Question

If the tangent and the normal to ${x^2} - {y^2} = 4$at a point cut off intercepts ${a_{1,}}{a_2}$on the x-axis respectively and ${b_1},{b_2}$ on the y-axis respectively, then the value of ${a_1}{a_2} + {b_1}{b_2}$is
(A) 1
(B) -1
(C) 0
(D) 4

Answer 1

Hint: This problem comes under analytical geometry. Here the given equation is in the form of hyperbola. We need to find the equation of tangent and normal of the hyperbola tangent means the line touches a point of the hyperbola. Normal means the perpendicular line on the tangent point of hyperbola and then with substituting equation and trigonometric values. The point cut off on x-axis and y-axis then we find the required value and complete step by step explanation.

Formula used: $\sec \theta = \dfrac{1}{{\cos \theta }},\tan \theta = \dfrac{1}{{\cot \theta }}$

Complete step-by-step answer:
The given equation is in the form of ${x^2} - {y^2} = 4$, it is like the equation of hyperbola is ${x^2} - {y^2} = {a^2}$
Therefore, the given equation is in the form of hyperbola.
Let any point on the given hyperbola is $P(a\sec \theta ,a\tan \theta )$
Therefore the equation of tangent to the hyperbola at point ‘p’ are given by
$x\sec \theta - y\tan \theta = a$
The equation of normal to the hyperbola at point ‘p’ are given by
$\Rightarrow$$\dfrac{x}{{\sec \theta }} + \dfrac{y}{{\tan \theta }} = 2a$
$\Rightarrow$$\therefore {a_1} = a\cos \theta ,{b_1} = - a\cot \theta $
$\Rightarrow$${a_2} = 2a\sec \theta ,{b_1} = 2a\tan \theta $
Now substituting the value ${a_1}{a_2} + {b_1}{b_2}$is
$=a\cos \theta .2a\sec \theta + ( - a\cot \theta .2a\tan \theta )$
$2{a^2}\cos \theta .\dfrac{1}{{\cos \theta }} - 2{a^2}\cot \theta .\dfrac{1}{{\cot \theta }}$
By using the formula mentioned in formula used, we get
=$2{a^2} - 2{a^2}$
= 0

Therefore, the answer is option (C).

Note: This kind of problem needs attention on tangent and normal equations. It mainly deals with the equation of hyperbola on this occasion. We need to assume the equation of hyperbola and then forming equation of tangent and normal and with that we get the points ${a_{1,}}{a_2}$ and ${b_1},{b_2}$ then with trigonometric identities we obtain the required value.