at the centre, then b 2 k h x 1 a 2 - 1 - 0 x 1 - 0 b 2 k h x 2 a 2 - 1 - 0 x 2 - 0 = - 1 ⇒ b 4 k 2 h x 1 a 2 - 1 h x 2 a 2 - 1 + x 1 x 2 = 0 ⇒ b 4 k 2 h 2 x 1 x 2 a 4 - h a 2 x 1 + x 2 + 1 + x 1 x 2 = 0 ⇒ b 4 k 2 h 2 a 4 × - b 2 k 2 - 1 1 a 2 - b 2 k 2 × h 2 a 4 - h a 2 - 2 h a 2 × b 2 k 2 1 a 2 - b 2 k 2 × h 2 a 4 + 1 + - b 2 k 2 - 1 1 a 2 - b 2 k 2 × h 2 a 4 = 0 ⇒ b 4 k 2 - h 2 a 4 × b 2 k 2 - h 2 a 4 + 2 h 2 a 4 × b 2 k 2 + 1 a 2 - b 2 k 2 × h 2 a 4 - b 2 k 2 - 1 = 0 ⇒ - b 4 k 2 × h 2 a 4 + 1 a 2 × b 4 k 2 - b 2 k 2 - 1 = 0 ⇒ b 4 h 2 + a 4 k 2 = a 2 b 4 - a 4 b 2 Replace h by x and k by y . ⇒ b 4 x 2 + a 4 y 2 = a 2 b 2 b 2 - a 2

If C is the centre of a hyperbola x 2 a 2 - y 2 b 2 = 1 , S , S ' its foci and P a point on it. Prove that S P · S ' P = C P 2 - a 2 + b 2 .

Let, the point P on the hyperbola x 2 a 2 - y 2 b 2 = 1 is P a sec θ , b tan θ . We know that the focal distance of a point x 1 , y 1 on the hyperbola are e x 1 ± a . Hence, we have S P · S ' P = a e sec θ - a × a e secθ + a ⇒ S P · S ' P = a 2 e 2 sec 2 θ - a 2 ⇒ S P · S ' P = a 2 1 + b 2 a 2 sec 2 θ - a 2 ⇒ S P · S ' P = a 2 + b 2 sec 2 θ - a 2 ⇒ S P · S ' P = a 2 sec 2 θ + b 2 sec 2 θ - a 2 Using 1 + tan 2 θ = sec 2 θ ⇒ S P · S ' P = a 2 sec 2 θ + b 2 1 + tan 2 θ - a 2 ⇒ S P · S ' P = a 2 sec 2 θ + b 2 tan 2 θ + b 2 - a 2 . . . 1 Now, the distance of the point P a sec θ , b tan θ from the centre C 0 , 0 is C P = a 2 sec 2 θ + b 2 tan 2 θ , ⇒ S P · S ' P = C P 2 + b 2 - a 2 .

E M B I B E

Eduwiser's Coordinate Geometry for JEE Main and Advanced>Chapter 7 - Conic Section>7.1>Q 153

MEDIUM

JEE Main

IMPORTANT

Earn 100

Prove that the portion of the tangent to the hyperbola intercepted between the asymptotes is bisected at the point of contact.

Important Questions on Conic Section

MEDIUM

JEE Main

IMPORTANT

Eduwiser's Coordinate Geometry for JEE Main and Advanced>Chapter 7 - Conic Section>7.1>Q 158

Show that the locus of the foot of the perpendicular drawn from focus to a tangent to the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

is

x^{2} + y^{2} = a^{2}

.

MEDIUM

JEE Main

IMPORTANT

Eduwiser's Coordinate Geometry for JEE Main and Advanced>Chapter 7 - Conic Section>7.1>Q 164

Chords of the hyperbola,

x^{2} - y^{2} = a^{2}

touch the parabola,

y^{2} = 4 a x

. Prove that the locus of their middle points is the curve,

y^{2} (x - a) = x^{3}

HARD

JEE Main

IMPORTANT

Eduwiser's Coordinate Geometry for JEE Main and Advanced>Chapter 7 - Conic Section>7.1>Q 165

The chord of contact of the tangents through

P

to the hyperbola

\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1

subtends a right angle at the centre. Prove that the locus of

P

is the ellipse

b^{4} x^{2} + a^{4} y^{2} = a^{2} b^{2} (b^{2} - a^{2})

.

HARD

JEE Main

IMPORTANT

Eduwiser's Coordinate Geometry for JEE Main and Advanced>Chapter 7 - Conic Section>7.1>Q 173

If $C$ is the centre of a hyperbola $\frac{x^{2}}{a^{2}} - \frac{y^{2}}{b^{2}} = 1, S$ , $S^{'}$ its foci and $P$ a point on it.

Prove that $S P \cdot S^{'} P = C P^{2} - a^{2} + b^{2} .$

HARD

JEE Main

IMPORTANT

Eduwiser's Coordinate Geometry for JEE Main and Advanced>Chapter 7 - Conic Section>7.1>Q 174

Find the point on the hyperbola

\frac{x^{2}}{24} - \frac{y^{2}}{18} = 1

which is nearest to the line

3 x + 2 y + 1 = 0

and compute the distance between the point and the line.

HARD

JEE Main

IMPORTANT

Eduwiser's Coordinate Geometry for JEE Main and Advanced>Chapter 7 - Conic Section>7.1>Q 176

Given the base of a triangle and the ratio of the tangent of half the base angles. Show that the vertex moves on a hyperbola whose foci are the extremities of the base.

MEDIUM

JEE Main

IMPORTANT

Eduwiser's Coordinate Geometry for JEE Main and Advanced>Chapter 7 - Conic Section>7.1>Q 1

The angle of intersection between the curves,