Question

$\huge\boxed{\dag\sf\red{ANSWER}\dag}$. → $\huge\fbox\pink{ The equation of the pair of tangents drawn from the pôint (1,2) to the ellipse 3x²+2y²=5}$​

Answer 1

Answer:

Answer:

$\Huge{\textbf{\textsf{{\color{Magenta}{⛄A}}{\red{n}}{\purple{s}}{\pink{w}}{\blue{e}}{\green{r⛄ }}{\color{pink}{:}}}}}$

tan

−1 ( 5 12 )

Equation of line through (1,2) is

y−2=m(x−1)

⇒y=mx+(2−m)

Equation for tangency is c

2 =a

2 m 2 +b 2

(2−m)

2 = 35m

2+ 254+m

2 −4m= 35 m

2 + 25 32 m 2 +4m−23

=0 m

1 +m 2

= 32−4 =−6 m 1 m 2 =−

2/3−3/2 =4−9

(m1 −m2 ) 2 =

(m 1 +m2 ) 2 −4m1 m 2

⇒m1 −m 2 −45

∴tanα= 1+m 1m 2m

1 −m2

= 5445 =53×4 =512

∴α=tan = −1 ( 512 )

~ ♛♡ ||MissButterfly|| ♡♛

Answer 2

Answer:

tan

tan−1 ( 5 12 )Equation of line through (1,2) isy−2=m(x−1)⇒y=mx+(2−m)Equation for tangency is c2 =a2 m 2 +b 2(2−m)2 = 35m2+ 254+m2 −4m= 35 m2 + 25 32 m 2 +4m−23=0 m1 +m 2= 32−4 =−6 m 1 m 2 =−2/3−3/2 =4−9(m1 −m2 ) 2 =(m 1 +m2 ) 2 −4m1 m 2⇒m1 −m 2 −45∴tanα= 1+m 1m 2m1 −m2= 5445 =53×4 =512∴α=tan = −1 ( 512 )

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