## The tangent at a point P on the hyperbola x^2/a^2 – y^2/b^2 = 1 meets one of its directrices in F.

Q: The tangent at a point P on the hyperbola $\frac{x^2}{a^2} – \frac{y^2}{b^2} = 1$ meets one of its directrices in F. If PF subtends an angle θ at the corresponding focus, then θ equals

(A) π/4

(B) π/2

(C) 3π/4

(D) π

Sol. Let directrix be x = a/e and focus be S(ae, 0). Let P( asecθ , btanθ) be any point on the curve,

Equation of tangent at ‘P’ is $\large \frac{x sec\theta}{a} – \frac{y tan\theta}{b} = 1$

Let ‘F’ be the intersection point of tangent of directrix, then

$\large F \equiv (a/e , \frac{b(sec\theta – e)}{e tan\theta})$

$\large m_{SF} = \frac{b(sec\theta – e)}{-a tan\theta (e^2 -1)}$

$\large m_{PS} = \frac{b tan\theta}{a(sec\theta -e)}$

⇒ mSF . mPS = –1

Hence (B) is the correct answer.

## The equation of a line passing through the centre of a rectangular hyperbola is x – y – 1 = 0. If one of its asymptotes is 3x – 4y – 6 = 0,

Q: The equation of a line passing through the centre of a rectangular hyperbola is x – y – 1 = 0. If one of its asymptotes is 3x – 4y – 6 = 0, the equation of the other asymptote is

(A) 4x – 3y + 17 = 0

(B) -4x – 3y + 17 = 0

(C) -4x + 3y + 1 = 0

(D) 4x + 3y + 17 = 0

Sol. We know that asymptotes of rectangular hyperbola are mutually perpendicular, thus other asymptote should be 4x + 3y + λ = 0

Intersection point of asymptotes is also the centre of the hyperbola.

Hence intersection point of 4x +3y + λ = 0 and 3x -4y -6 = 0 should lie on the line x- y -1= 0, using it λ can be easily obtained.

Hence (D) is the correct answer.

## For a hyperbola, the foci are at (±4 , 0) and vertices at (±2 , 0). Its equation is

Q: For a hyperbola, the foci are at (±4 , 0) and vertices at (±2 , 0). Its equation is

(A) $\frac{x^2}{4} – \frac{y^2}{12} = 1$

(B) $\frac{x^2}{12} – \frac{y^2}{4} = 1$

(C) $\frac{x^2}{16} – \frac{y^2}{4} = 1$

(D) $\frac{x^2}{4} – \frac{y^2}{16} = 1$

Sol. ae = 4, a = 2

⇒ e = 2

But b2 = a2(e2 – 1)

⇒ b2 = 12

Hyperbola is $\frac{x^2}{4} – \frac{y^2}{12} = 1$

Hence (A) is the correct answer.

## The point on the hyperbola x^2/24 – y^2/18 = 1 which is nearest to the line 3x + 2y+1 = 0 is

Q: The point on the hyperbola $\frac{x^2}{24} – \frac{y^2}{18} = 1$ which is nearest to the line 3x + 2y+1 = 0 is

(A) (6, 3)

(B) (–6, 3)

(C) (6, –3)

(D) (–6, –3)

Ans: (C)
Sol: Equation of tangent at $(\sqrt{24} sec\theta , \sqrt{18} tan\theta)$ is

$\large \frac{x sec\theta}{\sqrt{24}} – \frac{y tan\theta}{\sqrt{18}} =1$ ,

then point is nearest to the line 3x + 2y + 1 = 0,

so its slope = –3/2

$\large \frac{sec\theta}{\sqrt{24}} . \frac{\sqrt{18}}{tan\theta} = – \frac{3}{2}$

$\large sin\theta = -\frac{1}{\sqrt{3}}$

Hence the point is (6, –3).

Hence (C) is the correct answer.