Question

It y = mx + C is a common tangent of circle x2 + y2 = 3 and hyperbola x2/64 - y2/100 = 1 then which of the following statement is true :

Answer 1

ANSWER.

=> 61c² = 492

EXPLANATION.

$\sf \to \: y \: = mx + c \: is \: common \: to \: the \: circle \: = {x}^{2} + {y}^{2} = 3 \\ \\ \sf \to \: hyperbola \: = \dfrac{ {x}^{2} }{64} - \dfrac{ {y}^{2} }{100} = 1$

$\sf \to \: condition \: of \: tangent \: of \: circle\: \implies \: \: y = mx \: + \sqrt{r} \sqrt{1 + {m}^{2} } \\ \\ \sf \to \: condition \: of \: tangent \: of \: hyperbola \implies \: y \: = mx \: + \sqrt{ {a}^{2} {m}^{2} - {b}^{2} }$

$\sf \to \: \: we \: can \: equal \: the \: equation \: \\ \\ \sf \to \: \sqrt{3} \sqrt{1 + {m}^{2} } = \sqrt{64 {m}^{2} - 100} \\ \\ \sf \to \: 3(1 + {m}^{2} ) = 64 {m}^{2} - 100 \\ \\ \sf \to \: 3 + 3 {m}^{2} = 64 {m}^{2} - 100 \\ \\ \sf \to \: 61 {m}^{2} = 103 \\ \\ \sf \to \: {m}^{2} = \dfrac{103}{61}$

$\sf \to \: we \: can \: find \: c \: = \sqrt{r} \sqrt{1 + {m}^{2} } \\ \\ \sf \to \: {c}^{2} = 3(1 + {m}^{2} ) \\ \\ \sf \to \: {c}^{2} = 3(1 + \dfrac{103}{61} ) \\ \\ \sf \to \: {c}^{2} = 3( \frac{61 + 103}{61} ) \\ \\ \sf \to \: {c}^{2} = \frac{3 \times 164}{61} \\ \\ \sf \to \: 61{c}^{2} = 492$

Hence option [ 2 ] is correct.