Question

If the line $$2x + \sqrt 6 y = 2$$   touches the hyperbola $${x^2} - 2{y^2} = 4,$$   then the point of contact is-

A. $$\left( { - 2,\,\sqrt 6 } \right)$$
B. $$\left( { - 5,\,2\sqrt 6 } \right)$$
C. $$\left( {\frac{1}{2},\,\frac{1}{{\sqrt 6 }}} \right)$$
D. $$\left( {4,\, - \sqrt 6 } \right)$$  
Answer :   $$\left( {4,\, - \sqrt 6 } \right)$$
Solution :
Equation of tangent to hyperbola $${x^2} - 2{y^2} = 4$$   at any point $$\left( {{x_1},\,{y_1}} \right)$$  is $$x{x_1} - 2y{y_1} = 4$$
Comparing with $$2x + \sqrt 6 y = 2$$    or $$4x + 2\sqrt 6 y = 4$$
$$ \Rightarrow {x_1} = 4$$   and $$ - 2{y_1} = 2\sqrt 6 \Rightarrow \left( {4,\, - \sqrt 6 } \right)$$      is the required point.

Releted MCQ Question on
Geometry >> Hyperbola

Releted Question 1

Each of the four inequalities given below defines a region in the $$xy$$  plane. One of these four regions does not have the following property. For any two points $$\left( {{x_1},\,{y_1}} \right)$$  and $$\left( {{x_2},\,{y_2}} \right)$$  in the the region, the point $$\left( {\frac{{{x_1} + {x_2}}}{2},\,\frac{{{y_1} + {y_2}}}{2}} \right)$$    is also in the region. The inequality defining this region is :

A. $${x^2} + 2{y^2} \leqslant 1$$
B. $${\text{max }}\left\{ {\left| x \right|,\left| y \right|} \right\} \leqslant 1$$
C. $${x^2} - {y^2} \leqslant 1$$
D. $${y^2} - {x^2} \leqslant 0$$
View Answer
Releted Question 2

Let $$P\left( {a\,\sec \,\theta ,\,b\,\tan \,\theta } \right)$$    and $$Q\left( {a\,\sec \,\phi ,\,b\,\tan \,\phi } \right),$$    where $$\theta + \phi = \frac{\pi }{2},$$   be two points on the hyperbola $$\frac{{{x^2}}}{{{a^2}}} - \frac{{{y^2}}}{{{b^2}}} = 1.$$    If $$\left( {h,\,k} \right)$$  is the point of intersection of the normal at $$P$$ and $$Q,$$  then $$k$$ is equal to :

A. $$\frac{{{a^2} + {b^2}}}{a}$$
B. $$ - \left( {\frac{{{a^2} + {b^2}}}{a}} \right)$$
C. $$\frac{{{a^2} + {b^2}}}{b}$$
D. $$ - \left( {\frac{{{a^2} + {b^2}}}{b}} \right)$$
View Answer
Releted Question 3

If $$x=9$$  is the chord of contact of the hyperbola $${x^2} - {y^2} = 9,$$   then the equation of the corresponding pair of tangents is :

A. $$9{x^2} - 8{y^2} + 18x - 9 = 0$$
B. $$9{x^2} - 8{y^2} - 18x + 9 = 0$$
C. $$9{x^2} - 8{y^2} - 18x - 9 = 0$$
D. $$9{x^2} - 8{y^2} + 18x + 9 = 0$$
View Answer
Releted Question 4

For hyperbola $$\frac{{{x^2}}}{{{{\cos }^2}\alpha }} - \frac{{{y^2}}}{{{{\sin }^2}\alpha }} = 1,$$     which of the following remains constant with change in $$'\alpha \,'$$

A. abscissae of vertices
B. abscissae of foci
C. eccentricity
D. directrix
View Answer

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