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Given curve is y=mx+1 and curve y²=4x, solve both simlutaneously,
we get
⇒(mx+1)²=4x
⇒m²x²+(2m−4)x+1=0
As we know, tangent touches the curve,
∴roots of above quadratic will be equal
⇒D=b²−4ac=0
⇒(2m−4)²−4(m³)(1)=0
⇒4m²+16−16m−4m²=0
⇒m=1
Hope this may help you...
Asking questions is a sign of intelligency.
we get
⇒(mx+1)²=4x
⇒m²x²+(2m−4)x+1=0
As we know, tangent touches the curve,
∴roots of above quadratic will be equal
⇒D=b²−4ac=0
⇒(2m−4)²−4(m³)(1)=0
⇒4m²+16−16m−4m²=0
⇒m=1
Hope this may help you...
Asking questions is a sign of intelligency.
Kanayan:
I reached till eq. But how we will know if roots are equal
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