माना वक्रो 4left( x ^2+ y ^2right)=9 तथा y ^2 | vedclass

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Question

$x^{2}+y^{2}=\frac{9}{4} \quad y=4 x$

Equation tangent in slope form

$y=m x \pm \frac{3}{2} \sqrt{\left(1+m^{2}\right)}$      ...

Vedclass · Accepted Answer

$4$

Vedclass · Answer

$x^{2}+y^{2}=\frac{9}{4} \quad y=4 x$

Equation tangent in slope form

$y=m x \pm \frac{3}{2} \sqrt{\left(1+m^{2}ight)}$      .....$(1)$

$y=m x+\frac{1}{m}$      .....$(2)$

compare $(1)$ and $(2)$

$\pm \frac{3}{2} \sqrt{\left(1+m^{2}ight)}=\frac{1}{m^{2}}$

$9 m^{2}\left(1+m^{2}ight)=4$

$9 m^{4}+9 m^{2}-4=0$

$9 m^{4}+12 m^{2}-3 m^{2}-4=0$

$3 m^{2}\left(3 m^{2}+4ight)-\left(3 m^{2}+4ight)=0$

$m^{2}=-\frac{4}{3}(	ext { Rejected })$

$m^{2}=\frac{1}{3} \Rightarrow m=\pm \frac{1}{\sqrt{3}}$

Equation of common tangent

$y =\frac{1}{\sqrt{3}} x +\sqrt{3}$

$	ext { on } X 	ext { axis } y =0$

$OQ =-3$

$b =\mid OQ\mid =3$

$a =6$

$b ^{2}= a ^{2}\left(1- e ^{2}ight) \Rightarrow e ^{2}=1-\frac{9}{36}=\frac{3}{4} ~\ e =\frac{2 b ^{2}}{ a }=\frac{2 	imes 9}{6}=3 ~\ \frac{ e }{ e ^{2}}=\frac{3}{3 / 4}=4$

$\frac{ e }{ e ^{2}}=\frac{3}{3 / 4}=4$

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दो समुच्चय $A$ तथा $B$ निम्न प्रकार के हैं

$A=\{(a, b) \in R \times R:|a-5|< 1$ तथा $|b-5|< 1\}$

$B=\left\{(a, b) \in R \times R: 4(a-6)^{2}+9(b-5)^{2} \leq 36\right\}$ तो

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दो समुच्चय $A$ तथा $B$ निम्न प्रकार के हैं $A=\{(a, b) \in R \times R:|a-5|< 1$ तथा $|b-5|< 1\}$ $B=\left\{(a, b) \in R \times R: 4(a-6)^{2}+9(b-5)^{2} \leq 36\right\}$ तो

दो समुच्चय $A$ तथा $B$ निम्न प्रकार के हैं

$A=\{(a, b) \in R \times R:|a-5|< 1$ तथा $|b-5|< 1\}$

$B=\left\{(a, b) \in R \times R: 4(a-6)^{2}+9(b-5)^{2} \leq 36\right\}$ तो