Question

If the slope of one of the lines given by ax² + 2hxy + by² = 0 is square of the slope of the other line, then show that a²b + ab² + 8h³ = 6abh

Answer 1

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This is the equation of the line so ,

This is the equation of the line so ,\left ( y-mx \right )\left ( y-m^{2}x \right )=0(y−mx)(y−m2x)=0

This is the equation of the line so ,\left ( y-mx \right )\left ( y-m^{2}x \right )=0(y−mx)(y−m2x)=0m^{3}x^{2}-\left ( m^{2}+m \right )xy + y^{2} = 0m3x2−(m2+m)xy+y2=0

This is the equation of the line so ,\left ( y-mx \right )\left ( y-m^{2}x \right )=0(y−mx)(y−m2x)=0m^{3}x^{2}-\left ( m^{2}+m \right )xy + y^{2} = 0m3x2−(m2+m)xy+y2=0a=m^{3}a=m3

This is the equation of the line so ,\left ( y-mx \right )\left ( y-m^{2}x \right )=0(y−mx)(y−m2x)=0m^{3}x^{2}-\left ( m^{2}+m \right )xy + y^{2} = 0m3x2−(m2+m)xy+y2=0a=m^{3}a=m3h= -\left ( \dfrac{m^{2}+m}{2} \right )h=−(2m2+m)

This is the equation of the line so ,\left ( y-mx \right )\left ( y-m^{2}x \right )=0(y−mx)(y−m2x)=0m^{3}x^{2}-\left ( m^{2}+m \right )xy + y^{2} = 0m3x2−(m2+m)xy+y2=0a=m^{3}a=m3h= -\left ( \dfrac{m^{2}+m}{2} \right )h=−(2m2+m)b=1b=1

This is the equation of the line so ,\left ( y-mx \right )\left ( y-m^{2}x \right )=0(y−mx)(y−m2x)=0m^{3}x^{2}-\left ( m^{2}+m \right )xy + y^{2} = 0m3x2−(m2+m)xy+y2=0a=m^{3}a=m3h= -\left ( \dfrac{m^{2}+m}{2} \right )h=−(2m2+m)b=1b=1\left ( \dfrac{a+b}{h} \right ) + \left ( \dfrac{8h^{2}}{ab} \right ) =-\left ( \dfrac{2\left ( m^{2}-m+1 \right )}{m} \right )+\dfrac{2\left ( m+1 \right ) ^{2}}{m}(ha+b)+(ab8h2)=−(m2(m2−m+1))+m2(m+1)2

This is the equation of the line so ,\left ( y-mx \right )\left ( y-m^{2}x \right )=0(y−mx)(y−m2x)=0m^{3}x^{2}-\left ( m^{2}+m \right )xy + y^{2} = 0m3x2−(m2+m)xy+y2=0a=m^{3}a=m3h= -\left ( \dfrac{m^{2}+m}{2} \right )h=−(2m2+m)b=1b=1\left ( \dfrac{a+b}{h} \right ) + \left ( \dfrac{8h^{2}}{ab} \right ) =-\left ( \dfrac{2\left ( m^{2}-m+1 \right )}{m} \right )+\dfrac{2\left ( m+1 \right ) ^{2}}{m}(ha+b)+(ab8h2)=−(m2(m2−m+1))+m2(m+1)2\left ( \dfrac{a+b}{h} \right ) + \left ( \dfrac{8h^{2}}{ab} \right ) =6(ha+b)+(ab8h2)=6

This is the equation of the line so ,\left ( y-mx \right )\left ( y-m^{2}x \right )=0(y−mx)(y−m2x)=0m^{3}x^{2}-\left ( m^{2}+m \right )xy + y^{2} = 0m3x2−(m2+m)xy+y2=0a=m^{3}a=m3h= -\left ( \dfrac{m^{2}+m}{2} \right )h=−(2m2+m)b=1b=1\left ( \dfrac{a+b}{h} \right ) + \left ( \dfrac{8h^{2}}{ab} \right ) =-\left ( \dfrac{2\left ( m^{2}-m+1 \right )}{m} \right )+\dfrac{2\left ( m+1 \right ) ^{2}}{m}(ha+b)+(ab8h2)=−(m2(m2−m+1))+m2(m+1)2\left ( \dfrac{a+b}{h} \right ) + \left ( \dfrac{8h^{2}}{ab} \right ) =6(ha+b)+(ab8h2)=6a^{2}b +ab^{2}-6abh +8h^{3} = 0a2b+ab2−6abh+8h3=0

Answer 2

Answer:

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Step-by-step explanation:

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