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Question Number 81698 by peter frank last updated on 14/Feb/20

Commented by mr W last updated on 14/Feb/20

question is wrong!

questioniswrong!

Answered by mr W last updated on 14/Feb/20

y^2 =4x^2 ⇒y=±2x ⇒two lines y=mx+c=2x ⇒x_1 =(c/(2−m)), y_1 =((2c)/(2−m)) y=mx+c=−2x ⇒x_2 =−(c/(2+m)), y_2 =((2c)/(2+m)) distance=(√((x_2 −x_1 )^2 +(y_2 −y_1 )^2 )) =(√(((c/(2−m))+(c/(2+m)))^2 +(((2c)/(2−m))−((2c)/(2+m)))^2 )) =((4c(√(1+m^2 )))/(4−m^2 ))

y2=4x2⇒y=±2x⇒twolinesy=mx+c=2x⇒x1=c2−m,y1=2c2−my=mx+c=−2x⇒x2=−c2+m,y2=2c2+mdistance=(x2−x1)2+(y2−y1)2=(c2−m+c2+m)2+(2c2−m−2c2+m)2=4c1+m24−m2

Commented by peter frank last updated on 14/Feb/20

thank you sir.

thankyousir.

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