Question

Let A be the foot of the perpendicular from focus P of hyperbola `x^2/a^2 - y^2/b^2 = 1` on the line bx – ay = 0 and let C be the centre of hyperbola. Then the area of the rectangle whose sides are equal to that of PA and CA is, 

Options

• `a/2`

• `ab`

• `((a^2 + b^2))/2`

• `2ab`

Answer 1

`ab`

Explanation:

Let C : (0, 0) is the centre of the hyperbola assuming a > b. Let Focus p : (s, 0)

Then s = `sqrt(a^2 + b^2)`

Now, slope of the line `bx - ay = 0` is `b/a`

Hence slope of perpendicular from focus to line is `- a/b`

Hence equation of normal (perpendicular) is `y = - a/b x + k`

⇒ `bk = by + ax`

⇒ `bk = b.0 + asqrt(a^2 + b^2)` ......(As passes through the focus)

⇒ `bk = asqrt(a^2 + b^2)`

Hence coordiantes of A are `(a^2/sqrt(a^2 + b^2), (ab)/sqrt(a^2 + b^2))`

Hence, PA = `sqrt((sqrt(a^2 + b^2) - a^2/sqrt(a^2 + b^2))^2 + (0 - (ab)/sqrt(a^2 + b^2))^2`

PA = `sqrt(b^4/(a^2 + b^2) + (a^2b^2)/(a^2 + b^2)`

PA = `sqrt(b^2((a^2 + b^2)/(a^2 + b^2)) = b`

Similarly, CA = `sqrt((0 - a^2/sqrt(a^2 + b^2))^2 + (0 - (ab)/sqrt(a^2 + b^2))^2`

CA = `a`

Hence Area = `ab`.