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JEE Advanced 2010Paper 1

Exam: JEE Advanced
Year: 2010
Paper: Paper 1
Subject: Mathematics
Chapter: Hyperbola

JEE Advanced 2010 · Paper 1 · Mathematics · Hyperbola

Paragraph: The circle x^2+y^2-8 x=0 and hyperbola...

Paragraph: The circle

x^2+y^2-8 x=0

and hyperbola

\frac{x^2}{9}-\frac{y^2}{4}=1

intersect at the points

A

and

B

.Question: Equation of a common tangent with positive slope to the circle as well as to the hyperbola is

Options

A $2 x-\sqrt{5} y-20=0$
B $2 x-\sqrt{5} y+4=0$
C $3 x-4 y+8=0$
D $4 x-3 y+4=0$

Correct answer

B. $2 x-\sqrt{5} y+4=0$

Step-by-step solution

Equation of tangent to hyperbola having slope

m

is y=m x+

\sqrt{9 m^2-4}

Equation of tangent to circle is y=m(x-4)+

\sqrt{16 m^2+16}

Eqs. (i) and (ii) will be identical for

m=\frac{2}{\sqrt{5}}

satisfy.

\therefore

Equation of common tangent is

2 x-\sqrt{5} y+4=0

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