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If 2 x+3 y=7 \text { and } x-y=1 are two normals of a parabola from a point, then third normal may be ...........

Option: 1

\mathrm{x-2y+5=0}
 


Option: 2

\mathrm{x+3y=5}
 


Option: 3

\mathrm{x-2y+10=0}
 


Option: 4

\mathrm{x-5y-2=0}


Answers (1)

best_answer

Equations of the normals are 2 x+3 y=7, x-y=1 solving them we get,x=2, y=1.
Hence point of intersection of these normals is A(2,1). If the slope of three normals from the point A are

m_1, m_2, m_3

then m_1=-2 / 3, m_2=1

We know that, m_1+m_2+m_3=0

\Rightarrow \quad m_3=-m_1-m_2=\frac{2}{3}-1=-\frac{1}{3}

3rd normal is a line passing through A(2,1) with slope m_3\left(=-\frac{1}{3}\right). Thus its equation is y-1=-\frac{1}{3}(x-2)

\Rightarrow \; \; \; \; \; \; \; \; \; \; \quad x+3 y=5

Posted by

jitender.kumar

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