JEE MATHEMATICS MCQ: CIRCLES

1 Easy

1 of 1

Application +4 | -1

C

the centre of the hyperbola

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

. The tangents at any point

P

on this hyperbola meets the straight lines

b x-a y=0

and

b x+a y=0

in the points

Q

and

R

respectively. Then

C Q \cdot C R=

View answer and explanation

Correct option: A

a^2+b^2

Option A (

a^2+b^2

) is correct because the product of the distances CQ and CR simplifies to

a^2 + b^2

using the equation of the tangent to the hyperbola.

More Questions from CIRCLES

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{\left( {{\alpha ^2}{x^2} - 2\alpha \mathrm{ }x + 1} \right)^{51}},

x

\left( {\alpha ,2{\alpha ^2}} \right)

{x^2} + {y^2} = 4,

{x^2} + {y^2} - 2x + 4y - 4 = 0

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

\left( {{x_1},{y_1}} \right)

\quad \alpha, \beta

\gamma

P, Q

x^2+y^2=1

A

(-1,0)

A P, A Q

A R

G P

\cos \alpha / 2, \cos \beta / 2

\cos \gamma / 2

x+y=6

x+2 y=4

x+y=5

x-y=1

9 \pi