The circle x2 + y2 – 8x + 0 and hyperbola \frac{x^{2}}{9} \textrm{ } - \textrm{ } \frac{y^{2}}{4} = 1 intersect at the points A and B.

Engineering

Mathematics

Chord of Contact in Hyperbola

Question

The circle x² + y² – 8x + 0 and hyperbola $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$ intersect at the points A and B.

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Linked Question 1

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

4x – 3y + 4 = 0

$2 x - \sqrt{5} y - 20 = 0$

3x – 4y + 8 = 0

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

$y = mx + \sqrt{9 m^{2} - 4}$ , centre (4, 0)

$| \frac{4 m + \sqrt{9 m^{2} - 4}}{\sqrt{1 + m^{2}}} | = 4$ .....(1)

Cross check:

$\frac{\frac{8}{\sqrt{5}} + \sqrt{\frac{36}{5} - 4}}{\sqrt{1 + \frac{4}{5}}} = \frac{\frac{8}{\sqrt{5}} + \frac{4}{\sqrt{5}}}{\frac{3}{\sqrt{5}}} = 4$

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Linked Question 2

Equation of the circle with AB as its diameter is

x² + y² + 12x + 24 = 0

x² + y² + 24x – 12 = 0

x² + y² – 24x – 12 = 0

x² + y² – 12x + 24 = 0

x² + y² – 8x = 0 and $\frac{x^{2}}{9} - \frac{y^{2}}{4} = 1$

On solving, we get

13x2 – 72x – 36 = 0

On solving x = 6, $\frac{6}{5}$ (y imaginary)

y = $2 \sqrt{3}, - 2 \sqrt{3}$

$∴$ diametric ends $(6, 2 \sqrt{3}) and (6, - 2 \sqrt{3})$

$∴$ circle is (x – 6)(x – 6) + $(y - 2 \sqrt{3}) (y + 2 \sqrt{3})$ = 0 $\Rightarrow$ x² + y² – 12x + 24 = 0

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