Question

solve this............​

Answer 1

[tex]Let \: the \: point \: of \: contact \: of \: tangents \: on \\ \: the \: hyperbola \: have \: coordinates (x1,y1). \\ \\ So, \: the \: equation \: of \: a \: tangent \: is \: \\ \\

\frac{x x_{1}

}{9} − \frac{y y_{1}}{4} =1.....(1) \\ \\

The \: slope \: of \: the \: required \: tangent \ \\ to \: the \: hyperbola \: is  \: 2.  \\ \\

For \:  y=mx+c  \: to \: be \: a \\ tangent \: to \: the \: hyperbola \\

{c}^{2} = {a}^{2} {m}^{2} − {b}^{2} \\

\\

⇒c=± \sqrt{9 \times 4 - 4} \\

=±32 \\ =±4 \sqrt{2} \\

Hence, \: equations \: of \: the \: \\ tangents \: are \:

\\

y=2x±4 \sqrt{2} \\

⇒2x−y=±4 \sqrt{2} .....(2) \\

Equating \: (1) \: and \: (2) \: we \: get \\

\\

\frac{ x_{1} }{2 \times 9} = \frac{ y_{1}}{4 \times 1} \: = \: ± \frac{1}{4 \sqrt{2} } \\ \\

⇒ x_{1}=± \frac{9}{2 \sqrt{2} } \: , y_{1} =± \frac{1}{ \sqrt{2} } \\ \\

B \: is \: the \: right \: option

[tex]

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Answer 2

Hii akka eppadi irukinga..ippa laan on varathey illa poola.. neet preparation laan eppadi poguthu..

Tc..be safe and all the best akka ☺️...